/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


MAYBE
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Outermost Termination of the given OTRS could not be shown:

(0) OTRS
(1) Trivial-Transformation [SOUND, 0 ms]
(2) QTRS
    (3) DependencyPairsProof [EQUIVALENT, 0 ms]
    (4) QDP
    (5) DependencyGraphProof [EQUIVALENT, 1 ms]
    (6) AND
        (7) QDP
            (8) UsableRulesProof [EQUIVALENT, 0 ms]
            (9) QDP
            (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
            (11) YES
        (12) QDP
            (13) UsableRulesProof [EQUIVALENT, 0 ms]
            (14) QDP
            (15) NonTerminationLoopProof [COMPLETE, 0 ms]
            (16) NO
(17) Thiemann-SpecialC-Transformation [EQUIVALENT, 0 ms]
(18) QTRS
    (19) DependencyPairsProof [EQUIVALENT, 0 ms]
    (20) QDP
    (21) DependencyGraphProof [EQUIVALENT, 0 ms]
    (22) AND
        (23) QDP
            (24) UsableRulesProof [EQUIVALENT, 0 ms]
            (25) QDP
            (26) QReductionProof [EQUIVALENT, 0 ms]
            (27) QDP
            (28) MRRProof [EQUIVALENT, 10 ms]
            (29) QDP
            (30) UsableRulesProof [EQUIVALENT, 0 ms]
            (31) QDP
            (32) QReductionProof [EQUIVALENT, 0 ms]
            (33) QDP
            (34) UsableRulesReductionPairsProof [EQUIVALENT, 3 ms]
            (35) QDP
            (36) DependencyGraphProof [EQUIVALENT, 0 ms]
            (37) TRUE
        (38) QDP
            (39) UsableRulesProof [EQUIVALENT, 0 ms]
            (40) QDP
                (41) QReductionProof [EQUIVALENT, 0 ms]
                (42) QDP
                (43) TransformationProof [EQUIVALENT, 0 ms]
                (44) QDP
                (45) UsableRulesProof [EQUIVALENT, 0 ms]
                (46) QDP
                (47) QDPOrderProof [EQUIVALENT, 16 ms]
                (48) QDP
                (49) SplitQDPProof [EQUIVALENT, 0 ms]
                (50) AND
                    (51) QDP
                        (52) SemLabProof [SOUND, 0 ms]
                        (53) QDP
                        (54) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                        (55) QDP
                        (56) MRRProof [EQUIVALENT, 0 ms]
                        (57) QDP
                        (58) QDPOrderProof [EQUIVALENT, 0 ms]
                        (59) QDP
                        (60) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                        (61) QDP
                        (62) QDPOrderProof [EQUIVALENT, 0 ms]
                        (63) QDP
                        (64) QDPOrderProof [EQUIVALENT, 5 ms]
                        (65) QDP
                        (66) PisEmptyProof [SOUND, 0 ms]
                        (67) TRUE
                    (68) QDP
                        (69) TransformationProof [SOUND, 0 ms]
                        (70) QDP
                        (71) UsableRulesProof [EQUIVALENT, 0 ms]
                        (72) QDP
                        (73) QReductionProof [EQUIVALENT, 0 ms]
                        (74) QDP
            (75) UsableRulesProof [EQUIVALENT, 0 ms]
            (76) QDP
                (77) QReductionProof [EQUIVALENT, 0 ms]
                (78) QDP
(79) Raffelsieper-Zantema-Transformation [SOUND, 0 ms]
(80) QTRS
    (81) AAECC Innermost [EQUIVALENT, 14 ms]
    (82) QTRS
    (83) DependencyPairsProof [EQUIVALENT, 0 ms]
    (84) QDP
    (85) DependencyGraphProof [EQUIVALENT, 0 ms]
    (86) AND
        (87) QDP
            (88) UsableRulesProof [EQUIVALENT, 0 ms]
            (89) QDP
            (90) QReductionProof [EQUIVALENT, 0 ms]
            (91) QDP
            (92) QDPSizeChangeProof [EQUIVALENT, 0 ms]
            (93) YES
        (94) QDP
            (95) UsableRulesProof [EQUIVALENT, 0 ms]
            (96) QDP
            (97) QReductionProof [EQUIVALENT, 0 ms]
            (98) QDP
            (99) TransformationProof [EQUIVALENT, 14 ms]
            (100) QDP
            (101) DependencyGraphProof [EQUIVALENT, 0 ms]
            (102) QDP
            (103) UsableRulesProof [EQUIVALENT, 0 ms]
            (104) QDP
            (105) TransformationProof [EQUIVALENT, 0 ms]
            (106) QDP
            (107) TransformationProof [EQUIVALENT, 0 ms]
            (108) QDP
            (109) TransformationProof [EQUIVALENT, 0 ms]
            (110) QDP
            (111) TransformationProof [EQUIVALENT, 0 ms]
            (112) QDP
            (113) DependencyGraphProof [EQUIVALENT, 0 ms]
            (114) QDP
            (115) TransformationProof [EQUIVALENT, 0 ms]
            (116) QDP
            (117) TransformationProof [EQUIVALENT, 0 ms]
            (118) QDP
            (119) TransformationProof [EQUIVALENT, 0 ms]
            (120) QDP
            (121) QDPOrderProof [EQUIVALENT, 16 ms]
            (122) QDP
            (123) DependencyGraphProof [EQUIVALENT, 0 ms]
            (124) QDP
            (125) QDPOrderProof [EQUIVALENT, 14 ms]
            (126) QDP
            (127) DependencyGraphProof [EQUIVALENT, 0 ms]
            (128) QDP
            (129) SplitQDPProof [EQUIVALENT, 0 ms]
            (130) AND
                (131) QDP
                    (132) SemLabProof [SOUND, 0 ms]
                    (133) QDP
                    (134) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                    (135) QDP
                    (136) MRRProof [EQUIVALENT, 0 ms]
                    (137) QDP
                    (138) DependencyGraphProof [EQUIVALENT, 0 ms]
                    (139) QDP
                    (140) PisEmptyProof [SOUND, 0 ms]
                    (141) TRUE
                (142) QDP
                    (143) QReductionProof [EQUIVALENT, 0 ms]
                    (144) QDP
                    (145) SplitQDPProof [EQUIVALENT, 0 ms]
                    (146) AND
                        (147) QDP
                            (148) SemLabProof [SOUND, 0 ms]
                            (149) QDP
                            (150) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                            (151) QDP
                            (152) MRRProof [EQUIVALENT, 0 ms]
                            (153) QDP
                            (154) DependencyGraphProof [EQUIVALENT, 0 ms]
                            (155) QDP
                            (156) PisEmptyProof [SOUND, 0 ms]
                            (157) TRUE
                        (158) QDP
                            (159) QReductionProof [EQUIVALENT, 0 ms]
                            (160) QDP


----------------------------------------

(0)
Obligation:
Term rewrite system R:
The TRS R consists of the following rules:

   a -> f(a)
   f(f(x)) -> b
   0(f(x)) -> 1(x)
   1(f(x)) -> f(0(x))



Outermost Strategy.

----------------------------------------

(1) Trivial-Transformation (SOUND)
We applied the Trivial transformation  to transform the outermost TRS to a standard TRS.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a -> f(a)
   f(f(x)) -> b
   0(f(x)) -> 1(x)
   1(f(x)) -> f(0(x))

Q is empty.

----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A -> F(a)
   A -> A
   0^1(f(x)) -> 1^1(x)
   1^1(f(x)) -> F(0(x))
   1^1(f(x)) -> 0^1(x)

The TRS R consists of the following rules:

   a -> f(a)
   f(f(x)) -> b
   0(f(x)) -> 1(x)
   1(f(x)) -> f(0(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   1^1(f(x)) -> 0^1(x)
   0^1(f(x)) -> 1^1(x)

The TRS R consists of the following rules:

   a -> f(a)
   f(f(x)) -> b
   0(f(x)) -> 1(x)
   1(f(x)) -> f(0(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(8) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   1^1(f(x)) -> 0^1(x)
   0^1(f(x)) -> 1^1(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(10) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*0^1(f(x)) -> 1^1(x)
The graph contains the following edges 1 > 1


*1^1(f(x)) -> 0^1(x)
The graph contains the following edges 1 > 1


----------------------------------------

(11)
YES

----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A -> A

The TRS R consists of the following rules:

   a -> f(a)
   f(f(x)) -> b
   0(f(x)) -> 1(x)
   1(f(x)) -> f(0(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(13) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A -> A

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(15) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = A evaluates to  t =A

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [ ]

--------------------------------------------------------------------------------
Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from A to A.




----------------------------------------

(16)
NO

----------------------------------------

(17) Thiemann-SpecialC-Transformation (EQUIVALENT)
We applied the Thiemann-SpecialC transformation  to transform the outermost TRS to an innermost TRS.
----------------------------------------

(18)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   top(go_up(x)) -> top(reduce(x))
   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_f(f(x)) -> result_f(b)
   redex_0(f(x)) -> result_0(1(x))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_f(result_f(x)) -> go_up(x)
   check_0(result_0(x)) -> go_up(x)
   check_1(result_1(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))


----------------------------------------

(19) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(x)) -> TOP(reduce(x))
   TOP(go_up(x)) -> REDUCE(x)
   REDUCE(f(x_1)) -> CHECK_F(redex_f(x_1))
   REDUCE(f(x_1)) -> REDEX_F(x_1)
   REDUCE(0(x_1)) -> CHECK_0(redex_0(x_1))
   REDUCE(0(x_1)) -> REDEX_0(x_1)
   REDUCE(1(x_1)) -> CHECK_1(redex_1(x_1))
   REDUCE(1(x_1)) -> REDEX_1(x_1)
   CHECK_F(redex_f(x_1)) -> IN_F_1(reduce(x_1))
   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   CHECK_0(redex_0(x_1)) -> IN_0_1(reduce(x_1))
   CHECK_0(redex_0(x_1)) -> REDUCE(x_1)
   CHECK_1(redex_1(x_1)) -> IN_1_1(reduce(x_1))
   CHECK_1(redex_1(x_1)) -> REDUCE(x_1)

The TRS R consists of the following rules:

   top(go_up(x)) -> top(reduce(x))
   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_f(f(x)) -> result_f(b)
   redex_0(f(x)) -> result_0(1(x))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_f(result_f(x)) -> go_up(x)
   check_0(result_0(x)) -> go_up(x)
   check_1(result_1(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(21) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 9 less nodes.
----------------------------------------

(22)
Complex Obligation (AND)

----------------------------------------

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REDUCE(0(x_1)) -> CHECK_0(redex_0(x_1))
   CHECK_0(redex_0(x_1)) -> REDUCE(x_1)
   REDUCE(1(x_1)) -> CHECK_1(redex_1(x_1))
   CHECK_1(redex_1(x_1)) -> REDUCE(x_1)

The TRS R consists of the following rules:

   top(go_up(x)) -> top(reduce(x))
   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_f(f(x)) -> result_f(b)
   redex_0(f(x)) -> result_0(1(x))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_f(result_f(x)) -> go_up(x)
   check_0(result_0(x)) -> go_up(x)
   check_1(result_1(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(24) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REDUCE(0(x_1)) -> CHECK_0(redex_0(x_1))
   CHECK_0(redex_0(x_1)) -> REDUCE(x_1)
   REDUCE(1(x_1)) -> CHECK_1(redex_1(x_1))
   CHECK_1(redex_1(x_1)) -> REDUCE(x_1)

The TRS R consists of the following rules:

   redex_1(f(x)) -> result_1(f(0(x)))
   redex_0(f(x)) -> result_0(1(x))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(26) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   top(go_up(x0))
   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))


----------------------------------------

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REDUCE(0(x_1)) -> CHECK_0(redex_0(x_1))
   CHECK_0(redex_0(x_1)) -> REDUCE(x_1)
   REDUCE(1(x_1)) -> CHECK_1(redex_1(x_1))
   CHECK_1(redex_1(x_1)) -> REDUCE(x_1)

The TRS R consists of the following rules:

   redex_1(f(x)) -> result_1(f(0(x)))
   redex_0(f(x)) -> result_0(1(x))

The set Q consists of the following terms:

   redex_0(f(x0))
   redex_1(f(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(28) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   REDUCE(1(x_1)) -> CHECK_1(redex_1(x_1))
   CHECK_1(redex_1(x_1)) -> REDUCE(x_1)


Used ordering: Polynomial interpretation [POLO]:

   POL(0(x_1)) = x_1
   POL(1(x_1)) = 2 + 2*x_1
   POL(CHECK_0(x_1)) = 1 + x_1
   POL(CHECK_1(x_1)) = 2 + 2*x_1
   POL(REDUCE(x_1)) = 1 + 2*x_1
   POL(f(x_1)) = 2 + 2*x_1
   POL(redex_0(x_1)) = 2*x_1
   POL(redex_1(x_1)) = 2*x_1
   POL(result_0(x_1)) = 2*x_1
   POL(result_1(x_1)) = 2 + x_1


----------------------------------------

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REDUCE(0(x_1)) -> CHECK_0(redex_0(x_1))
   CHECK_0(redex_0(x_1)) -> REDUCE(x_1)

The TRS R consists of the following rules:

   redex_1(f(x)) -> result_1(f(0(x)))
   redex_0(f(x)) -> result_0(1(x))

The set Q consists of the following terms:

   redex_0(f(x0))
   redex_1(f(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(30) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REDUCE(0(x_1)) -> CHECK_0(redex_0(x_1))
   CHECK_0(redex_0(x_1)) -> REDUCE(x_1)

The TRS R consists of the following rules:

   redex_0(f(x)) -> result_0(1(x))

The set Q consists of the following terms:

   redex_0(f(x0))
   redex_1(f(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(32) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   redex_1(f(x0))


----------------------------------------

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REDUCE(0(x_1)) -> CHECK_0(redex_0(x_1))
   CHECK_0(redex_0(x_1)) -> REDUCE(x_1)

The TRS R consists of the following rules:

   redex_0(f(x)) -> result_0(1(x))

The set Q consists of the following terms:

   redex_0(f(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(34) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

   REDUCE(0(x_1)) -> CHECK_0(redex_0(x_1))
The following rules are removed from R:

   redex_0(f(x)) -> result_0(1(x))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0(x_1)) = 2*x_1
   POL(1(x_1)) = x_1
   POL(CHECK_0(x_1)) = x_1
   POL(REDUCE(x_1)) = 2*x_1
   POL(f(x_1)) = 2*x_1
   POL(redex_0(x_1)) = 2*x_1
   POL(result_0(x_1)) = 2*x_1


----------------------------------------

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHECK_0(redex_0(x_1)) -> REDUCE(x_1)

R is empty.
The set Q consists of the following terms:

   redex_0(f(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(36) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(37)
TRUE

----------------------------------------

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(x)) -> TOP(reduce(x))

The TRS R consists of the following rules:

   top(go_up(x)) -> top(reduce(x))
   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_f(f(x)) -> result_f(b)
   redex_0(f(x)) -> result_0(1(x))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_f(result_f(x)) -> go_up(x)
   check_0(result_0(x)) -> go_up(x)
   check_1(result_1(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(39) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(x)) -> TOP(reduce(x))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(41) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   top(go_up(x0))


----------------------------------------

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(x)) -> TOP(reduce(x))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(43) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TOP(go_up(x)) -> TOP(reduce(x)) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0))),TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0))))
   (TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0))),TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0))))
   (TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0))),TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0))))
   (TOP(go_up(a)) -> TOP(go_up(f(a))),TOP(go_up(a)) -> TOP(go_up(f(a))))


----------------------------------------

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0)))
   TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0)))
   TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0)))
   TOP(go_up(a)) -> TOP(go_up(f(a)))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(45) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0)))
   TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0)))
   TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0)))
   TOP(go_up(a)) -> TOP(go_up(f(a)))

The TRS R consists of the following rules:

   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(47) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   TOP(go_up(a)) -> TOP(go_up(f(a)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0(x_1)) = 0
   POL(1(x_1)) = 0
   POL(TOP(x_1)) = x_1
   POL(a) = 1
   POL(b) = 0
   POL(check_0(x_1)) = x_1
   POL(check_1(x_1)) = x_1
   POL(check_f(x_1)) = x_1
   POL(f(x_1)) = 0
   POL(go_up(x_1)) = x_1
   POL(in_0_1(x_1)) = 0
   POL(in_1_1(x_1)) = 0
   POL(in_f_1(x_1)) = 0
   POL(redex_0(x_1)) = 0
   POL(redex_1(x_1)) = 0
   POL(redex_f(x_1)) = 0
   POL(reduce(x_1)) = 0
   POL(result_0(x_1)) = x_1
   POL(result_1(x_1)) = x_1
   POL(result_f(x_1)) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))


----------------------------------------

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0)))
   TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0)))
   TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0)))

The TRS R consists of the following rules:

   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(49) SplitQDPProof (EQUIVALENT)
We show in the first subproof that some pairs and rules can be removed, afterwards, we continue with the remaining DP-Problem

----------------------------------------

(50)
Complex Obligation (AND)

----------------------------------------

(51)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0)))
   TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0)))
   TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0)))

The TRS R consists of the following rules:

   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(52) SemLabProof (SOUND)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
result_f: 0
a: 1
reduce: 0
redex_0: 0
check_1: 0
redex_f: 0
check_f: 0
in_1_1: 0
TOP: 0
redex_1: 0
go_up: 0
in_f_1: 0
b: 0
in_0_1: 0
result_0: 0
f: 0
0: 1
1: 1
check_0: 0
result_1: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

(53)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(go_up.0(f.0(x0))) -> TOP.0(check_f.0(redex_f.0(x0)))
   TOP.0(go_up.0(f.1(x0))) -> TOP.0(check_f.0(redex_f.1(x0)))
   TOP.0(go_up.1(0.0(x0))) -> TOP.0(check_0.0(redex_0.0(x0)))
   TOP.0(go_up.1(0.1(x0))) -> TOP.0(check_0.0(redex_0.1(x0)))
   TOP.0(go_up.1(1.0(x0))) -> TOP.0(check_1.0(redex_1.0(x0)))
   TOP.0(go_up.1(1.1(x0))) -> TOP.0(check_1.0(redex_1.1(x0)))

The TRS R consists of the following rules:

   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   check_1.0(result_1.1(x)) -> go_up.1(x)
   check_1.0(redex_1.0(x_1)) -> in_1_1.0(reduce.0(x_1))
   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   reduce.1(a.) -> go_up.0(f.1(a.))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(result_0.0(x)) -> go_up.0(x)
   check_0.0(result_0.1(x)) -> go_up.1(x)
   check_0.0(redex_0.0(x_1)) -> in_0_1.0(reduce.0(x_1))
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   redex_f.0(f.0(x)) -> result_f.0(b.)
   redex_f.0(f.1(x)) -> result_f.0(b.)
   check_f.0(result_f.0(x)) -> go_up.0(x)
   check_f.0(result_f.1(x)) -> go_up.1(x)
   check_f.0(redex_f.0(x_1)) -> in_f_1.0(reduce.0(x_1))
   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))

The set Q consists of the following terms:

   reduce.0(f.0(x0))
   reduce.0(f.1(x0))
   reduce.1(0.0(x0))
   reduce.1(0.1(x0))
   reduce.1(1.0(x0))
   reduce.1(1.1(x0))
   reduce.1(a.)
   redex_f.0(f.0(x0))
   redex_f.0(f.1(x0))
   redex_0.0(f.0(x0))
   redex_0.0(f.1(x0))
   redex_1.0(f.0(x0))
   redex_1.0(f.1(x0))
   check_f.0(result_f.0(x0))
   check_f.0(result_f.1(x0))
   check_0.0(result_0.0(x0))
   check_0.0(result_0.1(x0))
   check_1.0(result_1.0(x0))
   check_1.0(result_1.1(x0))
   check_f.0(redex_f.0(x0))
   check_f.0(redex_f.1(x0))
   check_0.0(redex_0.0(x0))
   check_0.0(redex_0.1(x0))
   check_1.0(redex_1.0(x0))
   check_1.0(redex_1.1(x0))
   in_f_1.0(go_up.0(x0))
   in_f_1.0(go_up.1(x0))
   in_0_1.0(go_up.0(x0))
   in_0_1.0(go_up.1(x0))
   in_1_1.0(go_up.0(x0))
   in_1_1.0(go_up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(54) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   check_1.0(result_1.1(x)) -> go_up.1(x)
   check_0.0(result_0.0(x)) -> go_up.0(x)
   check_f.0(result_f.1(x)) -> go_up.1(x)
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = x_1
   POL(0.1(x_1)) = x_1
   POL(1.0(x_1)) = x_1
   POL(1.1(x_1)) = x_1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(check_0.0(x_1)) = x_1
   POL(check_1.0(x_1)) = x_1
   POL(check_f.0(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(go_up.0(x_1)) = x_1
   POL(go_up.1(x_1)) = x_1
   POL(in_0_1.0(x_1)) = x_1
   POL(in_1_1.0(x_1)) = x_1
   POL(in_f_1.0(x_1)) = x_1
   POL(redex_0.0(x_1)) = x_1
   POL(redex_0.1(x_1)) = x_1
   POL(redex_1.0(x_1)) = x_1
   POL(redex_1.1(x_1)) = x_1
   POL(redex_f.0(x_1)) = x_1
   POL(redex_f.1(x_1)) = x_1
   POL(reduce.0(x_1)) = x_1
   POL(reduce.1(x_1)) = x_1
   POL(result_0.0(x_1)) = x_1
   POL(result_0.1(x_1)) = x_1
   POL(result_1.0(x_1)) = x_1
   POL(result_1.1(x_1)) = x_1
   POL(result_f.0(x_1)) = x_1
   POL(result_f.1(x_1)) = x_1


----------------------------------------

(55)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(go_up.0(f.0(x0))) -> TOP.0(check_f.0(redex_f.0(x0)))
   TOP.0(go_up.0(f.1(x0))) -> TOP.0(check_f.0(redex_f.1(x0)))
   TOP.0(go_up.1(0.0(x0))) -> TOP.0(check_0.0(redex_0.0(x0)))
   TOP.0(go_up.1(0.1(x0))) -> TOP.0(check_0.0(redex_0.1(x0)))
   TOP.0(go_up.1(1.0(x0))) -> TOP.0(check_1.0(redex_1.0(x0)))
   TOP.0(go_up.1(1.1(x0))) -> TOP.0(check_1.0(redex_1.1(x0)))

The TRS R consists of the following rules:

   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   reduce.1(a.) -> go_up.0(f.1(a.))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))
   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   check_1.0(redex_1.0(x_1)) -> in_1_1.0(reduce.0(x_1))
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   check_0.0(redex_0.0(x_1)) -> in_0_1.0(reduce.0(x_1))
   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))
   redex_f.0(f.0(x)) -> result_f.0(b.)
   redex_f.0(f.1(x)) -> result_f.0(b.)
   check_f.0(result_f.0(x)) -> go_up.0(x)
   check_f.0(redex_f.0(x_1)) -> in_f_1.0(reduce.0(x_1))

The set Q consists of the following terms:

   reduce.0(f.0(x0))
   reduce.0(f.1(x0))
   reduce.1(0.0(x0))
   reduce.1(0.1(x0))
   reduce.1(1.0(x0))
   reduce.1(1.1(x0))
   reduce.1(a.)
   redex_f.0(f.0(x0))
   redex_f.0(f.1(x0))
   redex_0.0(f.0(x0))
   redex_0.0(f.1(x0))
   redex_1.0(f.0(x0))
   redex_1.0(f.1(x0))
   check_f.0(result_f.0(x0))
   check_f.0(result_f.1(x0))
   check_0.0(result_0.0(x0))
   check_0.0(result_0.1(x0))
   check_1.0(result_1.0(x0))
   check_1.0(result_1.1(x0))
   check_f.0(redex_f.0(x0))
   check_f.0(redex_f.1(x0))
   check_0.0(redex_0.0(x0))
   check_0.0(redex_0.1(x0))
   check_1.0(redex_1.0(x0))
   check_1.0(redex_1.1(x0))
   in_f_1.0(go_up.0(x0))
   in_f_1.0(go_up.1(x0))
   in_0_1.0(go_up.0(x0))
   in_0_1.0(go_up.1(x0))
   in_1_1.0(go_up.0(x0))
   in_1_1.0(go_up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(56) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   check_1.0(redex_1.0(x_1)) -> in_1_1.0(reduce.0(x_1))
   check_0.0(redex_0.0(x_1)) -> in_0_1.0(reduce.0(x_1))
   check_f.0(redex_f.0(x_1)) -> in_f_1.0(reduce.0(x_1))

Used ordering: Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = x_1
   POL(0.1(x_1)) = x_1
   POL(1.0(x_1)) = x_1
   POL(1.1(x_1)) = x_1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(check_0.0(x_1)) = 1 + x_1
   POL(check_1.0(x_1)) = x_1
   POL(check_f.0(x_1)) = 1 + x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(go_up.0(x_1)) = 1 + x_1
   POL(go_up.1(x_1)) = 1 + x_1
   POL(in_0_1.0(x_1)) = x_1
   POL(in_1_1.0(x_1)) = x_1
   POL(in_f_1.0(x_1)) = x_1
   POL(redex_0.0(x_1)) = x_1
   POL(redex_0.1(x_1)) = x_1
   POL(redex_1.0(x_1)) = 1 + x_1
   POL(redex_1.1(x_1)) = 1 + x_1
   POL(redex_f.0(x_1)) = x_1
   POL(redex_f.1(x_1)) = x_1
   POL(reduce.0(x_1)) = x_1
   POL(reduce.1(x_1)) = 1 + x_1
   POL(result_0.1(x_1)) = x_1
   POL(result_1.0(x_1)) = 1 + x_1
   POL(result_f.0(x_1)) = x_1


----------------------------------------

(57)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(go_up.0(f.0(x0))) -> TOP.0(check_f.0(redex_f.0(x0)))
   TOP.0(go_up.0(f.1(x0))) -> TOP.0(check_f.0(redex_f.1(x0)))
   TOP.0(go_up.1(0.0(x0))) -> TOP.0(check_0.0(redex_0.0(x0)))
   TOP.0(go_up.1(0.1(x0))) -> TOP.0(check_0.0(redex_0.1(x0)))
   TOP.0(go_up.1(1.0(x0))) -> TOP.0(check_1.0(redex_1.0(x0)))
   TOP.0(go_up.1(1.1(x0))) -> TOP.0(check_1.0(redex_1.1(x0)))

The TRS R consists of the following rules:

   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   reduce.1(a.) -> go_up.0(f.1(a.))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))
   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))
   redex_f.0(f.0(x)) -> result_f.0(b.)
   redex_f.0(f.1(x)) -> result_f.0(b.)
   check_f.0(result_f.0(x)) -> go_up.0(x)

The set Q consists of the following terms:

   reduce.0(f.0(x0))
   reduce.0(f.1(x0))
   reduce.1(0.0(x0))
   reduce.1(0.1(x0))
   reduce.1(1.0(x0))
   reduce.1(1.1(x0))
   reduce.1(a.)
   redex_f.0(f.0(x0))
   redex_f.0(f.1(x0))
   redex_0.0(f.0(x0))
   redex_0.0(f.1(x0))
   redex_1.0(f.0(x0))
   redex_1.0(f.1(x0))
   check_f.0(result_f.0(x0))
   check_f.0(result_f.1(x0))
   check_0.0(result_0.0(x0))
   check_0.0(result_0.1(x0))
   check_1.0(result_1.0(x0))
   check_1.0(result_1.1(x0))
   check_f.0(redex_f.0(x0))
   check_f.0(redex_f.1(x0))
   check_0.0(redex_0.0(x0))
   check_0.0(redex_0.1(x0))
   check_1.0(redex_1.0(x0))
   check_1.0(redex_1.1(x0))
   in_f_1.0(go_up.0(x0))
   in_f_1.0(go_up.1(x0))
   in_0_1.0(go_up.0(x0))
   in_0_1.0(go_up.1(x0))
   in_1_1.0(go_up.0(x0))
   in_1_1.0(go_up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(58) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   TOP.0(go_up.0(f.0(x0))) -> TOP.0(check_f.0(redex_f.0(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = x_1
   POL(0.1(x_1)) = x_1
   POL(1.0(x_1)) = 0
   POL(1.1(x_1)) = x_1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(check_0.0(x_1)) = 1
   POL(check_1.0(x_1)) = x_1
   POL(check_f.0(x_1)) = x_1
   POL(f.0(x_1)) = 1
   POL(f.1(x_1)) = 1
   POL(go_up.0(x_1)) = x_1
   POL(go_up.1(x_1)) = 1
   POL(in_0_1.0(x_1)) = 1
   POL(in_1_1.0(x_1)) = 1
   POL(in_f_1.0(x_1)) = 1
   POL(redex_0.0(x_1)) = 0
   POL(redex_0.1(x_1)) = x_1
   POL(redex_1.0(x_1)) = 1
   POL(redex_1.1(x_1)) = 1
   POL(redex_f.0(x_1)) = 0
   POL(redex_f.1(x_1)) = 1
   POL(reduce.1(x_1)) = 1
   POL(result_0.1(x_1)) = 0
   POL(result_1.0(x_1)) = x_1
   POL(result_f.0(x_1)) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   redex_f.0(f.0(x)) -> result_f.0(b.)
   redex_f.0(f.1(x)) -> result_f.0(b.)
   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   check_f.0(result_f.0(x)) -> go_up.0(x)
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))


----------------------------------------

(59)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(go_up.0(f.1(x0))) -> TOP.0(check_f.0(redex_f.1(x0)))
   TOP.0(go_up.1(0.0(x0))) -> TOP.0(check_0.0(redex_0.0(x0)))
   TOP.0(go_up.1(0.1(x0))) -> TOP.0(check_0.0(redex_0.1(x0)))
   TOP.0(go_up.1(1.0(x0))) -> TOP.0(check_1.0(redex_1.0(x0)))
   TOP.0(go_up.1(1.1(x0))) -> TOP.0(check_1.0(redex_1.1(x0)))

The TRS R consists of the following rules:

   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   reduce.1(a.) -> go_up.0(f.1(a.))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))
   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))
   redex_f.0(f.0(x)) -> result_f.0(b.)
   redex_f.0(f.1(x)) -> result_f.0(b.)
   check_f.0(result_f.0(x)) -> go_up.0(x)

The set Q consists of the following terms:

   reduce.0(f.0(x0))
   reduce.0(f.1(x0))
   reduce.1(0.0(x0))
   reduce.1(0.1(x0))
   reduce.1(1.0(x0))
   reduce.1(1.1(x0))
   reduce.1(a.)
   redex_f.0(f.0(x0))
   redex_f.0(f.1(x0))
   redex_0.0(f.0(x0))
   redex_0.0(f.1(x0))
   redex_1.0(f.0(x0))
   redex_1.0(f.1(x0))
   check_f.0(result_f.0(x0))
   check_f.0(result_f.1(x0))
   check_0.0(result_0.0(x0))
   check_0.0(result_0.1(x0))
   check_1.0(result_1.0(x0))
   check_1.0(result_1.1(x0))
   check_f.0(redex_f.0(x0))
   check_f.0(redex_f.1(x0))
   check_0.0(redex_0.0(x0))
   check_0.0(redex_0.1(x0))
   check_1.0(redex_1.0(x0))
   check_1.0(redex_1.1(x0))
   in_f_1.0(go_up.0(x0))
   in_f_1.0(go_up.1(x0))
   in_0_1.0(go_up.0(x0))
   in_0_1.0(go_up.1(x0))
   in_1_1.0(go_up.0(x0))
   in_1_1.0(go_up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(60) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   redex_f.0(f.0(x)) -> result_f.0(b.)
   redex_f.0(f.1(x)) -> result_f.0(b.)
   check_f.0(result_f.0(x)) -> go_up.0(x)
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = x_1
   POL(0.1(x_1)) = x_1
   POL(1.0(x_1)) = x_1
   POL(1.1(x_1)) = x_1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(check_0.0(x_1)) = x_1
   POL(check_1.0(x_1)) = x_1
   POL(check_f.0(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(go_up.0(x_1)) = x_1
   POL(go_up.1(x_1)) = x_1
   POL(in_0_1.0(x_1)) = x_1
   POL(in_1_1.0(x_1)) = x_1
   POL(in_f_1.0(x_1)) = x_1
   POL(redex_0.0(x_1)) = x_1
   POL(redex_0.1(x_1)) = x_1
   POL(redex_1.0(x_1)) = x_1
   POL(redex_1.1(x_1)) = x_1
   POL(redex_f.1(x_1)) = x_1
   POL(reduce.1(x_1)) = x_1
   POL(result_0.1(x_1)) = x_1
   POL(result_1.0(x_1)) = x_1


----------------------------------------

(61)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(go_up.0(f.1(x0))) -> TOP.0(check_f.0(redex_f.1(x0)))
   TOP.0(go_up.1(0.0(x0))) -> TOP.0(check_0.0(redex_0.0(x0)))
   TOP.0(go_up.1(0.1(x0))) -> TOP.0(check_0.0(redex_0.1(x0)))
   TOP.0(go_up.1(1.0(x0))) -> TOP.0(check_1.0(redex_1.0(x0)))
   TOP.0(go_up.1(1.1(x0))) -> TOP.0(check_1.0(redex_1.1(x0)))

The TRS R consists of the following rules:

   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   reduce.1(a.) -> go_up.0(f.1(a.))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))
   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))

The set Q consists of the following terms:

   reduce.0(f.0(x0))
   reduce.0(f.1(x0))
   reduce.1(0.0(x0))
   reduce.1(0.1(x0))
   reduce.1(1.0(x0))
   reduce.1(1.1(x0))
   reduce.1(a.)
   redex_f.0(f.0(x0))
   redex_f.0(f.1(x0))
   redex_0.0(f.0(x0))
   redex_0.0(f.1(x0))
   redex_1.0(f.0(x0))
   redex_1.0(f.1(x0))
   check_f.0(result_f.0(x0))
   check_f.0(result_f.1(x0))
   check_0.0(result_0.0(x0))
   check_0.0(result_0.1(x0))
   check_1.0(result_1.0(x0))
   check_1.0(result_1.1(x0))
   check_f.0(redex_f.0(x0))
   check_f.0(redex_f.1(x0))
   check_0.0(redex_0.0(x0))
   check_0.0(redex_0.1(x0))
   check_1.0(redex_1.0(x0))
   check_1.0(redex_1.1(x0))
   in_f_1.0(go_up.0(x0))
   in_f_1.0(go_up.1(x0))
   in_0_1.0(go_up.0(x0))
   in_0_1.0(go_up.1(x0))
   in_1_1.0(go_up.0(x0))
   in_1_1.0(go_up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(62) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   TOP.0(go_up.1(0.0(x0))) -> TOP.0(check_0.0(redex_0.0(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = 1
   POL(0.1(x_1)) = 1
   POL(1.0(x_1)) = 0
   POL(1.1(x_1)) = 0
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(check_0.0(x_1)) = x_1
   POL(check_1.0(x_1)) = 0
   POL(check_f.0(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = 0
   POL(go_up.0(x_1)) = 0
   POL(go_up.1(x_1)) = x_1
   POL(in_0_1.0(x_1)) = 1
   POL(in_1_1.0(x_1)) = 0
   POL(in_f_1.0(x_1)) = 0
   POL(redex_0.0(x_1)) = 0
   POL(redex_0.1(x_1)) = 1
   POL(redex_1.0(x_1)) = 0
   POL(redex_1.1(x_1)) = x_1
   POL(redex_f.1(x_1)) = 0
   POL(reduce.1(x_1)) = x_1
   POL(result_0.1(x_1)) = x_1
   POL(result_1.0(x_1)) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))


----------------------------------------

(63)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(go_up.0(f.1(x0))) -> TOP.0(check_f.0(redex_f.1(x0)))
   TOP.0(go_up.1(0.1(x0))) -> TOP.0(check_0.0(redex_0.1(x0)))
   TOP.0(go_up.1(1.0(x0))) -> TOP.0(check_1.0(redex_1.0(x0)))
   TOP.0(go_up.1(1.1(x0))) -> TOP.0(check_1.0(redex_1.1(x0)))

The TRS R consists of the following rules:

   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   reduce.1(a.) -> go_up.0(f.1(a.))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))
   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))

The set Q consists of the following terms:

   reduce.0(f.0(x0))
   reduce.0(f.1(x0))
   reduce.1(0.0(x0))
   reduce.1(0.1(x0))
   reduce.1(1.0(x0))
   reduce.1(1.1(x0))
   reduce.1(a.)
   redex_f.0(f.0(x0))
   redex_f.0(f.1(x0))
   redex_0.0(f.0(x0))
   redex_0.0(f.1(x0))
   redex_1.0(f.0(x0))
   redex_1.0(f.1(x0))
   check_f.0(result_f.0(x0))
   check_f.0(result_f.1(x0))
   check_0.0(result_0.0(x0))
   check_0.0(result_0.1(x0))
   check_1.0(result_1.0(x0))
   check_1.0(result_1.1(x0))
   check_f.0(redex_f.0(x0))
   check_f.0(redex_f.1(x0))
   check_0.0(redex_0.0(x0))
   check_0.0(redex_0.1(x0))
   check_1.0(redex_1.0(x0))
   check_1.0(redex_1.1(x0))
   in_f_1.0(go_up.0(x0))
   in_f_1.0(go_up.1(x0))
   in_0_1.0(go_up.0(x0))
   in_0_1.0(go_up.1(x0))
   in_1_1.0(go_up.0(x0))
   in_1_1.0(go_up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(64) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   TOP.0(go_up.1(1.0(x0))) -> TOP.0(check_1.0(redex_1.0(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = 0
   POL(0.1(x_1)) = 0
   POL(1.0(x_1)) = 1
   POL(1.1(x_1)) = 1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(check_0.0(x_1)) = x_1
   POL(check_1.0(x_1)) = x_1
   POL(check_f.0(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = 0
   POL(go_up.0(x_1)) = 0
   POL(go_up.1(x_1)) = x_1
   POL(in_0_1.0(x_1)) = 0
   POL(in_1_1.0(x_1)) = 1
   POL(in_f_1.0(x_1)) = 0
   POL(redex_0.0(x_1)) = 1
   POL(redex_0.1(x_1)) = 0
   POL(redex_1.0(x_1)) = 0
   POL(redex_1.1(x_1)) = 1
   POL(redex_f.1(x_1)) = 0
   POL(reduce.1(x_1)) = 1
   POL(result_0.1(x_1)) = x_1
   POL(result_1.0(x_1)) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))


----------------------------------------

(65)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(go_up.0(f.1(x0))) -> TOP.0(check_f.0(redex_f.1(x0)))
   TOP.0(go_up.1(0.1(x0))) -> TOP.0(check_0.0(redex_0.1(x0)))
   TOP.0(go_up.1(1.1(x0))) -> TOP.0(check_1.0(redex_1.1(x0)))

The TRS R consists of the following rules:

   check_1.0(redex_1.1(x_1)) -> in_1_1.0(reduce.1(x_1))
   reduce.1(0.0(x_1)) -> check_0.0(redex_0.0(x_1))
   reduce.1(0.1(x_1)) -> check_0.0(redex_0.1(x_1))
   reduce.1(1.0(x_1)) -> check_1.0(redex_1.0(x_1))
   reduce.1(1.1(x_1)) -> check_1.0(redex_1.1(x_1))
   reduce.1(a.) -> go_up.0(f.1(a.))
   in_1_1.0(go_up.0(x_1)) -> go_up.1(1.0(x_1))
   in_1_1.0(go_up.1(x_1)) -> go_up.1(1.1(x_1))
   redex_1.0(f.0(x)) -> result_1.0(f.1(0.0(x)))
   redex_1.0(f.1(x)) -> result_1.0(f.1(0.1(x)))
   check_1.0(result_1.0(x)) -> go_up.0(x)
   check_0.0(redex_0.1(x_1)) -> in_0_1.0(reduce.1(x_1))
   in_0_1.0(go_up.0(x_1)) -> go_up.1(0.0(x_1))
   in_0_1.0(go_up.1(x_1)) -> go_up.1(0.1(x_1))
   redex_0.0(f.0(x)) -> result_0.1(1.0(x))
   redex_0.0(f.1(x)) -> result_0.1(1.1(x))
   check_0.0(result_0.1(x)) -> go_up.1(x)
   check_f.0(redex_f.1(x_1)) -> in_f_1.0(reduce.1(x_1))
   in_f_1.0(go_up.0(x_1)) -> go_up.0(f.0(x_1))
   in_f_1.0(go_up.1(x_1)) -> go_up.0(f.1(x_1))

The set Q consists of the following terms:

   reduce.0(f.0(x0))
   reduce.0(f.1(x0))
   reduce.1(0.0(x0))
   reduce.1(0.1(x0))
   reduce.1(1.0(x0))
   reduce.1(1.1(x0))
   reduce.1(a.)
   redex_f.0(f.0(x0))
   redex_f.0(f.1(x0))
   redex_0.0(f.0(x0))
   redex_0.0(f.1(x0))
   redex_1.0(f.0(x0))
   redex_1.0(f.1(x0))
   check_f.0(result_f.0(x0))
   check_f.0(result_f.1(x0))
   check_0.0(result_0.0(x0))
   check_0.0(result_0.1(x0))
   check_1.0(result_1.0(x0))
   check_1.0(result_1.1(x0))
   check_f.0(redex_f.0(x0))
   check_f.0(redex_f.1(x0))
   check_0.0(redex_0.0(x0))
   check_0.0(redex_0.1(x0))
   check_1.0(redex_1.0(x0))
   check_1.0(redex_1.1(x0))
   in_f_1.0(go_up.0(x0))
   in_f_1.0(go_up.1(x0))
   in_0_1.0(go_up.0(x0))
   in_0_1.0(go_up.1(x0))
   in_1_1.0(go_up.0(x0))
   in_1_1.0(go_up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(66) PisEmptyProof (SOUND)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(67)
TRUE

----------------------------------------

(68)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0)))
   TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0)))
   TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0)))

The TRS R consists of the following rules:

   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(69) TransformationProof (SOUND)
By rewriting [LPAR04] the rule TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(go_up(f(x0))) -> TOP(in_f_1(reduce(x0))),TOP(go_up(f(x0))) -> TOP(in_f_1(reduce(x0))))


----------------------------------------

(70)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0)))
   TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0)))
   TOP(go_up(f(x0))) -> TOP(in_f_1(reduce(x0)))

The TRS R consists of the following rules:

   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(71) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(72)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0)))
   TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0)))
   TOP(go_up(f(x0))) -> TOP(in_f_1(reduce(x0)))

The TRS R consists of the following rules:

   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   check_1(result_1(x)) -> go_up(x)
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   check_0(result_0(x)) -> go_up(x)
   in_0_1(go_up(x_1)) -> go_up(0(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(73) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   redex_f(f(x0))
   check_f(result_f(x0))
   check_f(redex_f(x0))


----------------------------------------

(74)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(0(x0))) -> TOP(check_0(redex_0(x0)))
   TOP(go_up(1(x0))) -> TOP(check_1(redex_1(x0)))
   TOP(go_up(f(x0))) -> TOP(in_f_1(reduce(x0)))

The TRS R consists of the following rules:

   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   check_1(result_1(x)) -> go_up(x)
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   check_0(result_0(x)) -> go_up(x)
   in_0_1(go_up(x_1)) -> go_up(0(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_0(f(x0))
   redex_1(f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(75) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(76)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(x)) -> TOP(reduce(x))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(77) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   top(go_up(x0))


----------------------------------------

(78)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(go_up(x)) -> TOP(reduce(x))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(0(x_1)) -> check_0(redex_0(x_1))
   reduce(1(x_1)) -> check_1(redex_1(x_1))
   reduce(a) -> go_up(f(a))
   redex_1(f(x)) -> result_1(f(0(x)))
   check_1(result_1(x)) -> go_up(x)
   check_1(redex_1(x_1)) -> in_1_1(reduce(x_1))
   in_1_1(go_up(x_1)) -> go_up(1(x_1))
   redex_0(f(x)) -> result_0(1(x))
   check_0(result_0(x)) -> go_up(x)
   check_0(redex_0(x_1)) -> in_0_1(reduce(x_1))
   in_0_1(go_up(x_1)) -> go_up(0(x_1))
   redex_f(f(x)) -> result_f(b)
   check_f(result_f(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))

The set Q consists of the following terms:

   reduce(f(x0))
   reduce(0(x0))
   reduce(1(x0))
   reduce(a)
   redex_f(f(x0))
   redex_0(f(x0))
   redex_1(f(x0))
   check_f(result_f(x0))
   check_0(result_0(x0))
   check_1(result_1(x0))
   check_f(redex_f(x0))
   check_0(redex_0(x0))
   check_1(redex_1(x0))
   in_f_1(go_up(x0))
   in_0_1(go_up(x0))
   in_1_1(go_up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(79) Raffelsieper-Zantema-Transformation (SOUND)
We applied the Raffelsieper-Zantema transformation  to transform the outermost TRS to a standard TRS.
----------------------------------------

(80)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   top(up(x)) -> top(down(x))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   1_flat(up(x_1)) -> up(1(x_1))

Q is empty.

----------------------------------------

(81) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))

The TRS R 2 is 
   top(up(x)) -> top(down(x))

The signature Sigma is {top_1}
----------------------------------------

(82)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   top(up(x)) -> top(down(x))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   1_flat(up(x_1)) -> up(1(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   top(up(x0))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))


----------------------------------------

(83) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(84)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(x)) -> TOP(down(x))
   TOP(up(x)) -> DOWN(x)
   DOWN(f(a)) -> F_FLAT(down(a))
   DOWN(f(a)) -> DOWN(a)
   DOWN(f(b)) -> F_FLAT(down(b))
   DOWN(f(b)) -> DOWN(b)
   DOWN(f(0(y5))) -> F_FLAT(down(0(y5)))
   DOWN(f(0(y5))) -> DOWN(0(y5))
   DOWN(f(1(y6))) -> F_FLAT(down(1(y6)))
   DOWN(f(1(y6))) -> DOWN(1(y6))
   DOWN(f(fresh_constant)) -> F_FLAT(down(fresh_constant))
   DOWN(f(fresh_constant)) -> DOWN(fresh_constant)
   DOWN(0(a)) -> 0_FLAT(down(a))
   DOWN(0(a)) -> DOWN(a)
   DOWN(0(b)) -> 0_FLAT(down(b))
   DOWN(0(b)) -> DOWN(b)
   DOWN(0(0(y9))) -> 0_FLAT(down(0(y9)))
   DOWN(0(0(y9))) -> DOWN(0(y9))
   DOWN(0(1(y10))) -> 0_FLAT(down(1(y10)))
   DOWN(0(1(y10))) -> DOWN(1(y10))
   DOWN(0(fresh_constant)) -> 0_FLAT(down(fresh_constant))
   DOWN(0(fresh_constant)) -> DOWN(fresh_constant)
   DOWN(1(a)) -> 1_FLAT(down(a))
   DOWN(1(a)) -> DOWN(a)
   DOWN(1(b)) -> 1_FLAT(down(b))
   DOWN(1(b)) -> DOWN(b)
   DOWN(1(0(y13))) -> 1_FLAT(down(0(y13)))
   DOWN(1(0(y13))) -> DOWN(0(y13))
   DOWN(1(1(y14))) -> 1_FLAT(down(1(y14)))
   DOWN(1(1(y14))) -> DOWN(1(y14))
   DOWN(1(fresh_constant)) -> 1_FLAT(down(fresh_constant))
   DOWN(1(fresh_constant)) -> DOWN(fresh_constant)

The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   top(up(x)) -> top(down(x))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   1_flat(up(x_1)) -> up(1(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   top(up(x0))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(85) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 27 less nodes.
----------------------------------------

(86)
Complex Obligation (AND)

----------------------------------------

(87)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DOWN(0(1(y10))) -> DOWN(1(y10))
   DOWN(1(0(y13))) -> DOWN(0(y13))
   DOWN(0(0(y9))) -> DOWN(0(y9))
   DOWN(1(1(y14))) -> DOWN(1(y14))

The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   top(up(x)) -> top(down(x))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   1_flat(up(x_1)) -> up(1(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   top(up(x0))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(88) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(89)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DOWN(0(1(y10))) -> DOWN(1(y10))
   DOWN(1(0(y13))) -> DOWN(0(y13))
   DOWN(0(0(y9))) -> DOWN(0(y9))
   DOWN(1(1(y14))) -> DOWN(1(y14))

R is empty.
The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   top(up(x0))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(90) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   top(up(x0))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))


----------------------------------------

(91)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DOWN(0(1(y10))) -> DOWN(1(y10))
   DOWN(1(0(y13))) -> DOWN(0(y13))
   DOWN(0(0(y9))) -> DOWN(0(y9))
   DOWN(1(1(y14))) -> DOWN(1(y14))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(92) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*DOWN(1(0(y13))) -> DOWN(0(y13))
The graph contains the following edges 1 > 1


*DOWN(1(1(y14))) -> DOWN(1(y14))
The graph contains the following edges 1 > 1


*DOWN(0(0(y9))) -> DOWN(0(y9))
The graph contains the following edges 1 > 1


*DOWN(0(1(y10))) -> DOWN(1(y10))
The graph contains the following edges 1 > 1


----------------------------------------

(93)
YES

----------------------------------------

(94)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(x)) -> TOP(down(x))

The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   top(up(x)) -> top(down(x))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   1_flat(up(x_1)) -> up(1(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   top(up(x0))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(95) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(96)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(x)) -> TOP(down(x))

The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   1_flat(up(x_1)) -> up(1(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   top(up(x0))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(97) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   top(up(x0))


----------------------------------------

(98)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(x)) -> TOP(down(x))

The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   1_flat(up(x_1)) -> up(1(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(99) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TOP(up(x)) -> TOP(down(x)) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(up(a)) -> TOP(up(f(a))),TOP(up(a)) -> TOP(up(f(a))))
   (TOP(up(f(f(x0)))) -> TOP(up(b)),TOP(up(f(f(x0)))) -> TOP(up(b)))
   (TOP(up(0(f(x0)))) -> TOP(up(1(x0))),TOP(up(0(f(x0)))) -> TOP(up(1(x0))))
   (TOP(up(1(f(x0)))) -> TOP(up(f(0(x0)))),TOP(up(1(f(x0)))) -> TOP(up(f(0(x0)))))
   (TOP(up(f(a))) -> TOP(f_flat(down(a))),TOP(up(f(a))) -> TOP(f_flat(down(a))))
   (TOP(up(f(b))) -> TOP(f_flat(down(b))),TOP(up(f(b))) -> TOP(f_flat(down(b))))
   (TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0)))),TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0)))))
   (TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0)))),TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0)))))
   (TOP(up(f(fresh_constant))) -> TOP(f_flat(down(fresh_constant))),TOP(up(f(fresh_constant))) -> TOP(f_flat(down(fresh_constant))))
   (TOP(up(0(a))) -> TOP(0_flat(down(a))),TOP(up(0(a))) -> TOP(0_flat(down(a))))
   (TOP(up(0(b))) -> TOP(0_flat(down(b))),TOP(up(0(b))) -> TOP(0_flat(down(b))))
   (TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0)))),TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0)))))
   (TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0)))),TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0)))))
   (TOP(up(0(fresh_constant))) -> TOP(0_flat(down(fresh_constant))),TOP(up(0(fresh_constant))) -> TOP(0_flat(down(fresh_constant))))
   (TOP(up(1(a))) -> TOP(1_flat(down(a))),TOP(up(1(a))) -> TOP(1_flat(down(a))))
   (TOP(up(1(b))) -> TOP(1_flat(down(b))),TOP(up(1(b))) -> TOP(1_flat(down(b))))
   (TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0)))),TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0)))))
   (TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0)))),TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0)))))
   (TOP(up(1(fresh_constant))) -> TOP(1_flat(down(fresh_constant))),TOP(up(1(fresh_constant))) -> TOP(1_flat(down(fresh_constant))))


----------------------------------------

(100)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(a)) -> TOP(up(f(a)))
   TOP(up(f(f(x0)))) -> TOP(up(b))
   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(f(a))) -> TOP(f_flat(down(a)))
   TOP(up(f(b))) -> TOP(f_flat(down(b)))
   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(f(fresh_constant))) -> TOP(f_flat(down(fresh_constant)))
   TOP(up(0(a))) -> TOP(0_flat(down(a)))
   TOP(up(0(b))) -> TOP(0_flat(down(b)))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(0(fresh_constant))) -> TOP(0_flat(down(fresh_constant)))
   TOP(up(1(a))) -> TOP(1_flat(down(a)))
   TOP(up(1(b))) -> TOP(1_flat(down(b)))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(1(fresh_constant))) -> TOP(1_flat(down(fresh_constant)))

The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   1_flat(up(x_1)) -> up(1(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(101) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 8 less nodes.
----------------------------------------

(102)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(a))) -> TOP(f_flat(down(a)))
   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(1(a))) -> TOP(1_flat(down(a)))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(0(a))) -> TOP(0_flat(down(a)))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))

The TRS R consists of the following rules:

   down(a) -> up(f(a))
   down(f(f(x))) -> up(b)
   down(0(f(x))) -> up(1(x))
   down(1(f(x))) -> up(f(0(x)))
   down(f(a)) -> f_flat(down(a))
   down(f(b)) -> f_flat(down(b))
   down(f(0(y5))) -> f_flat(down(0(y5)))
   down(f(1(y6))) -> f_flat(down(1(y6)))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   1_flat(up(x_1)) -> up(1(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(103) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(104)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(a))) -> TOP(f_flat(down(a)))
   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(1(a))) -> TOP(1_flat(down(a)))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(0(a))) -> TOP(0_flat(down(a)))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(105) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(a))) -> TOP(f_flat(down(a))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(a))) -> TOP(f_flat(up(f(a)))),TOP(up(f(a))) -> TOP(f_flat(up(f(a)))))


----------------------------------------

(106)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(1(a))) -> TOP(1_flat(down(a)))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(0(a))) -> TOP(0_flat(down(a)))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(f(a))) -> TOP(f_flat(up(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(107) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(1(a))) -> TOP(1_flat(down(a))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(up(1(a))) -> TOP(1_flat(up(f(a)))),TOP(up(1(a))) -> TOP(1_flat(up(f(a)))))


----------------------------------------

(108)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(0(a))) -> TOP(0_flat(down(a)))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(f(a))) -> TOP(f_flat(up(f(a))))
   TOP(up(1(a))) -> TOP(1_flat(up(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(109) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(0(a))) -> TOP(0_flat(down(a))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(up(0(a))) -> TOP(0_flat(up(f(a)))),TOP(up(0(a))) -> TOP(0_flat(up(f(a)))))


----------------------------------------

(110)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(f(a))) -> TOP(f_flat(up(f(a))))
   TOP(up(1(a))) -> TOP(1_flat(up(f(a))))
   TOP(up(0(a))) -> TOP(0_flat(up(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(111) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(a))) -> TOP(f_flat(up(f(a)))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(a))) -> TOP(up(f(f(a)))),TOP(up(f(a))) -> TOP(up(f(f(a)))))


----------------------------------------

(112)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(1(a))) -> TOP(1_flat(up(f(a))))
   TOP(up(0(a))) -> TOP(0_flat(up(f(a))))
   TOP(up(f(a))) -> TOP(up(f(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(113) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(114)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(1(a))) -> TOP(1_flat(up(f(a))))
   TOP(up(0(a))) -> TOP(0_flat(up(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(115) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(1(a))) -> TOP(1_flat(up(f(a)))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(up(1(a))) -> TOP(up(1(f(a)))),TOP(up(1(a))) -> TOP(up(1(f(a)))))


----------------------------------------

(116)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(0(a))) -> TOP(0_flat(up(f(a))))
   TOP(up(1(a))) -> TOP(up(1(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(117) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(0(a))) -> TOP(0_flat(up(f(a)))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(up(0(a))) -> TOP(up(0(f(a)))),TOP(up(0(a))) -> TOP(up(0(f(a)))))


----------------------------------------

(118)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(0(f(x0)))) -> TOP(up(1(x0)))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(1(a))) -> TOP(up(1(f(a))))
   TOP(up(0(a))) -> TOP(up(0(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(119) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule TOP(up(0(f(x0)))) -> TOP(up(1(x0))) we obtained the following new rules [LPAR04]:

   (TOP(up(0(f(f(y_0))))) -> TOP(up(1(f(y_0)))),TOP(up(0(f(f(y_0))))) -> TOP(up(1(f(y_0)))))
   (TOP(up(0(f(0(y_0))))) -> TOP(up(1(0(y_0)))),TOP(up(0(f(0(y_0))))) -> TOP(up(1(0(y_0)))))
   (TOP(up(0(f(1(y_0))))) -> TOP(up(1(1(y_0)))),TOP(up(0(f(1(y_0))))) -> TOP(up(1(1(y_0)))))
   (TOP(up(0(f(a)))) -> TOP(up(1(a))),TOP(up(0(f(a)))) -> TOP(up(1(a))))


----------------------------------------

(120)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(1(a))) -> TOP(up(1(f(a))))
   TOP(up(0(a))) -> TOP(up(0(f(a))))
   TOP(up(0(f(f(y_0))))) -> TOP(up(1(f(y_0))))
   TOP(up(0(f(0(y_0))))) -> TOP(up(1(0(y_0))))
   TOP(up(0(f(1(y_0))))) -> TOP(up(1(1(y_0))))
   TOP(up(0(f(a)))) -> TOP(up(1(a)))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(121) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   TOP(up(0(f(f(y_0))))) -> TOP(up(1(f(y_0))))
   TOP(up(0(f(0(y_0))))) -> TOP(up(1(0(y_0))))
   TOP(up(0(f(1(y_0))))) -> TOP(up(1(1(y_0))))
   TOP(up(0(f(a)))) -> TOP(up(1(a)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0(x_1)) = 1
   POL(0_flat(x_1)) = 1
   POL(1(x_1)) = 0
   POL(1_flat(x_1)) = 0
   POL(TOP(x_1)) = x_1
   POL(a) = 0
   POL(b) = 0
   POL(down(x_1)) = 0
   POL(f(x_1)) = 0
   POL(f_flat(x_1)) = 0
   POL(fresh_constant) = 0
   POL(up(x_1)) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   f_flat(up(x_1)) -> up(f(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   1_flat(up(x_1)) -> up(1(x_1))


----------------------------------------

(122)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(1(a))) -> TOP(up(1(f(a))))
   TOP(up(0(a))) -> TOP(up(0(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(123) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(124)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(1(a))) -> TOP(up(1(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(125) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   TOP(up(1(f(x0)))) -> TOP(up(f(0(x0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0(x_1)) = 0
   POL(0_flat(x_1)) = 0
   POL(1(x_1)) = 1
   POL(1_flat(x_1)) = 1
   POL(TOP(x_1)) = x_1
   POL(a) = 0
   POL(b) = 0
   POL(down(x_1)) = 0
   POL(f(x_1)) = 0
   POL(f_flat(x_1)) = 0
   POL(fresh_constant) = 0
   POL(up(x_1)) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   f_flat(up(x_1)) -> up(f(x_1))
   0_flat(up(x_1)) -> up(0(x_1))
   1_flat(up(x_1)) -> up(1(x_1))


----------------------------------------

(126)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))
   TOP(up(1(a))) -> TOP(up(1(f(a))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(127) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(128)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(129) SplitQDPProof (EQUIVALENT)
We show in the first subproof that some pairs and rules can be removed, afterwards, we continue with the remaining DP-Problem

----------------------------------------

(130)
Complex Obligation (AND)

----------------------------------------

(131)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))

The TRS R consists of the following rules:

   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(1(fresh_constant)) -> 1_flat(down(fresh_constant))
   f_flat(up(x_1)) -> up(f(x_1))
   1_flat(up(x_1)) -> up(1(x_1))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   down(0(fresh_constant)) -> 0_flat(down(fresh_constant))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(132) SemLabProof (SOUND)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
a: 0
b: 0
down: 0
f: 0
0: 0
fresh_constant: 1
up: 0
1: 0
1_flat: 0
0_flat: 0
f_flat: 0
TOP: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

(133)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(f.0(0.0(x0)))) -> TOP.0(f_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(f.0(0.1(x0)))) -> TOP.0(f_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(f.0(1.0(x0)))) -> TOP.0(f_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(f.0(1.1(x0)))) -> TOP.0(f_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(0.0(0.0(x0)))) -> TOP.0(0_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(0.0(0.1(x0)))) -> TOP.0(0_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(0.0(1.0(x0)))) -> TOP.0(0_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(0.0(1.1(x0)))) -> TOP.0(0_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(1.0(0.0(x0)))) -> TOP.0(1_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(1.0(0.1(x0)))) -> TOP.0(1_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(1.0(1.0(x0)))) -> TOP.0(1_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(1.0(1.1(x0)))) -> TOP.0(1_flat.0(down.0(1.1(x0))))

The TRS R consists of the following rules:

   down.0(1.0(f.0(x))) -> up.0(f.0(0.0(x)))
   down.0(1.0(f.1(x))) -> up.0(f.0(0.1(x)))
   down.0(1.0(a.)) -> 1_flat.0(down.0(a.))
   down.0(1.0(b.)) -> 1_flat.0(down.0(b.))
   down.0(1.0(0.0(y13))) -> 1_flat.0(down.0(0.0(y13)))
   down.0(1.0(0.1(y13))) -> 1_flat.0(down.0(0.1(y13)))
   down.0(1.0(1.0(y14))) -> 1_flat.0(down.0(1.0(y14)))
   down.0(1.0(1.1(y14))) -> 1_flat.0(down.0(1.1(y14)))
   down.0(1.1(fresh_constant.)) -> 1_flat.0(down.1(fresh_constant.))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))
   f_flat.0(up.1(x_1)) -> up.0(f.1(x_1))
   1_flat.0(up.0(x_1)) -> up.0(1.0(x_1))
   1_flat.0(up.1(x_1)) -> up.0(1.1(x_1))
   down.0(0.0(f.0(x))) -> up.0(1.0(x))
   down.0(0.0(f.1(x))) -> up.0(1.1(x))
   down.0(0.0(a.)) -> 0_flat.0(down.0(a.))
   down.0(0.0(b.)) -> 0_flat.0(down.0(b.))
   down.0(0.0(0.0(y9))) -> 0_flat.0(down.0(0.0(y9)))
   down.0(0.0(0.1(y9))) -> 0_flat.0(down.0(0.1(y9)))
   down.0(0.0(1.0(y10))) -> 0_flat.0(down.0(1.0(y10)))
   down.0(0.0(1.1(y10))) -> 0_flat.0(down.0(1.1(y10)))
   down.0(0.1(fresh_constant.)) -> 0_flat.0(down.1(fresh_constant.))
   0_flat.0(up.0(x_1)) -> up.0(0.0(x_1))
   0_flat.0(up.1(x_1)) -> up.0(0.1(x_1))
   down.0(a.) -> up.0(f.0(a.))

The set Q consists of the following terms:

   down.0(a.)
   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(0.0(f.0(x0)))
   down.0(0.0(f.1(x0)))
   down.0(1.0(f.0(x0)))
   down.0(1.0(f.1(x0)))
   down.0(f.0(a.))
   down.0(f.0(b.))
   down.0(f.0(0.0(x0)))
   down.0(f.0(0.1(x0)))
   down.0(f.0(1.0(x0)))
   down.0(f.0(1.1(x0)))
   down.0(f.1(fresh_constant.))
   down.0(0.0(a.))
   down.0(0.0(b.))
   down.0(0.0(0.0(x0)))
   down.0(0.0(0.1(x0)))
   down.0(0.0(1.0(x0)))
   down.0(0.0(1.1(x0)))
   down.0(0.1(fresh_constant.))
   down.0(1.0(a.))
   down.0(1.0(b.))
   down.0(1.0(0.0(x0)))
   down.0(1.0(0.1(x0)))
   down.0(1.0(1.0(x0)))
   down.0(1.0(1.1(x0)))
   down.0(1.1(fresh_constant.))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   0_flat.0(up.0(x0))
   0_flat.0(up.1(x0))
   1_flat.0(up.0(x0))
   1_flat.0(up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(134) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   down.0(1.0(f.1(x))) -> up.0(f.0(0.1(x)))
   f_flat.0(up.1(x_1)) -> up.0(f.1(x_1))
   1_flat.0(up.1(x_1)) -> up.0(1.1(x_1))
   down.0(0.0(f.1(x))) -> up.0(1.1(x))
   0_flat.0(up.1(x_1)) -> up.0(0.1(x_1))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = 1 + x_1
   POL(0.1(x_1)) = 1 + x_1
   POL(0_flat.0(x_1)) = 1 + x_1
   POL(1.0(x_1)) = 1 + x_1
   POL(1.1(x_1)) = 1 + x_1
   POL(1_flat.0(x_1)) = 1 + x_1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(x_1)) = x_1
   POL(down.1(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = 1 + x_1
   POL(f_flat.0(x_1)) = x_1
   POL(fresh_constant.) = 0
   POL(up.0(x_1)) = x_1
   POL(up.1(x_1)) = 1 + x_1


----------------------------------------

(135)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(f.0(0.0(x0)))) -> TOP.0(f_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(f.0(0.1(x0)))) -> TOP.0(f_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(f.0(1.0(x0)))) -> TOP.0(f_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(f.0(1.1(x0)))) -> TOP.0(f_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(0.0(0.0(x0)))) -> TOP.0(0_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(0.0(0.1(x0)))) -> TOP.0(0_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(0.0(1.0(x0)))) -> TOP.0(0_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(0.0(1.1(x0)))) -> TOP.0(0_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(1.0(0.0(x0)))) -> TOP.0(1_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(1.0(0.1(x0)))) -> TOP.0(1_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(1.0(1.0(x0)))) -> TOP.0(1_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(1.0(1.1(x0)))) -> TOP.0(1_flat.0(down.0(1.1(x0))))

The TRS R consists of the following rules:

   down.0(1.1(fresh_constant.)) -> 1_flat.0(down.1(fresh_constant.))
   1_flat.0(up.0(x_1)) -> up.0(1.0(x_1))
   down.0(1.0(f.0(x))) -> up.0(f.0(0.0(x)))
   down.0(1.0(a.)) -> 1_flat.0(down.0(a.))
   down.0(1.0(b.)) -> 1_flat.0(down.0(b.))
   down.0(1.0(0.0(y13))) -> 1_flat.0(down.0(0.0(y13)))
   down.0(1.0(0.1(y13))) -> 1_flat.0(down.0(0.1(y13)))
   down.0(1.0(1.0(y14))) -> 1_flat.0(down.0(1.0(y14)))
   down.0(1.0(1.1(y14))) -> 1_flat.0(down.0(1.1(y14)))
   down.0(0.1(fresh_constant.)) -> 0_flat.0(down.1(fresh_constant.))
   down.0(0.0(f.0(x))) -> up.0(1.0(x))
   down.0(0.0(a.)) -> 0_flat.0(down.0(a.))
   down.0(0.0(b.)) -> 0_flat.0(down.0(b.))
   down.0(0.0(0.0(y9))) -> 0_flat.0(down.0(0.0(y9)))
   down.0(0.0(0.1(y9))) -> 0_flat.0(down.0(0.1(y9)))
   down.0(0.0(1.0(y10))) -> 0_flat.0(down.0(1.0(y10)))
   down.0(0.0(1.1(y10))) -> 0_flat.0(down.0(1.1(y10)))
   0_flat.0(up.0(x_1)) -> up.0(0.0(x_1))
   down.0(a.) -> up.0(f.0(a.))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(a.)
   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(0.0(f.0(x0)))
   down.0(0.0(f.1(x0)))
   down.0(1.0(f.0(x0)))
   down.0(1.0(f.1(x0)))
   down.0(f.0(a.))
   down.0(f.0(b.))
   down.0(f.0(0.0(x0)))
   down.0(f.0(0.1(x0)))
   down.0(f.0(1.0(x0)))
   down.0(f.0(1.1(x0)))
   down.0(f.1(fresh_constant.))
   down.0(0.0(a.))
   down.0(0.0(b.))
   down.0(0.0(0.0(x0)))
   down.0(0.0(0.1(x0)))
   down.0(0.0(1.0(x0)))
   down.0(0.0(1.1(x0)))
   down.0(0.1(fresh_constant.))
   down.0(1.0(a.))
   down.0(1.0(b.))
   down.0(1.0(0.0(x0)))
   down.0(1.0(0.1(x0)))
   down.0(1.0(1.0(x0)))
   down.0(1.0(1.1(x0)))
   down.0(1.1(fresh_constant.))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   0_flat.0(up.0(x0))
   0_flat.0(up.1(x0))
   1_flat.0(up.0(x0))
   1_flat.0(up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(136) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   down.0(1.1(fresh_constant.)) -> 1_flat.0(down.1(fresh_constant.))
   down.0(0.1(fresh_constant.)) -> 0_flat.0(down.1(fresh_constant.))

Used ordering: Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = x_1
   POL(0.1(x_1)) = 1 + x_1
   POL(0_flat.0(x_1)) = x_1
   POL(1.0(x_1)) = x_1
   POL(1.1(x_1)) = 1 + x_1
   POL(1_flat.0(x_1)) = x_1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(x_1)) = x_1
   POL(down.1(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f_flat.0(x_1)) = x_1
   POL(fresh_constant.) = 0
   POL(up.0(x_1)) = x_1


----------------------------------------

(137)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(f.0(0.0(x0)))) -> TOP.0(f_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(f.0(0.1(x0)))) -> TOP.0(f_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(f.0(1.0(x0)))) -> TOP.0(f_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(f.0(1.1(x0)))) -> TOP.0(f_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(0.0(0.0(x0)))) -> TOP.0(0_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(0.0(0.1(x0)))) -> TOP.0(0_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(0.0(1.0(x0)))) -> TOP.0(0_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(0.0(1.1(x0)))) -> TOP.0(0_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(1.0(0.0(x0)))) -> TOP.0(1_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(1.0(0.1(x0)))) -> TOP.0(1_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(1.0(1.0(x0)))) -> TOP.0(1_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(1.0(1.1(x0)))) -> TOP.0(1_flat.0(down.0(1.1(x0))))

The TRS R consists of the following rules:

   1_flat.0(up.0(x_1)) -> up.0(1.0(x_1))
   down.0(1.0(f.0(x))) -> up.0(f.0(0.0(x)))
   down.0(1.0(a.)) -> 1_flat.0(down.0(a.))
   down.0(1.0(b.)) -> 1_flat.0(down.0(b.))
   down.0(1.0(0.0(y13))) -> 1_flat.0(down.0(0.0(y13)))
   down.0(1.0(0.1(y13))) -> 1_flat.0(down.0(0.1(y13)))
   down.0(1.0(1.0(y14))) -> 1_flat.0(down.0(1.0(y14)))
   down.0(1.0(1.1(y14))) -> 1_flat.0(down.0(1.1(y14)))
   down.0(0.0(f.0(x))) -> up.0(1.0(x))
   down.0(0.0(a.)) -> 0_flat.0(down.0(a.))
   down.0(0.0(b.)) -> 0_flat.0(down.0(b.))
   down.0(0.0(0.0(y9))) -> 0_flat.0(down.0(0.0(y9)))
   down.0(0.0(0.1(y9))) -> 0_flat.0(down.0(0.1(y9)))
   down.0(0.0(1.0(y10))) -> 0_flat.0(down.0(1.0(y10)))
   down.0(0.0(1.1(y10))) -> 0_flat.0(down.0(1.1(y10)))
   0_flat.0(up.0(x_1)) -> up.0(0.0(x_1))
   down.0(a.) -> up.0(f.0(a.))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(a.)
   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(0.0(f.0(x0)))
   down.0(0.0(f.1(x0)))
   down.0(1.0(f.0(x0)))
   down.0(1.0(f.1(x0)))
   down.0(f.0(a.))
   down.0(f.0(b.))
   down.0(f.0(0.0(x0)))
   down.0(f.0(0.1(x0)))
   down.0(f.0(1.0(x0)))
   down.0(f.0(1.1(x0)))
   down.0(f.1(fresh_constant.))
   down.0(0.0(a.))
   down.0(0.0(b.))
   down.0(0.0(0.0(x0)))
   down.0(0.0(0.1(x0)))
   down.0(0.0(1.0(x0)))
   down.0(0.0(1.1(x0)))
   down.0(0.1(fresh_constant.))
   down.0(1.0(a.))
   down.0(1.0(b.))
   down.0(1.0(0.0(x0)))
   down.0(1.0(0.1(x0)))
   down.0(1.0(1.0(x0)))
   down.0(1.0(1.1(x0)))
   down.0(1.1(fresh_constant.))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   0_flat.0(up.0(x0))
   0_flat.0(up.1(x0))
   1_flat.0(up.0(x0))
   1_flat.0(up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(138) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.
----------------------------------------

(139)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(f.0(0.0(x0)))) -> TOP.0(f_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(f.0(1.0(x0)))) -> TOP.0(f_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(0.0(0.0(x0)))) -> TOP.0(0_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(0.0(1.0(x0)))) -> TOP.0(0_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(1.0(0.0(x0)))) -> TOP.0(1_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(1.0(1.0(x0)))) -> TOP.0(1_flat.0(down.0(1.0(x0))))

The TRS R consists of the following rules:

   1_flat.0(up.0(x_1)) -> up.0(1.0(x_1))
   down.0(1.0(f.0(x))) -> up.0(f.0(0.0(x)))
   down.0(1.0(a.)) -> 1_flat.0(down.0(a.))
   down.0(1.0(b.)) -> 1_flat.0(down.0(b.))
   down.0(1.0(0.0(y13))) -> 1_flat.0(down.0(0.0(y13)))
   down.0(1.0(0.1(y13))) -> 1_flat.0(down.0(0.1(y13)))
   down.0(1.0(1.0(y14))) -> 1_flat.0(down.0(1.0(y14)))
   down.0(1.0(1.1(y14))) -> 1_flat.0(down.0(1.1(y14)))
   down.0(0.0(f.0(x))) -> up.0(1.0(x))
   down.0(0.0(a.)) -> 0_flat.0(down.0(a.))
   down.0(0.0(b.)) -> 0_flat.0(down.0(b.))
   down.0(0.0(0.0(y9))) -> 0_flat.0(down.0(0.0(y9)))
   down.0(0.0(0.1(y9))) -> 0_flat.0(down.0(0.1(y9)))
   down.0(0.0(1.0(y10))) -> 0_flat.0(down.0(1.0(y10)))
   down.0(0.0(1.1(y10))) -> 0_flat.0(down.0(1.1(y10)))
   0_flat.0(up.0(x_1)) -> up.0(0.0(x_1))
   down.0(a.) -> up.0(f.0(a.))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(a.)
   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(0.0(f.0(x0)))
   down.0(0.0(f.1(x0)))
   down.0(1.0(f.0(x0)))
   down.0(1.0(f.1(x0)))
   down.0(f.0(a.))
   down.0(f.0(b.))
   down.0(f.0(0.0(x0)))
   down.0(f.0(0.1(x0)))
   down.0(f.0(1.0(x0)))
   down.0(f.0(1.1(x0)))
   down.0(f.1(fresh_constant.))
   down.0(0.0(a.))
   down.0(0.0(b.))
   down.0(0.0(0.0(x0)))
   down.0(0.0(0.1(x0)))
   down.0(0.0(1.0(x0)))
   down.0(0.0(1.1(x0)))
   down.0(0.1(fresh_constant.))
   down.0(1.0(a.))
   down.0(1.0(b.))
   down.0(1.0(0.0(x0)))
   down.0(1.0(0.1(x0)))
   down.0(1.0(1.0(x0)))
   down.0(1.0(1.1(x0)))
   down.0(1.1(fresh_constant.))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   0_flat.0(up.0(x0))
   0_flat.0(up.1(x0))
   1_flat.0(up.0(x0))
   1_flat.0(up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(140) PisEmptyProof (SOUND)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(141)
TRUE

----------------------------------------

(142)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))

The TRS R consists of the following rules:

   1_flat(up(x_1)) -> up(1(x_1))
   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(143) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as they contain symbols which do neither occur in P nor in R.[THIEMANN].

   down(f(fresh_constant))
   down(0(fresh_constant))
   down(1(fresh_constant))


----------------------------------------

(144)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))

The TRS R consists of the following rules:

   1_flat(up(x_1)) -> up(1(x_1))
   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all (P,Q,R)-chains.
----------------------------------------

(145) SplitQDPProof (EQUIVALENT)
We show in the first subproof that some pairs and rules can be removed, afterwards, we continue with the remaining DP-Problem

----------------------------------------

(146)
Complex Obligation (AND)

----------------------------------------

(147)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))

The TRS R consists of the following rules:

   1_flat(up(x_1)) -> up(1(x_1))
   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(b)) -> 1_flat(down(b))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(b)) -> 0_flat(down(b))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(148) SemLabProof (SOUND)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
a: 0
b: 1
down: 0
f: 0
fresh_constant: 0
0: 0
up: 0
1: 0
1_flat: 0
0_flat: 0
f_flat: 0
TOP: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

(149)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(f.0(0.0(x0)))) -> TOP.0(f_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(f.0(0.1(x0)))) -> TOP.0(f_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(f.0(1.0(x0)))) -> TOP.0(f_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(f.0(1.1(x0)))) -> TOP.0(f_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(0.0(0.0(x0)))) -> TOP.0(0_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(0.0(0.1(x0)))) -> TOP.0(0_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(0.0(1.0(x0)))) -> TOP.0(0_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(0.0(1.1(x0)))) -> TOP.0(0_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(1.0(0.0(x0)))) -> TOP.0(1_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(1.0(0.1(x0)))) -> TOP.0(1_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(1.0(1.0(x0)))) -> TOP.0(1_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(1.0(1.1(x0)))) -> TOP.0(1_flat.0(down.0(1.1(x0))))

The TRS R consists of the following rules:

   1_flat.0(up.0(x_1)) -> up.0(1.0(x_1))
   1_flat.0(up.1(x_1)) -> up.0(1.1(x_1))
   down.0(1.0(f.0(x))) -> up.0(f.0(0.0(x)))
   down.0(1.0(f.1(x))) -> up.0(f.0(0.1(x)))
   down.0(1.0(a.)) -> 1_flat.0(down.0(a.))
   down.0(1.1(b.)) -> 1_flat.0(down.1(b.))
   down.0(1.0(0.0(y13))) -> 1_flat.0(down.0(0.0(y13)))
   down.0(1.0(0.1(y13))) -> 1_flat.0(down.0(0.1(y13)))
   down.0(1.0(1.0(y14))) -> 1_flat.0(down.0(1.0(y14)))
   down.0(1.0(1.1(y14))) -> 1_flat.0(down.0(1.1(y14)))
   down.0(0.0(f.0(x))) -> up.0(1.0(x))
   down.0(0.0(f.1(x))) -> up.0(1.1(x))
   down.0(0.0(a.)) -> 0_flat.0(down.0(a.))
   down.0(0.1(b.)) -> 0_flat.0(down.1(b.))
   down.0(0.0(0.0(y9))) -> 0_flat.0(down.0(0.0(y9)))
   down.0(0.0(0.1(y9))) -> 0_flat.0(down.0(0.1(y9)))
   down.0(0.0(1.0(y10))) -> 0_flat.0(down.0(1.0(y10)))
   down.0(0.0(1.1(y10))) -> 0_flat.0(down.0(1.1(y10)))
   0_flat.0(up.0(x_1)) -> up.0(0.0(x_1))
   0_flat.0(up.1(x_1)) -> up.0(0.1(x_1))
   down.0(a.) -> up.0(f.0(a.))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))
   f_flat.0(up.1(x_1)) -> up.0(f.1(x_1))

The set Q consists of the following terms:

   down.0(a.)
   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(0.0(f.0(x0)))
   down.0(0.0(f.1(x0)))
   down.0(1.0(f.0(x0)))
   down.0(1.0(f.1(x0)))
   down.0(f.0(a.))
   down.0(f.1(b.))
   down.0(f.0(0.0(x0)))
   down.0(f.0(0.1(x0)))
   down.0(f.0(1.0(x0)))
   down.0(f.0(1.1(x0)))
   down.0(f.0(fresh_constant.))
   down.0(0.0(a.))
   down.0(0.1(b.))
   down.0(0.0(0.0(x0)))
   down.0(0.0(0.1(x0)))
   down.0(0.0(1.0(x0)))
   down.0(0.0(1.1(x0)))
   down.0(0.0(fresh_constant.))
   down.0(1.0(a.))
   down.0(1.1(b.))
   down.0(1.0(0.0(x0)))
   down.0(1.0(0.1(x0)))
   down.0(1.0(1.0(x0)))
   down.0(1.0(1.1(x0)))
   down.0(1.0(fresh_constant.))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   0_flat.0(up.0(x0))
   0_flat.0(up.1(x0))
   1_flat.0(up.0(x0))
   1_flat.0(up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(150) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   1_flat.0(up.1(x_1)) -> up.0(1.1(x_1))
   down.0(1.0(f.1(x))) -> up.0(f.0(0.1(x)))
   down.0(0.0(f.1(x))) -> up.0(1.1(x))
   0_flat.0(up.1(x_1)) -> up.0(0.1(x_1))
   f_flat.0(up.1(x_1)) -> up.0(f.1(x_1))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = 1 + x_1
   POL(0.1(x_1)) = 1 + x_1
   POL(0_flat.0(x_1)) = 1 + x_1
   POL(1.0(x_1)) = 1 + x_1
   POL(1.1(x_1)) = 1 + x_1
   POL(1_flat.0(x_1)) = 1 + x_1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(x_1)) = x_1
   POL(down.1(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = 1 + x_1
   POL(f_flat.0(x_1)) = x_1
   POL(up.0(x_1)) = x_1
   POL(up.1(x_1)) = 1 + x_1


----------------------------------------

(151)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(f.0(0.0(x0)))) -> TOP.0(f_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(f.0(0.1(x0)))) -> TOP.0(f_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(f.0(1.0(x0)))) -> TOP.0(f_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(f.0(1.1(x0)))) -> TOP.0(f_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(0.0(0.0(x0)))) -> TOP.0(0_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(0.0(0.1(x0)))) -> TOP.0(0_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(0.0(1.0(x0)))) -> TOP.0(0_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(0.0(1.1(x0)))) -> TOP.0(0_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(1.0(0.0(x0)))) -> TOP.0(1_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(1.0(0.1(x0)))) -> TOP.0(1_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(1.0(1.0(x0)))) -> TOP.0(1_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(1.0(1.1(x0)))) -> TOP.0(1_flat.0(down.0(1.1(x0))))

The TRS R consists of the following rules:

   down.0(1.1(b.)) -> 1_flat.0(down.1(b.))
   1_flat.0(up.0(x_1)) -> up.0(1.0(x_1))
   down.0(1.0(f.0(x))) -> up.0(f.0(0.0(x)))
   down.0(1.0(a.)) -> 1_flat.0(down.0(a.))
   down.0(1.0(0.0(y13))) -> 1_flat.0(down.0(0.0(y13)))
   down.0(1.0(0.1(y13))) -> 1_flat.0(down.0(0.1(y13)))
   down.0(1.0(1.0(y14))) -> 1_flat.0(down.0(1.0(y14)))
   down.0(1.0(1.1(y14))) -> 1_flat.0(down.0(1.1(y14)))
   down.0(0.1(b.)) -> 0_flat.0(down.1(b.))
   down.0(0.0(f.0(x))) -> up.0(1.0(x))
   down.0(0.0(a.)) -> 0_flat.0(down.0(a.))
   down.0(0.0(0.0(y9))) -> 0_flat.0(down.0(0.0(y9)))
   down.0(0.0(0.1(y9))) -> 0_flat.0(down.0(0.1(y9)))
   down.0(0.0(1.0(y10))) -> 0_flat.0(down.0(1.0(y10)))
   down.0(0.0(1.1(y10))) -> 0_flat.0(down.0(1.1(y10)))
   0_flat.0(up.0(x_1)) -> up.0(0.0(x_1))
   down.0(a.) -> up.0(f.0(a.))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(a.)
   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(0.0(f.0(x0)))
   down.0(0.0(f.1(x0)))
   down.0(1.0(f.0(x0)))
   down.0(1.0(f.1(x0)))
   down.0(f.0(a.))
   down.0(f.1(b.))
   down.0(f.0(0.0(x0)))
   down.0(f.0(0.1(x0)))
   down.0(f.0(1.0(x0)))
   down.0(f.0(1.1(x0)))
   down.0(f.0(fresh_constant.))
   down.0(0.0(a.))
   down.0(0.1(b.))
   down.0(0.0(0.0(x0)))
   down.0(0.0(0.1(x0)))
   down.0(0.0(1.0(x0)))
   down.0(0.0(1.1(x0)))
   down.0(0.0(fresh_constant.))
   down.0(1.0(a.))
   down.0(1.1(b.))
   down.0(1.0(0.0(x0)))
   down.0(1.0(0.1(x0)))
   down.0(1.0(1.0(x0)))
   down.0(1.0(1.1(x0)))
   down.0(1.0(fresh_constant.))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   0_flat.0(up.0(x0))
   0_flat.0(up.1(x0))
   1_flat.0(up.0(x0))
   1_flat.0(up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(152) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   down.0(1.1(b.)) -> 1_flat.0(down.1(b.))
   down.0(0.1(b.)) -> 0_flat.0(down.1(b.))

Used ordering: Polynomial interpretation [POLO]:

   POL(0.0(x_1)) = x_1
   POL(0.1(x_1)) = 1 + x_1
   POL(0_flat.0(x_1)) = x_1
   POL(1.0(x_1)) = x_1
   POL(1.1(x_1)) = 1 + x_1
   POL(1_flat.0(x_1)) = x_1
   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(x_1)) = x_1
   POL(down.1(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f_flat.0(x_1)) = x_1
   POL(up.0(x_1)) = x_1


----------------------------------------

(153)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(f.0(0.0(x0)))) -> TOP.0(f_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(f.0(0.1(x0)))) -> TOP.0(f_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(f.0(1.0(x0)))) -> TOP.0(f_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(f.0(1.1(x0)))) -> TOP.0(f_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(0.0(0.0(x0)))) -> TOP.0(0_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(0.0(0.1(x0)))) -> TOP.0(0_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(0.0(1.0(x0)))) -> TOP.0(0_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(0.0(1.1(x0)))) -> TOP.0(0_flat.0(down.0(1.1(x0))))
   TOP.0(up.0(1.0(0.0(x0)))) -> TOP.0(1_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(1.0(0.1(x0)))) -> TOP.0(1_flat.0(down.0(0.1(x0))))
   TOP.0(up.0(1.0(1.0(x0)))) -> TOP.0(1_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(1.0(1.1(x0)))) -> TOP.0(1_flat.0(down.0(1.1(x0))))

The TRS R consists of the following rules:

   1_flat.0(up.0(x_1)) -> up.0(1.0(x_1))
   down.0(1.0(f.0(x))) -> up.0(f.0(0.0(x)))
   down.0(1.0(a.)) -> 1_flat.0(down.0(a.))
   down.0(1.0(0.0(y13))) -> 1_flat.0(down.0(0.0(y13)))
   down.0(1.0(0.1(y13))) -> 1_flat.0(down.0(0.1(y13)))
   down.0(1.0(1.0(y14))) -> 1_flat.0(down.0(1.0(y14)))
   down.0(1.0(1.1(y14))) -> 1_flat.0(down.0(1.1(y14)))
   down.0(0.0(f.0(x))) -> up.0(1.0(x))
   down.0(0.0(a.)) -> 0_flat.0(down.0(a.))
   down.0(0.0(0.0(y9))) -> 0_flat.0(down.0(0.0(y9)))
   down.0(0.0(0.1(y9))) -> 0_flat.0(down.0(0.1(y9)))
   down.0(0.0(1.0(y10))) -> 0_flat.0(down.0(1.0(y10)))
   down.0(0.0(1.1(y10))) -> 0_flat.0(down.0(1.1(y10)))
   0_flat.0(up.0(x_1)) -> up.0(0.0(x_1))
   down.0(a.) -> up.0(f.0(a.))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(a.)
   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(0.0(f.0(x0)))
   down.0(0.0(f.1(x0)))
   down.0(1.0(f.0(x0)))
   down.0(1.0(f.1(x0)))
   down.0(f.0(a.))
   down.0(f.1(b.))
   down.0(f.0(0.0(x0)))
   down.0(f.0(0.1(x0)))
   down.0(f.0(1.0(x0)))
   down.0(f.0(1.1(x0)))
   down.0(f.0(fresh_constant.))
   down.0(0.0(a.))
   down.0(0.1(b.))
   down.0(0.0(0.0(x0)))
   down.0(0.0(0.1(x0)))
   down.0(0.0(1.0(x0)))
   down.0(0.0(1.1(x0)))
   down.0(0.0(fresh_constant.))
   down.0(1.0(a.))
   down.0(1.1(b.))
   down.0(1.0(0.0(x0)))
   down.0(1.0(0.1(x0)))
   down.0(1.0(1.0(x0)))
   down.0(1.0(1.1(x0)))
   down.0(1.0(fresh_constant.))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   0_flat.0(up.0(x0))
   0_flat.0(up.1(x0))
   1_flat.0(up.0(x0))
   1_flat.0(up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(154) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.
----------------------------------------

(155)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(f.0(0.0(x0)))) -> TOP.0(f_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(f.0(1.0(x0)))) -> TOP.0(f_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(0.0(0.0(x0)))) -> TOP.0(0_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(0.0(1.0(x0)))) -> TOP.0(0_flat.0(down.0(1.0(x0))))
   TOP.0(up.0(1.0(0.0(x0)))) -> TOP.0(1_flat.0(down.0(0.0(x0))))
   TOP.0(up.0(1.0(1.0(x0)))) -> TOP.0(1_flat.0(down.0(1.0(x0))))

The TRS R consists of the following rules:

   1_flat.0(up.0(x_1)) -> up.0(1.0(x_1))
   down.0(1.0(f.0(x))) -> up.0(f.0(0.0(x)))
   down.0(1.0(a.)) -> 1_flat.0(down.0(a.))
   down.0(1.0(0.0(y13))) -> 1_flat.0(down.0(0.0(y13)))
   down.0(1.0(0.1(y13))) -> 1_flat.0(down.0(0.1(y13)))
   down.0(1.0(1.0(y14))) -> 1_flat.0(down.0(1.0(y14)))
   down.0(1.0(1.1(y14))) -> 1_flat.0(down.0(1.1(y14)))
   down.0(0.0(f.0(x))) -> up.0(1.0(x))
   down.0(0.0(a.)) -> 0_flat.0(down.0(a.))
   down.0(0.0(0.0(y9))) -> 0_flat.0(down.0(0.0(y9)))
   down.0(0.0(0.1(y9))) -> 0_flat.0(down.0(0.1(y9)))
   down.0(0.0(1.0(y10))) -> 0_flat.0(down.0(1.0(y10)))
   down.0(0.0(1.1(y10))) -> 0_flat.0(down.0(1.1(y10)))
   0_flat.0(up.0(x_1)) -> up.0(0.0(x_1))
   down.0(a.) -> up.0(f.0(a.))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(a.)
   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(0.0(f.0(x0)))
   down.0(0.0(f.1(x0)))
   down.0(1.0(f.0(x0)))
   down.0(1.0(f.1(x0)))
   down.0(f.0(a.))
   down.0(f.1(b.))
   down.0(f.0(0.0(x0)))
   down.0(f.0(0.1(x0)))
   down.0(f.0(1.0(x0)))
   down.0(f.0(1.1(x0)))
   down.0(f.0(fresh_constant.))
   down.0(0.0(a.))
   down.0(0.1(b.))
   down.0(0.0(0.0(x0)))
   down.0(0.0(0.1(x0)))
   down.0(0.0(1.0(x0)))
   down.0(0.0(1.1(x0)))
   down.0(0.0(fresh_constant.))
   down.0(1.0(a.))
   down.0(1.1(b.))
   down.0(1.0(0.0(x0)))
   down.0(1.0(0.1(x0)))
   down.0(1.0(1.0(x0)))
   down.0(1.0(1.1(x0)))
   down.0(1.0(fresh_constant.))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   0_flat.0(up.0(x0))
   0_flat.0(up.1(x0))
   1_flat.0(up.0(x0))
   1_flat.0(up.1(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(156) PisEmptyProof (SOUND)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(157)
TRUE

----------------------------------------

(158)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))

The TRS R consists of the following rules:

   1_flat(up(x_1)) -> up(1(x_1))
   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(b))
   down(f(0(x0)))
   down(f(1(x0)))
   down(f(fresh_constant))
   down(0(a))
   down(0(b))
   down(0(0(x0)))
   down(0(1(x0)))
   down(0(fresh_constant))
   down(1(a))
   down(1(b))
   down(1(0(x0)))
   down(1(1(x0)))
   down(1(fresh_constant))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all minimal (P,Q,R)-chains.
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(159) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as they contain symbols which do neither occur in P nor in R.[THIEMANN].

   down(f(b))
   down(f(fresh_constant))
   down(0(b))
   down(0(fresh_constant))
   down(1(b))
   down(1(fresh_constant))


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(160)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(f(0(x0)))) -> TOP(f_flat(down(0(x0))))
   TOP(up(f(1(x0)))) -> TOP(f_flat(down(1(x0))))
   TOP(up(0(0(x0)))) -> TOP(0_flat(down(0(x0))))
   TOP(up(0(1(x0)))) -> TOP(0_flat(down(1(x0))))
   TOP(up(1(0(x0)))) -> TOP(1_flat(down(0(x0))))
   TOP(up(1(1(x0)))) -> TOP(1_flat(down(1(x0))))

The TRS R consists of the following rules:

   1_flat(up(x_1)) -> up(1(x_1))
   down(1(f(x))) -> up(f(0(x)))
   down(1(a)) -> 1_flat(down(a))
   down(1(0(y13))) -> 1_flat(down(0(y13)))
   down(1(1(y14))) -> 1_flat(down(1(y14)))
   down(0(f(x))) -> up(1(x))
   down(0(a)) -> 0_flat(down(a))
   down(0(0(y9))) -> 0_flat(down(0(y9)))
   down(0(1(y10))) -> 0_flat(down(1(y10)))
   0_flat(up(x_1)) -> up(0(x_1))
   down(a) -> up(f(a))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(a)
   down(f(f(x0)))
   down(0(f(x0)))
   down(1(f(x0)))
   down(f(a))
   down(f(0(x0)))
   down(f(1(x0)))
   down(0(a))
   down(0(0(x0)))
   down(0(1(x0)))
   down(1(a))
   down(1(0(x0)))
   down(1(1(x0)))
   f_flat(up(x0))
   0_flat(up(x0))
   1_flat(up(x0))

We have to consider all (P,Q,R)-chains.