/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 42 ms]
    (4) QTRS
    (5) DependencyPairsProof [EQUIVALENT, 0 ms]
    (6) QDP
    (7) DependencyGraphProof [EQUIVALENT, 0 ms]
    (8) QDP
    (9) UsableRulesProof [EQUIVALENT, 0 ms]
    (10) QDP
    (11) NonTerminationLoopProof [COMPLETE, 0 ms]
    (12) NO
(13) Thiemann-SpecialC-Transformation [EQUIVALENT, 0 ms]
(14) QTRS
    (15) QTRSRRRProof [EQUIVALENT, 50 ms]
    (16) QTRS
    (17) DependencyPairsProof [EQUIVALENT, 0 ms]
    (18) QDP
    (19) DependencyGraphProof [EQUIVALENT, 0 ms]
    (20) AND
        (21) QDP
            (22) UsableRulesProof [EQUIVALENT, 0 ms]
            (23) QDP
            (24) QReductionProof [EQUIVALENT, 0 ms]
            (25) QDP
            (26) UsableRulesReductionPairsProof [EQUIVALENT, 5 ms]
            (27) QDP
            (28) DependencyGraphProof [EQUIVALENT, 0 ms]
            (29) TRUE
        (30) QDP
            (31) UsableRulesProof [EQUIVALENT, 0 ms]
            (32) QDP
                (33) QReductionProof [EQUIVALENT, 0 ms]
                (34) QDP
                (35) TransformationProof [SOUND, 0 ms]
                (36) QDP
                (37) TransformationProof [EQUIVALENT, 0 ms]
                (38) QDP
                (39) TransformationProof [EQUIVALENT, 0 ms]
                (40) QDP
                (41) QDPOrderProof [EQUIVALENT, 9 ms]
                (42) QDP
                (43) QReductionProof [EQUIVALENT, 0 ms]
                (44) QDP
            (45) UsableRulesProof [EQUIVALENT, 0 ms]
            (46) QDP
                (47) QReductionProof [EQUIVALENT, 0 ms]
                (48) QDP
                (49) QReductionProof [EQUIVALENT, 0 ms]
                (50) QDP
(51) Raffelsieper-Zantema-Transformation [SOUND, 0 ms]
(52) QTRS
    (53) QTRSRRRProof [EQUIVALENT, 40 ms]
    (54) QTRS
    (55) QTRSRRRProof [EQUIVALENT, 8 ms]
    (56) QTRS
    (57) AAECC Innermost [EQUIVALENT, 0 ms]
    (58) QTRS
    (59) DependencyPairsProof [EQUIVALENT, 9 ms]
    (60) QDP
    (61) DependencyGraphProof [EQUIVALENT, 0 ms]
    (62) AND
        (63) QDP
            (64) UsableRulesProof [EQUIVALENT, 0 ms]
            (65) QDP
            (66) QReductionProof [EQUIVALENT, 0 ms]
            (67) QDP
            (68) QDPSizeChangeProof [EQUIVALENT, 0 ms]
            (69) YES
        (70) QDP
            (71) UsableRulesProof [EQUIVALENT, 0 ms]
            (72) QDP
            (73) QReductionProof [EQUIVALENT, 0 ms]
            (74) QDP
            (75) TransformationProof [EQUIVALENT, 0 ms]
            (76) QDP
            (77) DependencyGraphProof [EQUIVALENT, 0 ms]
            (78) QDP
            (79) TransformationProof [EQUIVALENT, 0 ms]
            (80) QDP
            (81) TransformationProof [EQUIVALENT, 0 ms]
            (82) QDP
            (83) TransformationProof [EQUIVALENT, 0 ms]
            (84) QDP
            (85) TransformationProof [EQUIVALENT, 0 ms]
            (86) QDP
            (87) TransformationProof [EQUIVALENT, 0 ms]
            (88) QDP
            (89) TransformationProof [EQUIVALENT, 0 ms]
            (90) QDP
            (91) TransformationProof [EQUIVALENT, 0 ms]
            (92) QDP
            (93) DependencyGraphProof [EQUIVALENT, 0 ms]
            (94) QDP
            (95) TransformationProof [EQUIVALENT, 0 ms]
            (96) QDP
            (97) DependencyGraphProof [EQUIVALENT, 0 ms]
            (98) QDP
            (99) TransformationProof [EQUIVALENT, 0 ms]
            (100) QDP
            (101) DependencyGraphProof [EQUIVALENT, 0 ms]
            (102) QDP
            (103) TransformationProof [EQUIVALENT, 0 ms]
            (104) QDP
            (105) DependencyGraphProof [EQUIVALENT, 0 ms]
            (106) QDP
            (107) TransformationProof [EQUIVALENT, 0 ms]
            (108) QDP
            (109) DependencyGraphProof [EQUIVALENT, 0 ms]
            (110) QDP
            (111) MNOCProof [EQUIVALENT, 0 ms]
            (112) QDP
            (113) SplitQDPProof [EQUIVALENT, 0 ms]
            (114) AND
                (115) QDP
                    (116) SemLabProof [SOUND, 0 ms]
                    (117) QDP
                    (118) DependencyGraphProof [EQUIVALENT, 0 ms]
                    (119) QDP
                    (120) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                    (121) QDP
                    (122) MRRProof [EQUIVALENT, 7 ms]
                    (123) QDP
                    (124) MRRProof [EQUIVALENT, 0 ms]
                    (125) QDP
                    (126) PisEmptyProof [SOUND, 0 ms]
                    (127) TRUE
                (128) QDP
                    (129) SplitQDPProof [EQUIVALENT, 0 ms]
                    (130) AND
                        (131) QDP
                            (132) SemLabProof [SOUND, 0 ms]
                            (133) QDP
                            (134) DependencyGraphProof [EQUIVALENT, 0 ms]
                            (135) QDP
                            (136) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                            (137) QDP
                            (138) MRRProof [EQUIVALENT, 5 ms]
                            (139) QDP
                            (140) MRRProof [EQUIVALENT, 0 ms]
                            (141) QDP
                            (142) PisEmptyProof [SOUND, 0 ms]
                            (143) TRUE
                        (144) QDP
                            (145) QReductionProof [EQUIVALENT, 0 ms]
                            (146) QDP
                            (147) SplitQDPProof [EQUIVALENT, 0 ms]
                            (148) AND
                                (149) QDP
                                    (150) SemLabProof [SOUND, 0 ms]
                                    (151) QDP
                                    (152) DependencyGraphProof [EQUIVALENT, 0 ms]
                                    (153) QDP
                                    (154) MRRProof [EQUIVALENT, 0 ms]
                                    (155) QDP
                                    (156) MRRProof [EQUIVALENT, 0 ms]
                                    (157) QDP
                                    (158) PisEmptyProof [SOUND, 0 ms]
                                    (159) TRUE
                                (160) QDP
                                    (161) QReductionProof [EQUIVALENT, 0 ms]
                                    (162) QDP


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

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

   f(g(a)) -> f(s(g(b)))
   f(f(x)) -> b
   g(x) -> f(g(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:

   f(g(a)) -> f(s(g(b)))
   f(f(x)) -> b
   g(x) -> f(g(x))

Q is empty.

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

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 2
   POL(b) = 0
   POL(f(x_1)) = x_1
   POL(g(x_1)) = 2*x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(g(a)) -> f(s(g(b)))




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(f(x)) -> b
   g(x) -> f(g(x))

Q is empty.

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

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

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x) -> F(g(x))
   G(x) -> G(x)

The TRS R consists of the following rules:

   f(f(x)) -> b
   g(x) -> f(g(x))

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

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

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x) -> G(x)

The TRS R consists of the following rules:

   f(f(x)) -> b
   g(x) -> f(g(x))

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

(9) 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.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(x) -> G(x)

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

(11) 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 = G(x) evaluates to  t =G(x)

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 G(x) to G(x).




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

(12)
NO

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

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

(14)
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(g(x_1)) -> check_g(redex_g(x_1))
   redex_f(g(a)) -> result_f(f(s(g(b))))
   redex_f(f(x)) -> result_f(b)
   redex_g(x) -> result_g(f(g(x)))
   check_f(result_f(x)) -> go_up(x)
   check_g(result_g(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_g(redex_g(x_1)) -> in_g_1(reduce(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_g_1(go_up(x_1)) -> go_up(g(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))


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

(15) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 1
   POL(b) = 0
   POL(check_f(x_1)) = x_1
   POL(check_g(x_1)) = x_1
   POL(f(x_1)) = x_1
   POL(g(x_1)) = x_1
   POL(go_up(x_1)) = x_1
   POL(in_f_1(x_1)) = x_1
   POL(in_g_1(x_1)) = x_1
   POL(in_s_1(x_1)) = 2*x_1
   POL(redex_f(x_1)) = x_1
   POL(redex_g(x_1)) = x_1
   POL(reduce(x_1)) = x_1
   POL(result_f(x_1)) = x_1
   POL(result_g(x_1)) = x_1
   POL(s(x_1)) = 2*x_1
   POL(top(x_1)) = 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   redex_f(g(a)) -> result_f(f(s(g(b))))




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

(16)
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(g(x_1)) -> check_g(redex_g(x_1))
   redex_f(f(x)) -> result_f(b)
   redex_g(x) -> result_g(f(g(x)))
   check_f(result_f(x)) -> go_up(x)
   check_g(result_g(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_g(redex_g(x_1)) -> in_g_1(reduce(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_g_1(go_up(x_1)) -> go_up(g(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))


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

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

(18)
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(g(x_1)) -> CHECK_G(redex_g(x_1))
   REDUCE(g(x_1)) -> REDEX_G(x_1)
   CHECK_F(redex_f(x_1)) -> IN_F_1(reduce(x_1))
   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   CHECK_G(redex_g(x_1)) -> IN_G_1(reduce(x_1))
   CHECK_G(redex_g(x_1)) -> REDUCE(x_1)
   REDUCE(s(x_1)) -> IN_S_1(reduce(x_1))
   REDUCE(s(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(g(x_1)) -> check_g(redex_g(x_1))
   redex_f(f(x)) -> result_f(b)
   redex_g(x) -> result_g(f(g(x)))
   check_f(result_f(x)) -> go_up(x)
   check_g(result_g(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_g(redex_g(x_1)) -> in_g_1(reduce(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_g_1(go_up(x_1)) -> go_up(g(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))

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

(19) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes.
----------------------------------------

(20)
Complex Obligation (AND)

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

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   REDUCE(s(x_1)) -> REDUCE(x_1)
   REDUCE(f(x_1)) -> CHECK_F(redex_f(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(g(x_1)) -> check_g(redex_g(x_1))
   redex_f(f(x)) -> result_f(b)
   redex_g(x) -> result_g(f(g(x)))
   check_f(result_f(x)) -> go_up(x)
   check_g(result_g(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_g(redex_g(x_1)) -> in_g_1(reduce(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_g_1(go_up(x_1)) -> go_up(g(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))

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

(22) 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.
----------------------------------------

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

   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   REDUCE(s(x_1)) -> REDUCE(x_1)
   REDUCE(f(x_1)) -> CHECK_F(redex_f(x_1))

The TRS R consists of the following rules:

   redex_f(f(x)) -> result_f(b)

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))

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

(24) 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(g(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))


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

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

   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   REDUCE(s(x_1)) -> REDUCE(x_1)
   REDUCE(f(x_1)) -> CHECK_F(redex_f(x_1))

The TRS R consists of the following rules:

   redex_f(f(x)) -> result_f(b)

The set Q consists of the following terms:

   redex_f(g(a))
   redex_f(f(x0))

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

(26) 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(s(x_1)) -> REDUCE(x_1)
   REDUCE(f(x_1)) -> CHECK_F(redex_f(x_1))
The following rules are removed from R:

   redex_f(f(x)) -> result_f(b)
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(CHECK_F(x_1)) = x_1
   POL(REDUCE(x_1)) = 2*x_1
   POL(b) = 0
   POL(f(x_1)) = 2*x_1
   POL(redex_f(x_1)) = 2*x_1
   POL(result_f(x_1)) = x_1
   POL(s(x_1)) = 2*x_1


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

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

   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)

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

   redex_f(g(a))
   redex_f(f(x0))

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

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

(29)
TRUE

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

(30)
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(g(x_1)) -> check_g(redex_g(x_1))
   redex_f(f(x)) -> result_f(b)
   redex_g(x) -> result_g(f(g(x)))
   check_f(result_f(x)) -> go_up(x)
   check_g(result_g(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_g(redex_g(x_1)) -> in_g_1(reduce(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_g_1(go_up(x_1)) -> go_up(g(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))

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

(31) 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.
----------------------------------------

(32)
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(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))

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

(33) 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))
   in_g_1(go_up(x0))


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

(34)
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(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_s_1(go_up(x0))

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

(35) TransformationProof (SOUND)
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(g(x0))) -> TOP(check_g(redex_g(x0))),TOP(go_up(g(x0))) -> TOP(check_g(redex_g(x0))))
   (TOP(go_up(s(x0))) -> TOP(in_s_1(reduce(x0))),TOP(go_up(s(x0))) -> TOP(in_s_1(reduce(x0))))


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

(36)
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(g(x0))) -> TOP(check_g(redex_g(x0)))
   TOP(go_up(s(x0))) -> TOP(in_s_1(reduce(x0)))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_s_1(go_up(x0))

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

(37) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(go_up(g(x0))) -> TOP(check_g(redex_g(x0))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(go_up(g(x0))) -> TOP(check_g(result_g(f(g(x0))))),TOP(go_up(g(x0))) -> TOP(check_g(result_g(f(g(x0))))))


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

(38)
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(s(x0))) -> TOP(in_s_1(reduce(x0)))
   TOP(go_up(g(x0))) -> TOP(check_g(result_g(f(g(x0)))))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_s_1(go_up(x0))

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

(39) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(go_up(g(x0))) -> TOP(check_g(result_g(f(g(x0))))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(go_up(g(x0))) -> TOP(go_up(f(g(x0)))),TOP(go_up(g(x0))) -> TOP(go_up(f(g(x0)))))


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

(40)
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(s(x0))) -> TOP(in_s_1(reduce(x0)))
   TOP(go_up(g(x0))) -> TOP(go_up(f(g(x0))))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_s_1(go_up(x0))

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

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


The following pairs can be oriented strictly and are deleted.

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

   POL(TOP(x_1)) = x_1
   POL(b) = 0
   POL(check_f(x_1)) = x_1
   POL(check_g(x_1)) = 1
   POL(f(x_1)) = 0
   POL(g(x_1)) = 1 + x_1
   POL(go_up(x_1)) = x_1
   POL(in_f_1(x_1)) = 0
   POL(in_s_1(x_1)) = 0
   POL(redex_f(x_1)) = 0
   POL(redex_g(x_1)) = 1 + x_1
   POL(reduce(x_1)) = 0
   POL(result_f(x_1)) = x_1
   POL(result_g(x_1)) = 1
   POL(s(x_1)) = 0

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))
   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))


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

(42)
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(s(x0))) -> TOP(in_s_1(reduce(x0)))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_s_1(go_up(x0))

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

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

   redex_f(g(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(s(x0))) -> TOP(in_s_1(reduce(x0)))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_s_1(go_up(x0))

We have to consider all (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(x)) -> TOP(reduce(x))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_g_1(go_up(x0))
   in_s_1(go_up(x0))

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

(47) 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))
   in_g_1(go_up(x0))


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

(48)
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(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(g(a))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_s_1(go_up(x0))

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

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

   redex_f(g(a))


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

(50)
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(g(x_1)) -> check_g(redex_g(x_1))
   reduce(s(x_1)) -> in_s_1(reduce(x_1))
   in_s_1(go_up(x_1)) -> go_up(s(x_1))
   redex_g(x) -> result_g(f(g(x)))
   check_g(result_g(x)) -> go_up(x)
   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(g(x0))
   redex_f(f(x0))
   redex_g(x0)
   check_f(result_f(x0))
   check_g(result_g(x0))
   check_f(redex_f(x0))
   reduce(s(x0))
   in_f_1(go_up(x0))
   in_s_1(go_up(x0))

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

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

(52)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   down(f(g(a))) -> up(f(s(g(b))))
   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   top(up(x)) -> top(down(x))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   g_flat(up(x_1)) -> up(g(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

Q is empty.

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

(53) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 0
   POL(b) = 0
   POL(down(x_1)) = 2*x_1
   POL(f(x_1)) = x_1
   POL(f_flat(x_1)) = x_1
   POL(fresh_constant) = 0
   POL(g(x_1)) = 2*x_1
   POL(g_flat(x_1)) = 2 + 2*x_1
   POL(s(x_1)) = x_1
   POL(s_flat(x_1)) = x_1
   POL(top(x_1)) = x_1
   POL(up(x_1)) = 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   g_flat(up(x_1)) -> up(g(x_1))




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

(54)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   down(f(g(a))) -> up(f(s(g(b))))
   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   top(up(x)) -> top(down(x))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

Q is empty.

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

(55) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 1
   POL(b) = 0
   POL(down(x_1)) = 2*x_1
   POL(f(x_1)) = x_1
   POL(f_flat(x_1)) = x_1
   POL(fresh_constant) = 0
   POL(g(x_1)) = x_1
   POL(s(x_1)) = 2*x_1
   POL(s_flat(x_1)) = 2*x_1
   POL(top(x_1)) = 2*x_1
   POL(up(x_1)) = 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   down(f(g(a))) -> up(f(s(g(b))))




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

(56)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   top(up(x)) -> top(down(x))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

Q is empty.

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

(57) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))
   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))

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

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

(58)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   top(up(x)) -> top(down(x))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   top(up(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))


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

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

(60)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(x)) -> TOP(down(x))
   TOP(up(x)) -> DOWN(x)
   DOWN(s(y2)) -> S_FLAT(down(y2))
   DOWN(s(y2)) -> DOWN(y2)
   DOWN(f(a)) -> F_FLAT(down(a))
   DOWN(f(a)) -> DOWN(a)
   DOWN(f(s(y6))) -> F_FLAT(down(s(y6)))
   DOWN(f(s(y6))) -> DOWN(s(y6))
   DOWN(f(b)) -> F_FLAT(down(b))
   DOWN(f(b)) -> DOWN(b)
   DOWN(f(fresh_constant)) -> F_FLAT(down(fresh_constant))
   DOWN(f(fresh_constant)) -> DOWN(fresh_constant)
   DOWN(f(g(f(y8)))) -> F_FLAT(down(g(f(y8))))
   DOWN(f(g(f(y8)))) -> DOWN(g(f(y8)))
   DOWN(f(g(g(y9)))) -> F_FLAT(down(g(g(y9))))
   DOWN(f(g(g(y9)))) -> DOWN(g(g(y9)))
   DOWN(f(g(s(y10)))) -> F_FLAT(down(g(s(y10))))
   DOWN(f(g(s(y10)))) -> DOWN(g(s(y10)))
   DOWN(f(g(b))) -> F_FLAT(down(g(b)))
   DOWN(f(g(b))) -> DOWN(g(b))
   DOWN(f(g(fresh_constant))) -> F_FLAT(down(g(fresh_constant)))
   DOWN(f(g(fresh_constant))) -> DOWN(g(fresh_constant))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   top(up(x)) -> top(down(x))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   top(up(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(61) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 19 less nodes.
----------------------------------------

(62)
Complex Obligation (AND)

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

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

   DOWN(f(s(y6))) -> DOWN(s(y6))
   DOWN(s(y2)) -> DOWN(y2)

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   top(up(x)) -> top(down(x))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   top(up(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(64) 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.
----------------------------------------

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

   DOWN(f(s(y6))) -> DOWN(s(y6))
   DOWN(s(y2)) -> DOWN(y2)

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

   down(f(f(x0)))
   down(g(x0))
   top(up(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(66) 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(f(f(x0)))
   down(g(x0))
   top(up(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))


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

(67)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DOWN(f(s(y6))) -> DOWN(s(y6))
   DOWN(s(y2)) -> DOWN(y2)

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

(68) 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(s(y2)) -> DOWN(y2)
The graph contains the following edges 1 > 1


*DOWN(f(s(y6))) -> DOWN(s(y6))
The graph contains the following edges 1 > 1


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

(69)
YES

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

(70)
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(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   top(up(x)) -> top(down(x))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   top(up(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(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(up(x)) -> TOP(down(x))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   top(up(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(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].

   top(up(x0))


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

(74)
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(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(75) 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(f(f(x0)))) -> TOP(up(b)),TOP(up(f(f(x0)))) -> TOP(up(b)))
   (TOP(up(g(x0))) -> TOP(up(f(g(x0)))),TOP(up(g(x0))) -> TOP(up(f(g(x0)))))
   (TOP(up(s(x0))) -> TOP(s_flat(down(x0))),TOP(up(s(x0))) -> TOP(s_flat(down(x0))))
   (TOP(up(f(a))) -> TOP(f_flat(down(a))),TOP(up(f(a))) -> TOP(f_flat(down(a))))
   (TOP(up(f(s(x0)))) -> TOP(f_flat(down(s(x0)))),TOP(up(f(s(x0)))) -> TOP(f_flat(down(s(x0)))))
   (TOP(up(f(b))) -> TOP(f_flat(down(b))),TOP(up(f(b))) -> TOP(f_flat(down(b))))
   (TOP(up(f(fresh_constant))) -> TOP(f_flat(down(fresh_constant))),TOP(up(f(fresh_constant))) -> TOP(f_flat(down(fresh_constant))))
   (TOP(up(f(g(f(x0))))) -> TOP(f_flat(down(g(f(x0))))),TOP(up(f(g(f(x0))))) -> TOP(f_flat(down(g(f(x0))))))
   (TOP(up(f(g(g(x0))))) -> TOP(f_flat(down(g(g(x0))))),TOP(up(f(g(g(x0))))) -> TOP(f_flat(down(g(g(x0))))))
   (TOP(up(f(g(s(x0))))) -> TOP(f_flat(down(g(s(x0))))),TOP(up(f(g(s(x0))))) -> TOP(f_flat(down(g(s(x0))))))
   (TOP(up(f(g(b)))) -> TOP(f_flat(down(g(b)))),TOP(up(f(g(b)))) -> TOP(f_flat(down(g(b)))))
   (TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant)))),TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant)))))


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

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

   TOP(up(f(f(x0)))) -> TOP(up(b))
   TOP(up(g(x0))) -> TOP(up(f(g(x0))))
   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(a))) -> TOP(f_flat(down(a)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(down(s(x0))))
   TOP(up(f(b))) -> TOP(f_flat(down(b)))
   TOP(up(f(fresh_constant))) -> TOP(f_flat(down(fresh_constant)))
   TOP(up(f(g(f(x0))))) -> TOP(f_flat(down(g(f(x0)))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(down(g(g(x0)))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(down(g(s(x0)))))
   TOP(up(f(g(b)))) -> TOP(f_flat(down(g(b))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(77) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.
----------------------------------------

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(down(s(x0))))
   TOP(up(f(g(f(x0))))) -> TOP(f_flat(down(g(f(x0)))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(down(g(g(x0)))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(down(g(s(x0)))))
   TOP(up(f(g(b)))) -> TOP(f_flat(down(g(b))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(79) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(s(x0)))) -> TOP(f_flat(down(s(x0)))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0)))),TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0)))))


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

(80)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(g(f(x0))))) -> TOP(f_flat(down(g(f(x0)))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(down(g(g(x0)))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(down(g(s(x0)))))
   TOP(up(f(g(b)))) -> TOP(f_flat(down(g(b))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant))))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(81) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(g(f(x0))))) -> TOP(f_flat(down(g(f(x0))))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(g(f(x0))))) -> TOP(f_flat(up(f(g(f(x0)))))),TOP(up(f(g(f(x0))))) -> TOP(f_flat(up(f(g(f(x0)))))))


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

(82)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(down(g(g(x0)))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(down(g(s(x0)))))
   TOP(up(f(g(b)))) -> TOP(f_flat(down(g(b))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant))))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(f(x0))))) -> TOP(f_flat(up(f(g(f(x0))))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(83) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(g(g(x0))))) -> TOP(f_flat(down(g(g(x0))))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0)))))),TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0)))))))


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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(down(g(s(x0)))))
   TOP(up(f(g(b)))) -> TOP(f_flat(down(g(b))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant))))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(f(x0))))) -> TOP(f_flat(up(f(g(f(x0))))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0))))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(85) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(g(s(x0))))) -> TOP(f_flat(down(g(s(x0))))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0)))))),TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0)))))))


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

(86)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(g(b)))) -> TOP(f_flat(down(g(b))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant))))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(f(x0))))) -> TOP(f_flat(up(f(g(f(x0))))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0))))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0))))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

   (TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b))))),TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b))))))


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

(88)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(down(g(fresh_constant))))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(f(x0))))) -> TOP(f_flat(up(f(g(f(x0))))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0))))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0))))))
   TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b)))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

   (TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant))))),TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant))))))


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

(90)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(f(x0))))) -> TOP(f_flat(up(f(g(f(x0))))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0))))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0))))))
   TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b)))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(91) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(g(f(x0))))) -> TOP(f_flat(up(f(g(f(x0)))))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(g(f(x0))))) -> TOP(up(f(f(g(f(x0)))))),TOP(up(f(g(f(x0))))) -> TOP(up(f(f(g(f(x0)))))))


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

(92)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0))))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0))))))
   TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b)))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))
   TOP(up(f(g(f(x0))))) -> TOP(up(f(f(g(f(x0))))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0))))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0))))))
   TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b)))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(95) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(g(g(x0))))) -> TOP(f_flat(up(f(g(g(x0)))))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(g(g(x0))))) -> TOP(up(f(f(g(g(x0)))))),TOP(up(f(g(g(x0))))) -> TOP(up(f(f(g(g(x0)))))))


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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0))))))
   TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b)))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))
   TOP(up(f(g(g(x0))))) -> TOP(up(f(f(g(g(x0))))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0))))))
   TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b)))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(99) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(up(f(g(s(x0))))) -> TOP(f_flat(up(f(g(s(x0)))))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(up(f(g(s(x0))))) -> TOP(up(f(f(g(s(x0)))))),TOP(up(f(g(s(x0))))) -> TOP(up(f(f(g(s(x0)))))))


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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b)))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))
   TOP(up(f(g(s(x0))))) -> TOP(up(f(f(g(s(x0))))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_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 1 less node.
----------------------------------------

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(b)))) -> TOP(f_flat(up(f(g(b)))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

   (TOP(up(f(g(b)))) -> TOP(up(f(f(g(b))))),TOP(up(f(g(b)))) -> TOP(up(f(f(g(b))))))


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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))
   TOP(up(f(g(b)))) -> TOP(up(f(f(g(b)))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(fresh_constant)))) -> TOP(f_flat(up(f(g(fresh_constant)))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

   (TOP(up(f(g(fresh_constant)))) -> TOP(up(f(f(g(fresh_constant))))),TOP(up(f(g(fresh_constant)))) -> TOP(up(f(f(g(fresh_constant))))))


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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))
   TOP(up(f(g(fresh_constant)))) -> TOP(up(f(f(g(fresh_constant)))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(111) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
----------------------------------------

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

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

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

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

(114)
Complex Obligation (AND)

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

(115)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(fresh_constant)) -> f_flat(down(fresh_constant))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   f_flat(up(x_1)) -> up(f(x_1))
   s_flat(up(x_1)) -> up(s(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(116) 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.
s_flat: 0
s: 0
a: 0
b: 0
down: 0
f: 0
fresh_constant: 1
up: 0
f_flat: 0
TOP: 0
g: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

(117)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(s.1(x0))) -> TOP.0(s_flat.0(down.1(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))
   TOP.0(up.0(f.0(s.1(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.1(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(f.0(f.1(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.0(a.)) -> f_flat.0(down.0(a.))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.1(fresh_constant.)) -> f_flat.0(down.1(fresh_constant.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.))) -> f_flat.0(down.0(g.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))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))

The set Q consists of the following terms:

   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.1(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(118) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(119)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(f.0(f.1(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.0(a.)) -> f_flat.0(down.0(a.))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.1(fresh_constant.)) -> f_flat.0(down.1(fresh_constant.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.))) -> f_flat.0(down.0(g.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))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))

The set Q consists of the following terms:

   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.1(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(120) 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:

   f_flat.0(up.1(x_1)) -> up.0(f.1(x_1))
   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(x_1)) = 1 + x_1
   POL(down.1(x_1)) = 1 + x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(f_flat.0(x_1)) = x_1
   POL(fresh_constant.) = 0
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1
   POL(s.0(x_1)) = 1 + x_1
   POL(s.1(x_1)) = 1 + x_1
   POL(s_flat.0(x_1)) = 1 + x_1
   POL(up.0(x_1)) = 1 + x_1
   POL(up.1(x_1)) = 1 + x_1


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

(121)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(f.0(f.1(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.0(a.)) -> f_flat.0(down.0(a.))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.1(fresh_constant.)) -> f_flat.0(down.1(fresh_constant.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.))) -> f_flat.0(down.0(g.1(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.1(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(122) 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(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.1(fresh_constant.)) -> f_flat.0(down.1(fresh_constant.))

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(x_1)) = 1 + x_1
   POL(down.1(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(f_flat.0(x_1)) = x_1
   POL(fresh_constant.) = 0
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1
   POL(s.0(x_1)) = 1 + x_1
   POL(s.1(x_1)) = 1 + x_1
   POL(s_flat.0(x_1)) = 1 + x_1
   POL(up.0(x_1)) = 1 + x_1


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

(123)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(f.0(f.1(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(f.0(a.)) -> f_flat.0(down.0(a.))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.))) -> f_flat.0(down.0(g.1(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.1(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(124) 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(f.0(f.1(x))) -> up.0(b.)

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(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(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1
   POL(s.0(x_1)) = x_1
   POL(s.1(x_1)) = x_1
   POL(s_flat.0(x_1)) = x_1
   POL(up.0(x_1)) = x_1


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

(125)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(f.0(a.)) -> f_flat.0(down.0(a.))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.))) -> f_flat.0(down.0(g.1(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.1(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.1(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

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

(127)
TRUE

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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   s_flat(up(x_1)) -> up(s(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_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(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(a)) -> f_flat(down(a))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   s_flat(up(x_1)) -> up(s(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_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.
s_flat: 0
s: 0
a: 1
b: 0
down: 0
f: 0
fresh_constant: 0
up: 0
f_flat: 0
TOP: 0
g: 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(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(s.1(x0))) -> TOP.0(s_flat.0(down.1(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))
   TOP.0(up.0(f.0(s.1(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.1(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(f.0(f.1(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.1(a.)) -> f_flat.0(down.1(a.))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))
   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(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.1(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(134) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

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

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(f.0(f.1(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.1(a.)) -> f_flat.0(down.1(a.))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))
   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(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.1(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(136) 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:

   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))
   f_flat.0(up.1(x_1)) -> up.0(f.1(x_1))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(x_1)) = 1 + x_1
   POL(down.1(x_1)) = 1 + x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(f_flat.0(x_1)) = x_1
   POL(fresh_constant.) = 0
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1
   POL(s.0(x_1)) = 1 + x_1
   POL(s.1(x_1)) = 1 + x_1
   POL(s_flat.0(x_1)) = 1 + x_1
   POL(up.0(x_1)) = 1 + x_1
   POL(up.1(x_1)) = 1 + x_1


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

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

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(f.0(f.1(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.1(a.)) -> f_flat.0(down.1(a.))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.1(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(138) 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(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.1(a.)) -> f_flat.0(down.1(a.))

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP.0(x_1)) = x_1
   POL(a.) = 0
   POL(b.) = 0
   POL(down.0(x_1)) = 1 + x_1
   POL(down.1(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(f_flat.0(x_1)) = x_1
   POL(fresh_constant.) = 0
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1
   POL(s.0(x_1)) = 1 + x_1
   POL(s.1(x_1)) = 1 + x_1
   POL(s_flat.0(x_1)) = 1 + x_1
   POL(up.0(x_1)) = 1 + x_1


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

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

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(f.0(f.1(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.1(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(140) 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(f.0(f.1(x))) -> up.0(b.)

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP.0(x_1)) = x_1
   POL(b.) = 0
   POL(down.0(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(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1
   POL(s.0(x_1)) = x_1
   POL(s.1(x_1)) = x_1
   POL(s_flat.0(x_1)) = x_1
   POL(up.0(x_1)) = x_1


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

(141)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.0(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(b.)) -> f_flat.0(down.0(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.0(b.))) -> f_flat.0(down.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   f_flat.0(up.0(x_1)) -> up.0(f.0(x_1))

The set Q consists of the following terms:

   down.0(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.1(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.0(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.0(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

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

(143)
TRUE

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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   s_flat(up(x_1)) -> up(s(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(145) 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(a))


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

(146)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   s_flat(up(x_1)) -> up(s(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

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

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

(148)
Complex Obligation (AND)

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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(b)) -> f_flat(down(b))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   s_flat(up(x_1)) -> up(s(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(150) 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.
s_flat: 0
s: 0
a: 0
b: 1
down: 0
f: 0
fresh_constant: 0
up: 0
f_flat: 0
TOP: 0
g: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

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

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(s.1(x0))) -> TOP.0(s_flat.0(down.1(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))
   TOP.0(up.0(f.0(s.1(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.1(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.1(b.)
   down.0(f.0(f.1(x))) -> up.1(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.1(b.)) -> f_flat.0(down.1(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.1(b.))) -> f_flat.0(down.0(g.1(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))
   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(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.1(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.1(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(152) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

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

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.1(b.)
   down.0(f.0(f.1(x))) -> up.1(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.1(b.)) -> f_flat.0(down.1(b.))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.1(b.))) -> f_flat.0(down.0(g.1(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))
   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(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.1(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.1(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(154) 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(f.1(b.)) -> f_flat.0(down.1(b.))
   s_flat.0(up.1(x_1)) -> up.0(s.1(x_1))

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP.0(x_1)) = x_1
   POL(b.) = 0
   POL(down.0(x_1)) = 1 + x_1
   POL(down.1(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(f_flat.0(x_1)) = x_1
   POL(fresh_constant.) = 0
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1
   POL(s.0(x_1)) = 1 + x_1
   POL(s.1(x_1)) = x_1
   POL(s_flat.0(x_1)) = 1 + x_1
   POL(up.0(x_1)) = 1 + x_1
   POL(up.1(x_1)) = 1 + x_1


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

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

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.1(b.)
   down.0(f.0(f.1(x))) -> up.1(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(s.1(y2)) -> s_flat.0(down.1(y2))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.1(b.))) -> f_flat.0(down.0(g.1(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   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(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.1(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.1(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

(156) 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(s.1(y2)) -> s_flat.0(down.1(y2))

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP.0(x_1)) = x_1
   POL(b.) = 0
   POL(down.0(x_1)) = 1 + x_1
   POL(down.1(x_1)) = x_1
   POL(f.0(x_1)) = x_1
   POL(f.1(x_1)) = x_1
   POL(f_flat.0(x_1)) = x_1
   POL(fresh_constant.) = 0
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1
   POL(s.0(x_1)) = x_1
   POL(s.1(x_1)) = 1 + x_1
   POL(s_flat.0(x_1)) = x_1
   POL(up.0(x_1)) = 1 + x_1
   POL(up.1(x_1)) = 1 + x_1


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

(157)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP.0(up.0(s.0(x0))) -> TOP.0(s_flat.0(down.0(x0)))
   TOP.0(up.0(f.0(s.0(x0)))) -> TOP.0(f_flat.0(s_flat.0(down.0(x0))))

The TRS R consists of the following rules:

   down.0(f.0(f.0(x))) -> up.1(b.)
   down.0(f.0(f.1(x))) -> up.1(b.)
   down.0(g.0(x)) -> up.0(f.0(g.0(x)))
   down.0(g.1(x)) -> up.0(f.0(g.1(x)))
   down.0(s.0(y2)) -> s_flat.0(down.0(y2))
   down.0(f.0(s.0(y6))) -> f_flat.0(down.0(s.0(y6)))
   down.0(f.0(s.1(y6))) -> f_flat.0(down.0(s.1(y6)))
   down.0(f.0(g.0(f.0(y8)))) -> f_flat.0(down.0(g.0(f.0(y8))))
   down.0(f.0(g.0(f.1(y8)))) -> f_flat.0(down.0(g.0(f.1(y8))))
   down.0(f.0(g.0(g.0(y9)))) -> f_flat.0(down.0(g.0(g.0(y9))))
   down.0(f.0(g.0(g.1(y9)))) -> f_flat.0(down.0(g.0(g.1(y9))))
   down.0(f.0(g.0(s.0(y10)))) -> f_flat.0(down.0(g.0(s.0(y10))))
   down.0(f.0(g.0(s.1(y10)))) -> f_flat.0(down.0(g.0(s.1(y10))))
   down.0(f.0(g.1(b.))) -> f_flat.0(down.0(g.1(b.)))
   down.0(f.0(g.0(fresh_constant.))) -> f_flat.0(down.0(g.0(fresh_constant.)))
   s_flat.0(up.0(x_1)) -> up.0(s.0(x_1))
   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(f.0(f.0(x0)))
   down.0(f.0(f.1(x0)))
   down.0(g.0(x0))
   down.0(g.1(x0))
   down.0(s.0(x0))
   down.0(s.1(x0))
   down.0(f.0(a.))
   down.0(f.0(s.0(x0)))
   down.0(f.0(s.1(x0)))
   down.0(f.1(b.))
   down.0(f.0(fresh_constant.))
   down.0(f.0(g.0(f.0(x0))))
   down.0(f.0(g.0(f.1(x0))))
   down.0(f.0(g.0(g.0(x0))))
   down.0(f.0(g.0(g.1(x0))))
   down.0(f.0(g.0(s.0(x0))))
   down.0(f.0(g.0(s.1(x0))))
   down.0(f.0(g.1(b.)))
   down.0(f.0(g.0(fresh_constant.)))
   f_flat.0(up.0(x0))
   f_flat.0(up.1(x0))
   s_flat.0(up.0(x0))
   s_flat.0(up.1(x0))

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

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

(159)
TRUE

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

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

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   s_flat(up(x_1)) -> up(s(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(a))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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

(161) 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(a))


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

(162)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(up(s(x0))) -> TOP(s_flat(down(x0)))
   TOP(up(f(s(x0)))) -> TOP(f_flat(s_flat(down(x0))))

The TRS R consists of the following rules:

   down(f(f(x))) -> up(b)
   down(g(x)) -> up(f(g(x)))
   down(s(y2)) -> s_flat(down(y2))
   down(f(s(y6))) -> f_flat(down(s(y6)))
   down(f(g(f(y8)))) -> f_flat(down(g(f(y8))))
   down(f(g(g(y9)))) -> f_flat(down(g(g(y9))))
   down(f(g(s(y10)))) -> f_flat(down(g(s(y10))))
   down(f(g(b))) -> f_flat(down(g(b)))
   down(f(g(fresh_constant))) -> f_flat(down(g(fresh_constant)))
   s_flat(up(x_1)) -> up(s(x_1))
   f_flat(up(x_1)) -> up(f(x_1))

The set Q consists of the following terms:

   down(f(f(x0)))
   down(g(x0))
   down(s(x0))
   down(f(s(x0)))
   down(f(b))
   down(f(fresh_constant))
   down(f(g(f(x0))))
   down(f(g(g(x0))))
   down(f(g(s(x0))))
   down(f(g(b)))
   down(f(g(fresh_constant)))
   f_flat(up(x0))
   s_flat(up(x0))

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