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


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


MAYBE
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 50 ms]
    (4) QTRS
    (5) AAECC Innermost [EQUIVALENT, 0 ms]
    (6) QTRS
    (7) DependencyPairsProof [EQUIVALENT, 0 ms]
    (8) QDP
    (9) DependencyGraphProof [EQUIVALENT, 0 ms]
    (10) QDP
    (11) UsableRulesProof [EQUIVALENT, 0 ms]
    (12) QDP
    (13) QReductionProof [EQUIVALENT, 0 ms]
    (14) QDP
    (15) TransformationProof [EQUIVALENT, 0 ms]
    (16) QDP
    (17) UsableRulesProof [EQUIVALENT, 0 ms]
    (18) QDP
    (19) QReductionProof [EQUIVALENT, 0 ms]
    (20) QDP
    (21) NonTerminationLoopProof [COMPLETE, 0 ms]
    (22) NO
(23) Thiemann-SpecialC-Transformation [EQUIVALENT, 0 ms]
(24) QTRS
    (25) QTRSRRRProof [EQUIVALENT, 80 ms]
    (26) QTRS
    (27) DependencyPairsProof [EQUIVALENT, 0 ms]
    (28) QDP
    (29) DependencyGraphProof [EQUIVALENT, 0 ms]
    (30) AND
        (31) QDP
            (32) UsableRulesProof [EQUIVALENT, 0 ms]
            (33) QDP
            (34) QReductionProof [EQUIVALENT, 0 ms]
            (35) QDP
            (36) UsableRulesReductionPairsProof [EQUIVALENT, 8 ms]
            (37) QDP
            (38) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
            (39) QDP
            (40) DependencyGraphProof [EQUIVALENT, 0 ms]
            (41) TRUE
        (42) QDP
            (43) UsableRulesProof [EQUIVALENT, 0 ms]
            (44) QDP
                (45) QReductionProof [EQUIVALENT, 0 ms]
                (46) QDP
                (47) TransformationProof [SOUND, 0 ms]
                (48) QDP
                (49) TransformationProof [EQUIVALENT, 0 ms]
                (50) QDP
                (51) TransformationProof [EQUIVALENT, 0 ms]
                (52) QDP
                (53) QDPOrderProof [EQUIVALENT, 0 ms]
                (54) QDP
            (55) UsableRulesProof [EQUIVALENT, 0 ms]
            (56) QDP
                (57) QReductionProof [EQUIVALENT, 0 ms]
                (58) QDP


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

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

   f(h(x, x)) -> f(i(x))
   f(i(x)) -> a
   i(x) -> h(x, 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(h(x, x)) -> f(i(x))
   f(i(x)) -> a
   i(x) -> h(x, x)

Q is empty.

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

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

   POL(a) = 2
   POL(f(x_1)) = 1 + 2*x_1
   POL(h(x_1, x_2)) = 1 + x_1 + x_2
   POL(i(x_1)) = 1 + 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(i(x)) -> a




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

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

   f(h(x, x)) -> f(i(x))
   i(x) -> h(x, x)

Q is empty.

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

(5) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   i(x) -> h(x, x)

The TRS R 2 is 
   f(h(x, x)) -> f(i(x))

The signature Sigma is {f_1}
----------------------------------------

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(h(x, x)) -> f(i(x))
   i(x) -> h(x, x)

The set Q consists of the following terms:

   f(h(x0, x0))
   i(x0)


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

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

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

   F(h(x, x)) -> F(i(x))
   F(h(x, x)) -> I(x)

The TRS R consists of the following rules:

   f(h(x, x)) -> f(i(x))
   i(x) -> h(x, x)

The set Q consists of the following terms:

   f(h(x0, x0))
   i(x0)

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

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

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

   F(h(x, x)) -> F(i(x))

The TRS R consists of the following rules:

   f(h(x, x)) -> f(i(x))
   i(x) -> h(x, x)

The set Q consists of the following terms:

   f(h(x0, x0))
   i(x0)

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

(11) 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.
----------------------------------------

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

   F(h(x, x)) -> F(i(x))

The TRS R consists of the following rules:

   i(x) -> h(x, x)

The set Q consists of the following terms:

   f(h(x0, x0))
   i(x0)

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

(13) 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].

   f(h(x0, x0))


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

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

   F(h(x, x)) -> F(i(x))

The TRS R consists of the following rules:

   i(x) -> h(x, x)

The set Q consists of the following terms:

   i(x0)

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

(15) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule F(h(x, x)) -> F(i(x)) at position [0] we obtained the following new rules [LPAR04]:

   (F(h(x, x)) -> F(h(x, x)),F(h(x, x)) -> F(h(x, x)))


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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(x, x)) -> F(h(x, x))

The TRS R consists of the following rules:

   i(x) -> h(x, x)

The set Q consists of the following terms:

   i(x0)

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

(17) 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.
----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(x, x)) -> F(h(x, x))

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

   i(x0)

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

(19) 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].

   i(x0)


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

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

   F(h(x, x)) -> F(h(x, x))

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

(21) 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 = F(h(x, x)) evaluates to  t =F(h(x, 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 F(h(x, x)) to F(h(x, x)).




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

(22)
NO

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

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

(24)
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(i(x_1)) -> check_i(redex_i(x_1))
   redex_f(h(x, x)) -> result_f(f(i(x)))
   redex_f(i(x)) -> result_f(a)
   redex_i(x) -> result_i(h(x, x))
   check_f(result_f(x)) -> go_up(x)
   check_i(result_i(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_i(redex_i(x_1)) -> in_i_1(reduce(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_i_1(go_up(x_1)) -> go_up(i(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))


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

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

   POL(a) = 0
   POL(check_f(x_1)) = x_1
   POL(check_i(x_1)) = x_1
   POL(f(x_1)) = 1 + x_1
   POL(go_up(x_1)) = x_1
   POL(h(x_1, x_2)) = x_1 + x_2
   POL(i(x_1)) = 2*x_1
   POL(in_f_1(x_1)) = 1 + x_1
   POL(in_h_1(x_1, x_2)) = x_1 + x_2
   POL(in_h_2(x_1, x_2)) = x_1 + x_2
   POL(in_i_1(x_1)) = 2*x_1
   POL(redex_f(x_1)) = 1 + x_1
   POL(redex_i(x_1)) = 2*x_1
   POL(reduce(x_1)) = x_1
   POL(result_f(x_1)) = x_1
   POL(result_i(x_1)) = 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(i(x)) -> result_f(a)




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

(26)
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(i(x_1)) -> check_i(redex_i(x_1))
   redex_f(h(x, x)) -> result_f(f(i(x)))
   redex_i(x) -> result_i(h(x, x))
   check_f(result_f(x)) -> go_up(x)
   check_i(result_i(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_i(redex_i(x_1)) -> in_i_1(reduce(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_i_1(go_up(x_1)) -> go_up(i(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))


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

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

(28)
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(i(x_1)) -> CHECK_I(redex_i(x_1))
   REDUCE(i(x_1)) -> REDEX_I(x_1)
   CHECK_F(redex_f(x_1)) -> IN_F_1(reduce(x_1))
   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   CHECK_I(redex_i(x_1)) -> IN_I_1(reduce(x_1))
   CHECK_I(redex_i(x_1)) -> REDUCE(x_1)
   REDUCE(h(x_1, x_2)) -> IN_H_1(reduce(x_1), x_2)
   REDUCE(h(x_1, x_2)) -> REDUCE(x_1)
   REDUCE(h(x_1, x_2)) -> IN_H_2(x_1, reduce(x_2))
   REDUCE(h(x_1, x_2)) -> REDUCE(x_2)

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(i(x_1)) -> check_i(redex_i(x_1))
   redex_f(h(x, x)) -> result_f(f(i(x)))
   redex_i(x) -> result_i(h(x, x))
   check_f(result_f(x)) -> go_up(x)
   check_i(result_i(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_i(redex_i(x_1)) -> in_i_1(reduce(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_i_1(go_up(x_1)) -> go_up(i(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))

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

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

(30)
Complex Obligation (AND)

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

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

   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   REDUCE(f(x_1)) -> CHECK_F(redex_f(x_1))
   REDUCE(h(x_1, x_2)) -> REDUCE(x_1)
   REDUCE(h(x_1, x_2)) -> REDUCE(x_2)

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(i(x_1)) -> check_i(redex_i(x_1))
   redex_f(h(x, x)) -> result_f(f(i(x)))
   redex_i(x) -> result_i(h(x, x))
   check_f(result_f(x)) -> go_up(x)
   check_i(result_i(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_i(redex_i(x_1)) -> in_i_1(reduce(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_i_1(go_up(x_1)) -> go_up(i(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))

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

(32) 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.
----------------------------------------

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

   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   REDUCE(f(x_1)) -> CHECK_F(redex_f(x_1))
   REDUCE(h(x_1, x_2)) -> REDUCE(x_1)
   REDUCE(h(x_1, x_2)) -> REDUCE(x_2)

The TRS R consists of the following rules:

   redex_f(h(x, x)) -> result_f(f(i(x)))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))

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

(34) 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(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))


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

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

   CHECK_F(redex_f(x_1)) -> REDUCE(x_1)
   REDUCE(f(x_1)) -> CHECK_F(redex_f(x_1))
   REDUCE(h(x_1, x_2)) -> REDUCE(x_1)
   REDUCE(h(x_1, x_2)) -> REDUCE(x_2)

The TRS R consists of the following rules:

   redex_f(h(x, x)) -> result_f(f(i(x)))

The set Q consists of the following terms:

   redex_f(h(x0, x0))
   redex_f(i(x0))

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

(36) 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(h(x_1, x_2)) -> REDUCE(x_1)
   REDUCE(h(x_1, x_2)) -> REDUCE(x_2)
The following rules are removed from R:

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

   POL(CHECK_F(x_1)) = x_1
   POL(REDUCE(x_1)) = 2*x_1
   POL(f(x_1)) = 2*x_1
   POL(h(x_1, x_2)) = 2*x_1 + x_2
   POL(i(x_1)) = x_1
   POL(redex_f(x_1)) = 2*x_1
   POL(result_f(x_1)) = 2*x_1


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

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   redex_f(h(x0, x0))
   redex_f(i(x0))

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

(38) 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(f(x_1)) -> CHECK_F(redex_f(x_1))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(CHECK_F(x_1)) = 2*x_1
   POL(REDUCE(x_1)) = 2*x_1
   POL(f(x_1)) = 2*x_1
   POL(redex_f(x_1)) = x_1


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

(39)
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(h(x0, x0))
   redex_f(i(x0))

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

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

(41)
TRUE

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

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

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

The TRS R consists of the following rules:

   top(go_up(x)) -> top(reduce(x))
   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(i(x_1)) -> check_i(redex_i(x_1))
   redex_f(h(x, x)) -> result_f(f(i(x)))
   redex_i(x) -> result_i(h(x, x))
   check_f(result_f(x)) -> go_up(x)
   check_i(result_i(x)) -> go_up(x)
   check_f(redex_f(x_1)) -> in_f_1(reduce(x_1))
   check_i(redex_i(x_1)) -> in_i_1(reduce(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_i_1(go_up(x_1)) -> go_up(i(x_1))

The set Q consists of the following terms:

   top(go_up(x0))
   reduce(f(x0))
   reduce(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))

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

(43) 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.
----------------------------------------

(44)
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(i(x_1)) -> check_i(redex_i(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   redex_i(x) -> result_i(h(x, x))
   check_i(result_i(x)) -> go_up(x)
   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))

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

(45) 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_i_1(go_up(x0))


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

(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(i(x_1)) -> check_i(redex_i(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   redex_i(x) -> result_i(h(x, x))
   check_i(result_i(x)) -> go_up(x)
   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))

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

(47) 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(i(x0))) -> TOP(check_i(redex_i(x0))),TOP(go_up(i(x0))) -> TOP(check_i(redex_i(x0))))
   (TOP(go_up(h(x0, x1))) -> TOP(in_h_1(reduce(x0), x1)),TOP(go_up(h(x0, x1))) -> TOP(in_h_1(reduce(x0), x1)))
   (TOP(go_up(h(x0, x1))) -> TOP(in_h_2(x0, reduce(x1))),TOP(go_up(h(x0, x1))) -> TOP(in_h_2(x0, reduce(x1))))


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

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

   TOP(go_up(f(x0))) -> TOP(check_f(redex_f(x0)))
   TOP(go_up(i(x0))) -> TOP(check_i(redex_i(x0)))
   TOP(go_up(h(x0, x1))) -> TOP(in_h_1(reduce(x0), x1))
   TOP(go_up(h(x0, x1))) -> TOP(in_h_2(x0, reduce(x1)))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(i(x_1)) -> check_i(redex_i(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   redex_i(x) -> result_i(h(x, x))
   check_i(result_i(x)) -> go_up(x)
   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))

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

(49) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(go_up(i(x0))) -> TOP(check_i(redex_i(x0))) at position [0,0] we obtained the following new rules [LPAR04]:

   (TOP(go_up(i(x0))) -> TOP(check_i(result_i(h(x0, x0)))),TOP(go_up(i(x0))) -> TOP(check_i(result_i(h(x0, x0)))))


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

(50)
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(h(x0, x1))) -> TOP(in_h_1(reduce(x0), x1))
   TOP(go_up(h(x0, x1))) -> TOP(in_h_2(x0, reduce(x1)))
   TOP(go_up(i(x0))) -> TOP(check_i(result_i(h(x0, x0))))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(i(x_1)) -> check_i(redex_i(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   redex_i(x) -> result_i(h(x, x))
   check_i(result_i(x)) -> go_up(x)
   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))

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

(51) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule TOP(go_up(i(x0))) -> TOP(check_i(result_i(h(x0, x0)))) at position [0] we obtained the following new rules [LPAR04]:

   (TOP(go_up(i(x0))) -> TOP(go_up(h(x0, x0))),TOP(go_up(i(x0))) -> TOP(go_up(h(x0, x0))))


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

(52)
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(h(x0, x1))) -> TOP(in_h_1(reduce(x0), x1))
   TOP(go_up(h(x0, x1))) -> TOP(in_h_2(x0, reduce(x1)))
   TOP(go_up(i(x0))) -> TOP(go_up(h(x0, x0)))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(i(x_1)) -> check_i(redex_i(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   redex_i(x) -> result_i(h(x, x))
   check_i(result_i(x)) -> go_up(x)
   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))

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

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


The following pairs can be oriented strictly and are deleted.

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

   POL(TOP(x_1)) = x_1
   POL(check_f(x_1)) = x_1
   POL(check_i(x_1)) = 1
   POL(f(x_1)) = 0
   POL(go_up(x_1)) = x_1
   POL(h(x_1, x_2)) = 0
   POL(i(x_1)) = 1 + x_1
   POL(in_f_1(x_1)) = 0
   POL(in_h_1(x_1, x_2)) = 0
   POL(in_h_2(x_1, x_2)) = 0
   POL(redex_f(x_1)) = 0
   POL(redex_i(x_1)) = 1 + x_1
   POL(reduce(x_1)) = 0
   POL(result_f(x_1)) = x_1
   POL(result_i(x_1)) = 1 + x_1

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

   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_f_1(go_up(x_1)) -> go_up(f(x_1))


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

(54)
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(h(x0, x1))) -> TOP(in_h_1(reduce(x0), x1))
   TOP(go_up(h(x0, x1))) -> TOP(in_h_2(x0, reduce(x1)))

The TRS R consists of the following rules:

   reduce(f(x_1)) -> check_f(redex_f(x_1))
   reduce(i(x_1)) -> check_i(redex_i(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   redex_i(x) -> result_i(h(x, x))
   check_i(result_i(x)) -> go_up(x)
   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))

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

(55) 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.
----------------------------------------

(56)
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(i(x_1)) -> check_i(redex_i(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   redex_i(x) -> result_i(h(x, x))
   check_i(result_i(x)) -> go_up(x)
   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))
   in_i_1(go_up(x0))

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

(57) 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_i_1(go_up(x0))


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

(58)
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(i(x_1)) -> check_i(redex_i(x_1))
   reduce(h(x_1, x_2)) -> in_h_1(reduce(x_1), x_2)
   reduce(h(x_1, x_2)) -> in_h_2(x_1, reduce(x_2))
   in_h_2(x_1, go_up(x_2)) -> go_up(h(x_1, x_2))
   in_h_1(go_up(x_1), x_2) -> go_up(h(x_1, x_2))
   redex_i(x) -> result_i(h(x, x))
   check_i(result_i(x)) -> go_up(x)
   redex_f(h(x, x)) -> result_f(f(i(x)))
   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(i(x0))
   redex_f(h(x0, x0))
   redex_f(i(x0))
   redex_i(x0)
   check_f(result_f(x0))
   check_i(result_i(x0))
   check_f(redex_f(x0))
   reduce(h(x0, x1))
   in_f_1(go_up(x0))
   in_h_1(go_up(x0), x1)
   in_h_2(x0, go_up(x1))

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