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


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NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be disproven:

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QReductionProof [EQUIVALENT, 0 ms]
(8) QDP
(9) NonTerminationLoopProof [COMPLETE, 0 ms]
(10) NO


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

   f(x, y) -> f(x, x)
   f(s(x), y) -> f(y, x)

Q is empty.

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(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is 
   f(x, y) -> f(x, x)
   f(s(x), y) -> f(y, x)

The signature Sigma is {f_2}
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x, y) -> f(x, x)
   f(s(x), y) -> f(y, x)

The set Q consists of the following terms:

   f(x0, x1)


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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, y) -> F(x, x)
   F(s(x), y) -> F(y, x)

The TRS R consists of the following rules:

   f(x, y) -> f(x, x)
   f(s(x), y) -> f(y, x)

The set Q consists of the following terms:

   f(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
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(5) 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.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, y) -> F(x, x)
   F(s(x), y) -> F(y, x)

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

   f(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
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(7) 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(x0, x1)


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

   F(x, y) -> F(x, x)
   F(s(x), y) -> F(y, x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) 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(x, y) evaluates to  t =F(x, x)

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

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(x, y) to F(x, x).




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(10)
NO