/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) NonTerminationLoopProof [COMPLETE, 43 ms]
(6) NO


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

   f(0, 1, X) -> f(g(X, X), X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, 1, X) -> F(g(X, X), X, X)
   F(0, 1, X) -> G(X, X)

The TRS R consists of the following rules:

   f(0, 1, X) -> f(g(X, X), X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, 1, X) -> F(g(X, X), X, X)

The TRS R consists of the following rules:

   f(0, 1, X) -> f(g(X, X), X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F(g(g(0, Y), Y'), g(X', 1), X) evaluates to  t =F(g(X, X), X, X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [Y / 1, Y' / g(0, 1), X' / 0, X / g(0, 1)]

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Rewriting sequence

F(g(g(0, 1), g(0, 1)), g(0, 1), g(0, 1)) -> F(g(g(0, 1), g(0, 1)), 1, g(0, 1))
with rule g(X', Y'') -> Y'' at position [1] and matcher [X' / 0, Y'' / 1]

F(g(g(0, 1), g(0, 1)), 1, g(0, 1)) -> F(g(0, 1), 1, g(0, 1))
with rule g(X', Y') -> X' at position [0] and matcher [X' / g(0, 1), Y' / g(0, 1)]

F(g(0, 1), 1, g(0, 1)) -> F(0, 1, g(0, 1))
with rule g(X', Y) -> X' at position [0] and matcher [X' / 0, Y / 1]

F(0, 1, g(0, 1)) -> F(g(g(0, 1), g(0, 1)), g(0, 1), g(0, 1))
with rule F(0, 1, X) -> F(g(X, X), X, X)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(6)
NO