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


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NO
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(6) NO


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

   f(a, n__b, X) -> f(X, X, X)
   c -> a
   c -> b
   b -> n__b
   activate(n__b) -> b
   activate(X) -> X

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(a, n__b, X) -> F(X, X, X)
   C -> B
   ACTIVATE(n__b) -> B

The TRS R consists of the following rules:

   f(a, n__b, X) -> f(X, X, X)
   c -> a
   c -> b
   b -> n__b
   activate(n__b) -> b
   activate(X) -> X

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

   F(a, n__b, X) -> F(X, X, X)

The TRS R consists of the following rules:

   f(a, n__b, X) -> f(X, X, X)
   c -> a
   c -> b
   b -> n__b
   activate(n__b) -> b
   activate(X) -> X

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(c, c, X) evaluates to  t =F(X, X, X)

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

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Rewriting sequence

F(c, c, c) -> F(c, b, c)
with rule c -> b at position [1] and matcher [ ]

F(c, b, c) -> F(c, n__b, c)
with rule b -> n__b at position [1] and matcher [ ]

F(c, n__b, c) -> F(a, n__b, c)
with rule c -> a at position [0] and matcher [ ]

F(a, n__b, c) -> F(c, c, c)
with rule F(a, n__b, X) -> F(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