/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QReductionProof [EQUIVALENT, 0 ms]
(10) QDP
(11) TransformationProof [EQUIVALENT, 0 ms]
(12) QDP
(13) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) NonTerminationLoopProof [COMPLETE, 0 ms]
(16) NO


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

   2nd(cons1(X, cons(Y, Z))) -> Y
   2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
   from(X) -> cons(X, from(s(X)))

Q is empty.

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(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   2nd(cons1(X, cons(Y, Z))) -> Y
   2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
   from(X) -> cons(X, from(s(X)))

The set Q consists of the following terms:

   2nd(cons1(x0, cons(x1, x2)))
   2nd(cons(x0, x1))
   from(x0)


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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:

   2ND(cons(X, X1)) -> 2ND(cons1(X, X1))
   FROM(X) -> FROM(s(X))

The TRS R consists of the following rules:

   2nd(cons1(X, cons(Y, Z))) -> Y
   2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
   from(X) -> cons(X, from(s(X)))

The set Q consists of the following terms:

   2nd(cons1(x0, cons(x1, x2)))
   2nd(cons(x0, x1))
   from(x0)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FROM(X) -> FROM(s(X))

The TRS R consists of the following rules:

   2nd(cons1(X, cons(Y, Z))) -> Y
   2nd(cons(X, X1)) -> 2nd(cons1(X, X1))
   from(X) -> cons(X, from(s(X)))

The set Q consists of the following terms:

   2nd(cons1(x0, cons(x1, x2)))
   2nd(cons(x0, x1))
   from(x0)

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

(7) 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.
----------------------------------------

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

   FROM(X) -> FROM(s(X))

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

   2nd(cons1(x0, cons(x1, x2)))
   2nd(cons(x0, x1))
   from(x0)

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

(9) 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].

   2nd(cons1(x0, cons(x1, x2)))
   2nd(cons(x0, x1))
   from(x0)


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

   FROM(X) -> FROM(s(X))

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

(11) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule FROM(X) -> FROM(s(X)) we obtained the following new rules [LPAR04]:

   (FROM(s(z0)) -> FROM(s(s(z0))),FROM(s(z0)) -> FROM(s(s(z0))))


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

   FROM(s(z0)) -> FROM(s(s(z0)))

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

(13) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule FROM(s(z0)) -> FROM(s(s(z0))) we obtained the following new rules [LPAR04]:

   (FROM(s(s(z0))) -> FROM(s(s(s(z0)))),FROM(s(s(z0))) -> FROM(s(s(s(z0)))))


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

   FROM(s(s(z0))) -> FROM(s(s(s(z0))))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(15) 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 = FROM(s(s(z0))) evaluates to  t =FROM(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [z0 / s(z0)]
* Semiunifier: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FROM(s(s(z0))) to FROM(s(s(s(z0)))).




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(16)
NO