/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) TransformationProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) NonTerminationLoopProof [COMPLETE, 0 ms]
(14) NO


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

   *(X, +(Y, 1)) -> +(*(X, +(Y, *(1, 0))), X)
   *(X, 1) -> X
   *(X, 0) -> X
   *(X, 0) -> 0

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:

   *^1(X, +(Y, 1)) -> *^1(X, +(Y, *(1, 0)))
   *^1(X, +(Y, 1)) -> *^1(1, 0)

The TRS R consists of the following rules:

   *(X, +(Y, 1)) -> +(*(X, +(Y, *(1, 0))), X)
   *(X, 1) -> X
   *(X, 0) -> X
   *(X, 0) -> 0

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:

   *^1(X, +(Y, 1)) -> *^1(X, +(Y, *(1, 0)))

The TRS R consists of the following rules:

   *(X, +(Y, 1)) -> +(*(X, +(Y, *(1, 0))), X)
   *(X, 1) -> X
   *(X, 0) -> X
   *(X, 0) -> 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   *^1(X, +(Y, 1)) -> *^1(X, +(Y, *(1, 0)))

The TRS R consists of the following rules:

   *(X, 0) -> X
   *(X, 0) -> 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule *^1(X, +(Y, 1)) -> *^1(X, +(Y, *(1, 0))) at position [1,1] we obtained the following new rules [LPAR04]:

   (*^1(y0, +(y1, 1)) -> *^1(y0, +(y1, 1)),*^1(y0, +(y1, 1)) -> *^1(y0, +(y1, 1)))
   (*^1(y0, +(y1, 1)) -> *^1(y0, +(y1, 0)),*^1(y0, +(y1, 1)) -> *^1(y0, +(y1, 0)))


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

   *^1(y0, +(y1, 1)) -> *^1(y0, +(y1, 1))
   *^1(y0, +(y1, 1)) -> *^1(y0, +(y1, 0))

The TRS R consists of the following rules:

   *(X, 0) -> X
   *(X, 0) -> 0

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

   *^1(y0, +(y1, 1)) -> *^1(y0, +(y1, 1))

The TRS R consists of the following rules:

   *(X, 0) -> X
   *(X, 0) -> 0

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

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   *^1(y0, +(y1, 1)) -> *^1(y0, +(y1, 1))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(13) 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 = *^1(y0, +(y1, 1)) evaluates to  t =*^1(y0, +(y1, 1))

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

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from *^1(y0, +(y1, 1)) to *^1(y0, +(y1, 1)).




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(14)
NO