/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) TransformationProof [EQUIVALENT, 1 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QReductionProof [EQUIVALENT, 0 ms]
(14) QDP
(15) ATransformationProof [EQUIVALENT, 0 ms]
(16) QDP
(17) NonTerminationLoopProof [COMPLETE, 0 ms]
(18) NO


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

   +(1, x) -> +(+(0, 1), x)
   +(0, x) -> 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:

   +(1, x) -> +(+(0, 1), x)
   +(0, x) -> x

The set Q consists of the following terms:

   +(1, x0)
   +(0, 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:

   +^1(1, x) -> +^1(+(0, 1), x)
   +^1(1, x) -> +^1(0, 1)

The TRS R consists of the following rules:

   +(1, x) -> +(+(0, 1), x)
   +(0, x) -> x

The set Q consists of the following terms:

   +(1, x0)
   +(0, 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.
----------------------------------------

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

   +^1(1, x) -> +^1(+(0, 1), x)

The TRS R consists of the following rules:

   +(1, x) -> +(+(0, 1), x)
   +(0, x) -> x

The set Q consists of the following terms:

   +(1, x0)
   +(0, 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:

   +^1(1, x) -> +^1(+(0, 1), x)

The TRS R consists of the following rules:

   +(0, x) -> x

The set Q consists of the following terms:

   +(1, x0)
   +(0, x0)

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

(9) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule +^1(1, x) -> +^1(+(0, 1), x) at position [0] we obtained the following new rules [LPAR04]:

   (+^1(1, x) -> +^1(1, x),+^1(1, x) -> +^1(1, x))


----------------------------------------

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

   +^1(1, x) -> +^1(1, x)

The TRS R consists of the following rules:

   +(0, x) -> x

The set Q consists of the following terms:

   +(1, x0)
   +(0, x0)

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

(11) 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.
----------------------------------------

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

   +^1(1, x) -> +^1(1, x)

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

   +(1, x0)
   +(0, x0)

We have to consider all minimal (P,Q,R)-chains.
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(13) 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].

   +(1, x0)
   +(0, x0)


----------------------------------------

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

   +^1(1, x) -> +^1(1, x)

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

(15) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   1(x) -> 1(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(17) 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(x) evaluates to  t =1(x)

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(x) to 1(x).




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(18)
NO