/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QDPOrderProof [EQUIVALENT, 30 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) NonTerminationLoopProof [COMPLETE, 130 ms]
(10) NO


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

   a__f(a, b, X) -> a__f(X, X, mark(X))
   a__c -> a
   a__c -> b
   mark(f(X1, X2, X3)) -> a__f(X1, X2, mark(X3))
   mark(c) -> a__c
   mark(a) -> a
   mark(b) -> b
   a__f(X1, X2, X3) -> f(X1, X2, X3)
   a__c -> c

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:

   A__F(a, b, X) -> A__F(X, X, mark(X))
   A__F(a, b, X) -> MARK(X)
   MARK(f(X1, X2, X3)) -> A__F(X1, X2, mark(X3))
   MARK(f(X1, X2, X3)) -> MARK(X3)
   MARK(c) -> A__C

The TRS R consists of the following rules:

   a__f(a, b, X) -> a__f(X, X, mark(X))
   a__c -> a
   a__c -> b
   mark(f(X1, X2, X3)) -> a__f(X1, X2, mark(X3))
   mark(c) -> a__c
   mark(a) -> a
   mark(b) -> b
   a__f(X1, X2, X3) -> f(X1, X2, X3)
   a__c -> c

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:

   A__F(a, b, X) -> MARK(X)
   MARK(f(X1, X2, X3)) -> A__F(X1, X2, mark(X3))
   A__F(a, b, X) -> A__F(X, X, mark(X))
   MARK(f(X1, X2, X3)) -> MARK(X3)

The TRS R consists of the following rules:

   a__f(a, b, X) -> a__f(X, X, mark(X))
   a__c -> a
   a__c -> b
   mark(f(X1, X2, X3)) -> a__f(X1, X2, mark(X3))
   mark(c) -> a__c
   mark(a) -> a
   mark(b) -> b
   a__f(X1, X2, X3) -> f(X1, X2, X3)
   a__c -> c

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   MARK(f(X1, X2, X3)) -> A__F(X1, X2, mark(X3))
   MARK(f(X1, X2, X3)) -> MARK(X3)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A__F_3(x_1, ..., x_3) ) = x_3
POL( mark_1(x_1) ) = x_1
POL( f_3(x_1, ..., x_3) ) = x_3 + 1
POL( a__f_3(x_1, ..., x_3) ) = x_3 + 1
POL( c ) = 0
POL( a__c ) = 0
POL( a ) = 0
POL( b ) = 0
POL( MARK_1(x_1) ) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   mark(f(X1, X2, X3)) -> a__f(X1, X2, mark(X3))
   mark(c) -> a__c
   mark(a) -> a
   mark(b) -> b
   a__f(X1, X2, X3) -> f(X1, X2, X3)
   a__f(a, b, X) -> a__f(X, X, mark(X))
   a__c -> a
   a__c -> b
   a__c -> c


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

   A__F(a, b, X) -> MARK(X)
   A__F(a, b, X) -> A__F(X, X, mark(X))

The TRS R consists of the following rules:

   a__f(a, b, X) -> a__f(X, X, mark(X))
   a__c -> a
   a__c -> b
   mark(f(X1, X2, X3)) -> a__f(X1, X2, mark(X3))
   mark(c) -> a__c
   mark(a) -> a
   mark(b) -> b
   a__f(X1, X2, X3) -> f(X1, X2, X3)
   a__c -> c

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__F(a, b, X) -> A__F(X, X, mark(X))

The TRS R consists of the following rules:

   a__f(a, b, X) -> a__f(X, X, mark(X))
   a__c -> a
   a__c -> b
   mark(f(X1, X2, X3)) -> a__f(X1, X2, mark(X3))
   mark(c) -> a__c
   mark(a) -> a
   mark(b) -> b
   a__f(X1, X2, X3) -> f(X1, X2, X3)
   a__c -> c

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) 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 = A__F(mark(a__c), mark(a__c), mark(mark(a__c))) evaluates to  t =A__F(mark(a__c), mark(a__c), mark(mark(a__c)))

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

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Rewriting sequence

A__F(mark(a__c), mark(a__c), mark(mark(a__c))) -> A__F(mark(a__c), mark(a__c), mark(mark(c)))
with rule a__c -> c at position [2,0,0] and matcher [ ]

A__F(mark(a__c), mark(a__c), mark(mark(c))) -> A__F(mark(a__c), mark(a__c), mark(a__c))
with rule mark(c) -> a__c at position [2,0] and matcher [ ]

A__F(mark(a__c), mark(a__c), mark(a__c)) -> A__F(mark(a__c), mark(a__c), mark(c))
with rule a__c -> c at position [2,0] and matcher [ ]

A__F(mark(a__c), mark(a__c), mark(c)) -> A__F(mark(a__c), mark(a__c), a__c)
with rule mark(c) -> a__c at position [2] and matcher [ ]

A__F(mark(a__c), mark(a__c), a__c) -> A__F(mark(a__c), mark(b), a__c)
with rule a__c -> b at position [1,0] and matcher [ ]

A__F(mark(a__c), mark(b), a__c) -> A__F(mark(a__c), b, a__c)
with rule mark(b) -> b at position [1] and matcher [ ]

A__F(mark(a__c), b, a__c) -> A__F(mark(a), b, a__c)
with rule a__c -> a at position [0,0] and matcher [ ]

A__F(mark(a), b, a__c) -> A__F(a, b, a__c)
with rule mark(a) -> a at position [0] and matcher [ ]

A__F(a, b, a__c) -> A__F(a__c, a__c, mark(a__c))
with rule A__F(a, b, X') -> A__F(X', X', mark(X')) at position [] and matcher [X' / a__c]

A__F(a__c, a__c, mark(a__c)) -> A__F(a__c, b, mark(a__c))
with rule a__c -> b at position [1] and matcher [ ]

A__F(a__c, b, mark(a__c)) -> A__F(a, b, mark(a__c))
with rule a__c -> a at position [0] and matcher [ ]

A__F(a, b, mark(a__c)) -> A__F(mark(a__c), mark(a__c), mark(mark(a__c)))
with rule A__F(a, b, X) -> A__F(X, X, mark(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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(10)
NO