/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 26 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) MNOCProof [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:

   g(X) -> h(X)
   c -> d
   h(d) -> g(c)

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   c -> d

The TRS R 2 is 
   g(X) -> h(X)
   h(d) -> g(c)

The signature Sigma is {g_1, h_1}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(X) -> h(X)
   c -> d
   h(d) -> g(c)

The set Q consists of the following terms:

   g(x0)
   c
   h(d)


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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

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

   G(X) -> H(X)
   H(d) -> G(c)
   H(d) -> C

The TRS R consists of the following rules:

   g(X) -> h(X)
   c -> d
   h(d) -> g(c)

The set Q consists of the following terms:

   g(x0)
   c
   h(d)

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:

   H(d) -> G(c)
   G(X) -> H(X)

The TRS R consists of the following rules:

   g(X) -> h(X)
   c -> d
   h(d) -> g(c)

The set Q consists of the following terms:

   g(x0)
   c
   h(d)

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:

   H(d) -> G(c)
   G(X) -> H(X)

The TRS R consists of the following rules:

   c -> d

The set Q consists of the following terms:

   g(x0)
   c
   h(d)

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].

   g(x0)
   h(d)


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

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

   H(d) -> G(c)
   G(X) -> H(X)

The TRS R consists of the following rules:

   c -> d

The set Q consists of the following terms:

   c

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

(11) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
----------------------------------------

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

   H(d) -> G(c)
   G(X) -> H(X)

The TRS R consists of the following rules:

   c -> d

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

(13) 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 = G(c) evaluates to  t =G(c)

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

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Rewriting sequence

G(c) -> G(d)
with rule c -> d at position [0] and matcher [ ]

G(d) -> H(d)
with rule G(X) -> H(X) at position [] and matcher [X / d]

H(d) -> G(c)
with rule H(d) -> G(c)

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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(14)
NO