/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 17 ms]
(2) QTRS
(3) AAECC Innermost [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 17 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QReductionProof [EQUIVALENT, 0 ms]
(10) QDP
(11) NonTerminationLoopProof [COMPLETE, 6 ms]
(12) NO


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

   h(X) -> g(X)
   g(a) -> f(b)
   f(X) -> h(a)
   a -> b

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 1
   POL(b) = 0
   POL(f(x_1)) = 1 + x_1
   POL(g(x_1)) = x_1
   POL(h(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a -> b




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

   h(X) -> g(X)
   g(a) -> f(b)
   f(X) -> h(a)

Q is empty.

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(3) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is 
   h(X) -> g(X)
   g(a) -> f(b)
   f(X) -> h(a)

The signature Sigma is {h_1, g_1, f_1}
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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   h(X) -> g(X)
   g(a) -> f(b)
   f(X) -> h(a)

The set Q consists of the following terms:

   h(x0)
   g(a)
   f(x0)


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(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(X) -> G(X)
   G(a) -> F(b)
   F(X) -> H(a)

The TRS R consists of the following rules:

   h(X) -> g(X)
   g(a) -> f(b)
   f(X) -> h(a)

The set Q consists of the following terms:

   h(x0)
   g(a)
   f(x0)

We have to consider all minimal (P,Q,R)-chains.
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(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.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(X) -> G(X)
   G(a) -> F(b)
   F(X) -> H(a)

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

   h(x0)
   g(a)
   f(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].

   h(x0)
   g(a)
   f(x0)


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

   H(X) -> G(X)
   G(a) -> F(b)
   F(X) -> H(a)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(11) 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(a) evaluates to  t =G(a)

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

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Rewriting sequence

G(a) -> F(b)
with rule G(a) -> F(b) at position [] and matcher [ ]

F(b) -> H(a)
with rule F(X') -> H(a) at position [] and matcher [X' / b]

H(a) -> G(a)
with rule H(X) -> G(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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(12)
NO