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


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YES
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 0 ms]
(6) QDP
(7) PisEmptyProof [EQUIVALENT, 0 ms]
(8) YES


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

   f(f(x)) -> f(g(f(x), x))
   f(f(x)) -> f(h(f(x), f(x)))
   g(x, y) -> y
   h(x, x) -> g(x, 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:

   F(f(x)) -> F(g(f(x), x))
   F(f(x)) -> G(f(x), x)
   F(f(x)) -> F(h(f(x), f(x)))
   F(f(x)) -> H(f(x), f(x))
   H(x, x) -> G(x, 0)

The TRS R consists of the following rules:

   f(f(x)) -> f(g(f(x), x))
   f(f(x)) -> f(h(f(x), f(x)))
   g(x, y) -> y
   h(x, x) -> g(x, 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 3 less nodes.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(x)) -> F(h(f(x), f(x)))
   F(f(x)) -> F(g(f(x), x))

The TRS R consists of the following rules:

   f(f(x)) -> f(g(f(x), x))
   f(f(x)) -> f(h(f(x), f(x)))
   g(x, y) -> y
   h(x, x) -> g(x, 0)

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.

   F(f(x)) -> F(h(f(x), f(x)))
   F(f(x)) -> F(g(f(x), x))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
F(x1)  =  x1

f(x1)  =  f(x1)

h(x1, x2)  =  h

g(x1, x2)  =  x2

0  =  0


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   0=1
   h=2
   f_1=2

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

   h(x, x) -> g(x, 0)
   g(x, y) -> y


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

   f(f(x)) -> f(g(f(x), x))
   f(f(x)) -> f(h(f(x), f(x)))
   g(x, y) -> y
   h(x, x) -> g(x, 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(8)
YES