/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) TransformationProof [EQUIVALENT, 0 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 5 ms]
(8) QDP
(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(10) YES


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

   f(f(x, a), y) -> f(y, f(x, y))

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, a), y) -> F(y, f(x, y))
   F(f(x, a), y) -> F(x, y)

The TRS R consists of the following rules:

   f(f(x, a), y) -> f(y, f(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule F(f(x, a), y) -> F(y, f(x, y)) we obtained the following new rules [LPAR04]:

   (F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a))),F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a))))


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

   F(f(x, a), y) -> F(x, y)
   F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a)))

The TRS R consists of the following rules:

   f(f(x, a), y) -> f(y, f(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule F(f(x, a), y) -> F(x, y) we obtained the following new rules [LPAR04]:

   (F(f(f(y_0, a), a), x1) -> F(f(y_0, a), x1),F(f(f(y_0, a), a), x1) -> F(f(y_0, a), x1))
   (F(f(f(y_0, a), a), f(y_1, a)) -> F(f(y_0, a), f(y_1, a)),F(f(f(y_0, a), a), f(y_1, a)) -> F(f(y_0, a), f(y_1, a)))


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

   F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a)))
   F(f(f(y_0, a), a), x1) -> F(f(y_0, a), x1)
   F(f(f(y_0, a), a), f(y_1, a)) -> F(f(y_0, a), f(y_1, a))

The TRS R consists of the following rules:

   f(f(x, a), y) -> f(y, f(x, y))

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


The following pairs can be oriented strictly and are deleted.

   F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(F(x_1, x_2)) = [1/4]x_2
   POL(a) = [1/4]
   POL(f(x_1, x_2)) = [1/4]x_2
The value of delta used in the strict ordering is 3/256.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   f(f(x, a), y) -> f(y, f(x, y))


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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(f(y_0, a), a), x1) -> F(f(y_0, a), x1)
   F(f(f(y_0, a), a), f(y_1, a)) -> F(f(y_0, a), f(y_1, a))

The TRS R consists of the following rules:

   f(f(x, a), y) -> f(y, f(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*F(f(f(y_0, a), a), x1) -> F(f(y_0, a), x1)
The graph contains the following edges 1 > 1, 2 >= 2


*F(f(f(y_0, a), a), f(y_1, a)) -> F(f(y_0, a), f(y_1, a))
The graph contains the following edges 1 > 1, 2 >= 2


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(10)
YES