/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 24 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 19 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

   p(a(x0), p(b(x1), p(a(x2), x3))) -> p(x2, p(a(a(x0)), p(b(x1), x3)))

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:

   P(a(x0), p(b(x1), p(a(x2), x3))) -> P(x2, p(a(a(x0)), p(b(x1), x3)))
   P(a(x0), p(b(x1), p(a(x2), x3))) -> P(a(a(x0)), p(b(x1), x3))
   P(a(x0), p(b(x1), p(a(x2), x3))) -> P(b(x1), x3)

The TRS R consists of the following rules:

   p(a(x0), p(b(x1), p(a(x2), x3))) -> p(x2, p(a(a(x0)), p(b(x1), x3)))

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:

   P(a(x0), p(b(x1), p(a(x2), x3))) -> P(a(a(x0)), p(b(x1), x3))
   P(a(x0), p(b(x1), p(a(x2), x3))) -> P(x2, p(a(a(x0)), p(b(x1), x3)))

The TRS R consists of the following rules:

   p(a(x0), p(b(x1), p(a(x2), x3))) -> p(x2, p(a(a(x0)), p(b(x1), x3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   P(a(x0), p(b(x1), p(a(x2), x3))) -> P(a(a(x0)), p(b(x1), x3))


Used ordering: Polynomial interpretation [POLO]:

   POL(P(x_1, x_2)) = 2*x_1 + x_2
   POL(a(x_1)) = x_1
   POL(b(x_1)) = 2*x_1
   POL(p(x_1, x_2)) = 2 + 2*x_1 + x_2


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

   P(a(x0), p(b(x1), p(a(x2), x3))) -> P(x2, p(a(a(x0)), p(b(x1), x3)))

The TRS R consists of the following rules:

   p(a(x0), p(b(x1), p(a(x2), x3))) -> p(x2, p(a(a(x0)), p(b(x1), x3)))

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.

   P(a(x0), p(b(x1), p(a(x2), x3))) -> P(x2, p(a(a(x0)), p(b(x1), x3)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(P(x_1, x_2)) = [4]x_1 + [4]x_2
   POL(a(x_1)) = [4] + [1/4]x_1
   POL(b(x_1)) = 0
   POL(p(x_1, x_2)) = [4]x_1 + [4]x_2
The value of delta used in the strict ordering is 192.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   p(a(x0), p(b(x1), p(a(x2), x3))) -> p(x2, p(a(a(x0)), p(b(x1), x3)))


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

   p(a(x0), p(b(x1), p(a(x2), x3))) -> p(x2, p(a(a(x0)), p(b(x1), x3)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(10)
YES