YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 17 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:

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

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

The TRS R consists of the following rules:

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

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


The following pairs can be oriented strictly and are deleted.

   P(a(x0), p(a(a(a(x1))), x2)) -> P(a(a(b(x0))), x2)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
P(x1, x2)  =  x2

p(x1, x2)  =  p(x2)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   p_1=1

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

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


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

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

The TRS R consists of the following rules:

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

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.

   P(a(x0), p(a(a(a(x1))), x2)) -> P(a(x2), p(a(a(b(x0))), x2))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
P(x1, x2)  =  x2

a(x1)  =  a(x1)

p(x1, x2)  =  p(x1, x2)

b(x1)  =  b


Recursive path order with status [RPO].
Quasi-Precedence: p_2 > a_1 > b

Status: a_1: multiset status
p_2: [2,1]
b: multiset status


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

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


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

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

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