YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 26 ms]
(4) QDP
(5) PisEmptyProof [EQUIVALENT, 4 ms]
(6) YES


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

   a(b(b(x1))) -> b(a(a(x1)))
   a(a(b(x1))) -> b(b(a(x1)))

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:

   A(b(b(x1))) -> A(a(x1))
   A(b(b(x1))) -> A(x1)
   A(a(b(x1))) -> A(x1)

The TRS R consists of the following rules:

   a(b(b(x1))) -> b(a(a(x1)))
   a(a(b(x1))) -> b(b(a(x1)))

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.

   A(b(b(x1))) -> A(a(x1))
   A(b(b(x1))) -> A(x1)
   A(a(b(x1))) -> A(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A_1(x_1) ) = x_1
POL( a_1(x_1) ) = 2x_1
POL( b_1(x_1) ) = 2x_1 + 1

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

   a(b(b(x1))) -> b(a(a(x1)))
   a(a(b(x1))) -> b(b(a(x1)))


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

   a(b(b(x1))) -> b(a(a(x1)))
   a(a(b(x1))) -> b(b(a(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(6)
YES