/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) MRRProof [EQUIVALENT, 39 ms]
(4) QDP
(5) PisEmptyProof [EQUIVALENT, 0 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) 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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = 1 + x_1


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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