/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 8 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 109 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 22 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:

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

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(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:

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


Used ordering: Polynomial interpretation [POLO]:

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


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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) 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:

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

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

   POL(B(x_1)) = 2*x_1
   POL(a(x_1)) = 2*x_1
   POL(b(x_1)) = 1 + 2*x_1


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(8)
Obligation:
Q DP problem:
P is empty.
R is empty.
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