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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 70 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:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   b(c(a(x1))) -> a(b(c(x1)))
   c(b(x1)) -> b(a(x1))
   a(c(b(x1))) -> c(b(a(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:

   a(a(b(x1))) -> c(b(a(x1)))
   a(c(x1)) -> c(a(x1))
   a(c(b(x1))) -> c(b(a(x1)))
   b(c(x1)) -> a(b(x1))
   b(c(a(x1))) -> a(b(c(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:

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

The TRS R consists of the following rules:

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

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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2 + 3*x_1
   POL(B(x_1)) = 3 + 2*x_1
   POL(a(x_1)) = 2 + 2*x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = 2 + 2*x_1


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

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

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