/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 45 ms]
(4) QTRS
(5) Overlay + Local Confluence [EQUIVALENT, 4 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 28 ms]
(8) QDP
(9) MRRProof [EQUIVALENT, 145 ms]
(10) QDP
(11) PisEmptyProof [EQUIVALENT, 0 ms]
(12) YES


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

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

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(A(x_1)) = 3 + x_1
   POL(B(x_1)) = 3 + x_1
   POL(C(x_1)) = 3 + x_1
   POL(a(x_1)) = 2 + x_1
   POL(b(x_1)) = 2 + x_1
   POL(c(x_1)) = 2 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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




----------------------------------------

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

Q is empty.

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(5) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   a(a(a(a(B(x0)))))
   b(A(A(A(x0))))
   b(b(b(b(C(x0)))))
   c(B(B(B(x0))))
   c(c(c(c(A(x0)))))
   a(C(C(C(x0))))


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(7) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A^1(a(a(a(B(x1))))) -> C^1(x1)
   B^1(A(A(A(x1)))) -> A^1(a(a(a(x1))))
   B^1(A(A(A(x1)))) -> A^1(a(a(x1)))
   B^1(A(A(A(x1)))) -> A^1(a(x1))
   B^1(A(A(A(x1)))) -> A^1(x1)
   B^1(b(b(b(C(x1))))) -> A^1(x1)
   C^1(B(B(B(x1)))) -> B^1(b(b(b(x1))))
   C^1(B(B(B(x1)))) -> B^1(b(b(x1)))
   C^1(B(B(B(x1)))) -> B^1(b(x1))
   C^1(B(B(B(x1)))) -> B^1(x1)
   C^1(c(c(c(A(x1))))) -> B^1(x1)
   A^1(C(C(C(x1)))) -> C^1(c(c(c(x1))))
   A^1(C(C(C(x1)))) -> C^1(c(c(x1)))
   A^1(C(C(C(x1)))) -> C^1(c(x1))
   A^1(C(C(C(x1)))) -> C^1(x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   a(a(a(a(B(x0)))))
   b(A(A(A(x0))))
   b(b(b(b(C(x0)))))
   c(B(B(B(x0))))
   c(c(c(c(A(x0)))))
   a(C(C(C(x0))))

We have to consider all minimal (P,Q,R)-chains.
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(9) 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^1(a(a(a(B(x1))))) -> C^1(x1)
   B^1(A(A(A(x1)))) -> A^1(a(a(a(x1))))
   B^1(A(A(A(x1)))) -> A^1(a(a(x1)))
   B^1(A(A(A(x1)))) -> A^1(a(x1))
   B^1(A(A(A(x1)))) -> A^1(x1)
   B^1(b(b(b(C(x1))))) -> A^1(x1)
   C^1(B(B(B(x1)))) -> B^1(b(b(b(x1))))
   C^1(B(B(B(x1)))) -> B^1(b(b(x1)))
   C^1(B(B(B(x1)))) -> B^1(b(x1))
   C^1(B(B(B(x1)))) -> B^1(x1)
   C^1(c(c(c(A(x1))))) -> B^1(x1)
   A^1(C(C(C(x1)))) -> C^1(c(c(c(x1))))
   A^1(C(C(C(x1)))) -> C^1(c(c(x1)))
   A^1(C(C(C(x1)))) -> C^1(c(x1))
   A^1(C(C(C(x1)))) -> C^1(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 3 + x_1
   POL(A^1(x_1)) = 2*x_1
   POL(B(x_1)) = 3 + x_1
   POL(B^1(x_1)) = 2*x_1
   POL(C(x_1)) = 3 + x_1
   POL(C^1(x_1)) = 2*x_1
   POL(a(x_1)) = 2 + x_1
   POL(b(x_1)) = 2 + x_1
   POL(c(x_1)) = 2 + x_1


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(10)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   a(a(a(a(B(x0)))))
   b(A(A(A(x0))))
   b(b(b(b(C(x0)))))
   c(B(B(B(x0))))
   c(c(c(c(A(x0)))))
   a(C(C(C(x0))))

We have to consider all minimal (P,Q,R)-chains.
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(11) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(12)
YES