/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 19 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 40 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 16 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(14) YES


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

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

The TRS R consists of the following rules:

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

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


Used ordering: Polynomial interpretation [POLO]:

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


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

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

The TRS R consists of the following rules:

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

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

(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:

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

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

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


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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> C(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
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(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> C(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(13) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*C(b(x1)) -> C(x1)
The graph contains the following edges 1 > 1


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(14)
YES