/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) MRRProof [EQUIVALENT, 88 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) TRUE


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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) 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:

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

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.
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:

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

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

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


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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
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(6)
TRUE