/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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, 23 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 43 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) TRUE


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

   a(p(x1)) -> p(a(A(x1)))
   a(A(x1)) -> A(a(x1))
   p(A(A(x1))) -> a(p(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:

   p(a(x1)) -> A(a(p(x1)))
   A(a(x1)) -> a(A(x1))
   A(A(p(x1))) -> p(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:

   P(a(x1)) -> A^1(a(p(x1)))
   P(a(x1)) -> P(x1)
   A^1(a(x1)) -> A^1(x1)
   A^1(A(p(x1))) -> P(a(x1))

The TRS R consists of the following rules:

   p(a(x1)) -> A(a(p(x1)))
   A(a(x1)) -> a(A(x1))
   A(A(p(x1))) -> p(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:

   P(a(x1)) -> P(x1)
   A^1(a(x1)) -> A^1(x1)
   A^1(A(p(x1))) -> P(a(x1))

Strictly oriented rules of the TRS R:

   A(A(p(x1))) -> p(a(x1))

Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2 + x_1
   POL(A^1(x_1)) = 2 + x_1
   POL(P(x_1)) = 3*x_1
   POL(a(x_1)) = 1 + x_1
   POL(p(x_1)) = 3*x_1


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

   P(a(x1)) -> A^1(a(p(x1)))

The TRS R consists of the following rules:

   p(a(x1)) -> A(a(p(x1)))
   A(a(x1)) -> a(A(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(8)
TRUE