/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 10 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (7) YES
    (8) QDP
        (9) QDPOrderProof [EQUIVALENT, 51 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(b(b(x1)))) -> b(a(a(b(x1))))
   b(a(a(a(x1)))) -> b(b(a(a(x1))))
   a(b(a(b(x1)))) -> b(b(a(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:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
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(4)
Complex Obligation (AND)

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(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(6) 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:
*B(a(a(a(x1)))) -> B(a(a(x1)))
The graph contains the following edges 1 > 1


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(7)
YES

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   A(b(b(b(x1)))) -> A(b(x1))
   A(b(b(b(x1)))) -> A(a(b(x1)))
   A(b(a(b(x1)))) -> A(a(x1))
   A(b(a(b(x1)))) -> A(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = 1 + x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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


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

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

Q is empty.
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