/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 20 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [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:

   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   *(+(y, z), x) -> +(*(x, y), *(x, z))
   *(*(x, y), z) -> *(x, *(y, z))
   +(+(x, y), z) -> +(x, +(y, z))

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:

   *^1(x, +(y, z)) -> +^1(*(x, y), *(x, z))
   *^1(x, +(y, z)) -> *^1(x, y)
   *^1(x, +(y, z)) -> *^1(x, z)
   *^1(+(y, z), x) -> +^1(*(x, y), *(x, z))
   *^1(+(y, z), x) -> *^1(x, y)
   *^1(+(y, z), x) -> *^1(x, z)
   *^1(*(x, y), z) -> *^1(x, *(y, z))
   *^1(*(x, y), z) -> *^1(y, z)
   +^1(+(x, y), z) -> +^1(x, +(y, z))
   +^1(+(x, y), z) -> +^1(y, z)

The TRS R consists of the following rules:

   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   *(+(y, z), x) -> +(*(x, y), *(x, z))
   *(*(x, y), z) -> *(x, *(y, z))
   +(+(x, y), z) -> +(x, +(y, z))

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 2 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:

   +^1(+(x, y), z) -> +^1(y, z)
   +^1(+(x, y), z) -> +^1(x, +(y, z))

The TRS R consists of the following rules:

   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   *(+(y, z), x) -> +(*(x, y), *(x, z))
   *(*(x, y), z) -> *(x, *(y, z))
   +(+(x, y), z) -> +(x, +(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(6) 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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(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   +^1(+(x, y), z) -> +^1(y, z)
   +^1(+(x, y), z) -> +^1(x, +(y, z))

The TRS R consists of the following rules:

   +(+(x, y), z) -> +(x, +(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(8) 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:
*+^1(+(x, y), z) -> +^1(y, z)
The graph contains the following edges 1 > 1, 2 >= 2


*+^1(+(x, y), z) -> +^1(x, +(y, z))
The graph contains the following edges 1 > 1


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(9)
YES

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

   *^1(x, +(y, z)) -> *^1(x, z)
   *^1(x, +(y, z)) -> *^1(x, y)
   *^1(+(y, z), x) -> *^1(x, y)
   *^1(+(y, z), x) -> *^1(x, z)
   *^1(*(x, y), z) -> *^1(x, *(y, z))
   *^1(*(x, y), z) -> *^1(y, z)

The TRS R consists of the following rules:

   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   *(+(y, z), x) -> +(*(x, y), *(x, z))
   *(*(x, y), z) -> *(x, *(y, z))
   +(+(x, y), z) -> +(x, +(y, z))

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


The following pairs can be oriented strictly and are deleted.

   *^1(x, +(y, z)) -> *^1(x, z)
   *^1(x, +(y, z)) -> *^1(x, y)
   *^1(+(y, z), x) -> *^1(x, y)
   *^1(+(y, z), x) -> *^1(x, z)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(*(x_1, x_2)) = x_1 + x_1*x_2 + x_2
   POL(*^1(x_1, x_2)) = x_1 + x_1*x_2 + x_2
   POL(+(x_1, x_2)) = 1 + x_1 + x_2

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

   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   *(+(y, z), x) -> +(*(x, y), *(x, z))
   *(*(x, y), z) -> *(x, *(y, z))
   +(+(x, y), z) -> +(x, +(y, z))


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(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   *^1(*(x, y), z) -> *^1(x, *(y, z))
   *^1(*(x, y), z) -> *^1(y, z)

The TRS R consists of the following rules:

   *(x, +(y, z)) -> +(*(x, y), *(x, z))
   *(+(y, z), x) -> +(*(x, y), *(x, z))
   *(*(x, y), z) -> *(x, *(y, z))
   +(+(x, y), z) -> +(x, +(y, z))

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:
**^1(*(x, y), z) -> *^1(x, *(y, z))
The graph contains the following edges 1 > 1


**^1(*(x, y), z) -> *^1(y, z)
The graph contains the following edges 1 > 1, 2 >= 2


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(14)
YES