/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 28 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) MRRProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES


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

   O(0) -> 0
   +(0, x) -> x
   +(x, 0) -> x
   +(O(x), O(y)) -> O(+(x, y))
   +(O(x), I(y)) -> I(+(x, y))
   +(I(x), O(y)) -> I(+(x, y))
   +(I(x), I(y)) -> O(+(+(x, y), I(0)))
   *(0, x) -> 0
   *(x, 0) -> 0
   *(O(x), y) -> O(*(x, y))
   *(I(x), y) -> +(O(*(x, y)), y)

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(O(x), O(y)) -> O^1(+(x, y))
   +^1(O(x), O(y)) -> +^1(x, y)
   +^1(O(x), I(y)) -> +^1(x, y)
   +^1(I(x), O(y)) -> +^1(x, y)
   +^1(I(x), I(y)) -> O^1(+(+(x, y), I(0)))
   +^1(I(x), I(y)) -> +^1(+(x, y), I(0))
   +^1(I(x), I(y)) -> +^1(x, y)
   *^1(O(x), y) -> O^1(*(x, y))
   *^1(O(x), y) -> *^1(x, y)
   *^1(I(x), y) -> +^1(O(*(x, y)), y)
   *^1(I(x), y) -> O^1(*(x, y))
   *^1(I(x), y) -> *^1(x, y)

The TRS R consists of the following rules:

   O(0) -> 0
   +(0, x) -> x
   +(x, 0) -> x
   +(O(x), O(y)) -> O(+(x, y))
   +(O(x), I(y)) -> I(+(x, y))
   +(I(x), O(y)) -> I(+(x, y))
   +(I(x), I(y)) -> O(+(+(x, y), I(0)))
   *(0, x) -> 0
   *(x, 0) -> 0
   *(O(x), y) -> O(*(x, y))
   *(I(x), y) -> +(O(*(x, y)), y)

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 5 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(O(x), I(y)) -> +^1(x, y)
   +^1(O(x), O(y)) -> +^1(x, y)
   +^1(I(x), O(y)) -> +^1(x, y)
   +^1(I(x), I(y)) -> +^1(+(x, y), I(0))
   +^1(I(x), I(y)) -> +^1(x, y)

The TRS R consists of the following rules:

   O(0) -> 0
   +(0, x) -> x
   +(x, 0) -> x
   +(O(x), O(y)) -> O(+(x, y))
   +(O(x), I(y)) -> I(+(x, y))
   +(I(x), O(y)) -> I(+(x, y))
   +(I(x), I(y)) -> O(+(+(x, y), I(0)))
   *(0, x) -> 0
   *(x, 0) -> 0
   *(O(x), y) -> O(*(x, y))
   *(I(x), y) -> +(O(*(x, y)), y)

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(O(x), I(y)) -> +^1(x, y)
   +^1(O(x), O(y)) -> +^1(x, y)
   +^1(I(x), O(y)) -> +^1(x, y)
   +^1(I(x), I(y)) -> +^1(+(x, y), I(0))
   +^1(I(x), I(y)) -> +^1(x, y)

The TRS R consists of the following rules:

   +(0, x) -> x
   +(x, 0) -> x
   +(O(x), O(y)) -> O(+(x, y))
   +(O(x), I(y)) -> I(+(x, y))
   +(I(x), O(y)) -> I(+(x, y))
   +(I(x), I(y)) -> O(+(+(x, y), I(0)))
   O(0) -> 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(8) 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:

   +^1(O(x), I(y)) -> +^1(x, y)
   +^1(O(x), O(y)) -> +^1(x, y)
   +^1(I(x), O(y)) -> +^1(x, y)
   +^1(I(x), I(y)) -> +^1(+(x, y), I(0))
   +^1(I(x), I(y)) -> +^1(x, y)

Strictly oriented rules of the TRS R:

   +(0, x) -> x
   +(x, 0) -> x
   +(O(x), O(y)) -> O(+(x, y))
   +(O(x), I(y)) -> I(+(x, y))
   +(I(x), O(y)) -> I(+(x, y))
   +(I(x), I(y)) -> O(+(+(x, y), I(0)))
   O(0) -> 0

Used ordering: Knuth-Bendix order [KBO] with precedence:0 > +^1_2 > +_2 > I_1 > O_1

and weight map:

   0=2
   O_1=1
   I_1=3
   +_2=0
   +^1_2=0

The variable weight is 1

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(9)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(10) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(11)
YES

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

   *^1(I(x), y) -> *^1(x, y)
   *^1(O(x), y) -> *^1(x, y)

The TRS R consists of the following rules:

   O(0) -> 0
   +(0, x) -> x
   +(x, 0) -> x
   +(O(x), O(y)) -> O(+(x, y))
   +(O(x), I(y)) -> I(+(x, y))
   +(I(x), O(y)) -> I(+(x, y))
   +(I(x), I(y)) -> O(+(+(x, y), I(0)))
   *(0, x) -> 0
   *(x, 0) -> 0
   *(O(x), y) -> O(*(x, y))
   *(I(x), y) -> +(O(*(x, y)), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(13) 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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(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   *^1(I(x), y) -> *^1(x, y)
   *^1(O(x), y) -> *^1(x, y)

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


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


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(16)
YES