/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 11 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QReductionProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(12) YES


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

   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Q is empty.

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(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)


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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:

   PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
   F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
   F(s(x), s(y), z, u) -> F(x, u, z, u)

The TRS R consists of the following rules:

   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

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 1 SCC with 2 less nodes.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(x), s(y), z, u) -> F(x, u, z, u)
   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)

The TRS R consists of the following rules:

   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
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(7) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   F(s(x), s(y), z, u) -> F(x, u, z, u)
   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)

R is empty.
The set Q consists of the following terms:

   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
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(9) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)


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

   F(s(x), s(y), z, u) -> F(x, u, z, u)
   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(11) 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:
*F(s(x), s(y), z, u) -> F(x, u, z, u)
The graph contains the following edges 1 > 1, 4 >= 2, 3 >= 3, 4 >= 4


*F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
The graph contains the following edges 1 > 1, 4 >= 2, 4 >= 4


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(12)
YES