/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 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) MRRProof [EQUIVALENT, 0 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) TRUE


----------------------------------------

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

   p(s(x)) -> x
   fac(0) -> s(0)
   fac(s(x)) -> times(s(x), fac(p(s(x))))

Q is empty.

----------------------------------------

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   p(s(x)) -> x
   fac(0) -> s(0)
   fac(s(x)) -> times(s(x), fac(p(s(x))))

The set Q consists of the following terms:

   p(s(x0))
   fac(0)
   fac(s(x0))


----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FAC(s(x)) -> FAC(p(s(x)))
   FAC(s(x)) -> P(s(x))

The TRS R consists of the following rules:

   p(s(x)) -> x
   fac(0) -> s(0)
   fac(s(x)) -> times(s(x), fac(p(s(x))))

The set Q consists of the following terms:

   p(s(x0))
   fac(0)
   fac(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

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

   FAC(s(x)) -> FAC(p(s(x)))

The TRS R consists of the following rules:

   p(s(x)) -> x
   fac(0) -> s(0)
   fac(s(x)) -> times(s(x), fac(p(s(x))))

The set Q consists of the following terms:

   p(s(x0))
   fac(0)
   fac(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(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:

   FAC(s(x)) -> FAC(p(s(x)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   p(s(x0))
   fac(0)
   fac(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(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].

   fac(0)
   fac(s(x0))


----------------------------------------

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

   FAC(s(x)) -> FAC(p(s(x)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) 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 rules of the TRS R:

   p(s(x)) -> x

Used ordering: Polynomial interpretation [POLO]:

   POL(FAC(x_1)) = 2*x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 2 + x_1


----------------------------------------

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

   FAC(s(x)) -> FAC(p(s(x)))

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

   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(13) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(14)
TRUE