(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
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:
fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
fac(s(x0))
p(s(0))
p(s(s(x0)))
(3) DependencyPairsProof (EQUIVALENT transformation)
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))
P(s(s(x))) → P(s(x))
The TRS R consists of the following rules:
fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
fac(s(x0))
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(x))
The TRS R consists of the following rules:
fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
fac(s(x0))
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(x))
R is empty.
The set Q consists of the following terms:
fac(s(x0))
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
fac(s(x0))
p(s(0))
p(s(s(x0)))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(x))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) QDPSizeChangeProof (EQUIVALENT transformation)
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:
- P(s(s(x))) → P(s(x))
The graph contains the following edges 1 > 1
(13) YES
(14) 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:
fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
fac(s(x0))
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
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.
(16) 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(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
fac(s(x0))
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(17) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
fac(s(x0))
(18) 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(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(19) MRRProof (EQUIVALENT transformation)
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(0)) → 0
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(FAC(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 2 + 2·x1
(20) 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(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(21) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
FAC(s(x)) → FAC(p(s(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( p(x1) ) = max{0, x1 - 1} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
p(s(s(x))) → s(p(s(x)))
(22) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(23) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(24) YES