(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
s(f(x)) → s(p(f(s(s(x)))))
0'(f(x)) → 0'(x)
s(p(x)) → x
Q is empty.
(3) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
s(f(x)) → s(p(f(s(s(x)))))
0'(f(x)) → 0'(x)
s(p(x)) → x
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(f(x1)) = 1 + x1
POL(p(x1)) = x1
POL(s(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(0) → 0
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x
Q is empty.
(7) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(8) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x
The set Q consists of the following terms:
f(s(x0))
p(s(x0))
(9) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(p(s(x)))
F(s(x)) → P(s(x))
The TRS R consists of the following rules:
f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x
The set Q consists of the following terms:
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(11) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(p(s(x)))
The TRS R consists of the following rules:
f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x
The set Q consists of the following terms:
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(13) 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.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(p(s(x)))
The TRS R consists of the following rules:
p(s(x)) → x
The set Q consists of the following terms:
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(15) 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].
f(s(x0))
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(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.
(17) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(p(s(x)))
The TRS R consists of the following rules:
p(s(x)) → x
Q is empty.
We have to consider all (P,Q,R)-chains.
(19) 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.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(p(s(x)))
The TRS R consists of the following rules:
p(s(x)) → x
The set Q consists of the following terms:
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(21) 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].
f(s(x0))
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(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.
(23) 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(x)) → x
Used ordering: Polynomial interpretation [POLO]:
POL(F(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 1 + x1
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(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.
(25) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(26) TRUE