(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0(x)) → s(0(x))
d(0(x)) → 0(x)
d(s(x)) → s(s(d(p(s(x)))))
f(s(x)) → d(f(p(s(x))))
p(s(x)) → x
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = x1
POL(d(x1)) = x1
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(x)) → s(0(x))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
d(0(x)) → 0(x)
d(s(x)) → s(s(d(p(s(x)))))
f(s(x)) → d(f(p(s(x))))
p(s(x)) → x
Q is empty.
(3) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
d(0(x)) → 0(x)
d(s(x)) → s(s(d(p(s(x)))))
f(s(x)) → d(f(p(s(x))))
p(s(x)) → x
The set Q consists of the following terms:
d(0(x0))
d(s(x0))
f(s(x0))
p(s(x0))
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(s(x)) → D(p(s(x)))
D(s(x)) → P(s(x))
F(s(x)) → D(f(p(s(x))))
F(s(x)) → F(p(s(x)))
F(s(x)) → P(s(x))
The TRS R consists of the following rules:
d(0(x)) → 0(x)
d(s(x)) → s(s(d(p(s(x)))))
f(s(x)) → d(f(p(s(x))))
p(s(x)) → x
The set Q consists of the following terms:
d(0(x0))
d(s(x0))
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(s(x)) → D(p(s(x)))
The TRS R consists of the following rules:
d(0(x)) → 0(x)
d(s(x)) → s(s(d(p(s(x)))))
f(s(x)) → d(f(p(s(x))))
p(s(x)) → x
The set Q consists of the following terms:
d(0(x0))
d(s(x0))
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) 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.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(s(x)) → D(p(s(x)))
The TRS R consists of the following rules:
p(s(x)) → x
The set Q consists of the following terms:
d(0(x0))
d(s(x0))
f(s(x0))
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(12) 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].
d(0(x0))
d(s(x0))
f(s(x0))
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(s(x)) → D(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.
(14) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
D(s(x)) → D(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(x)) → x
(15) Obligation:
Q DP problem:
P is empty.
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.
(16) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(17) YES
(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:
d(0(x)) → 0(x)
d(s(x)) → s(s(d(p(s(x)))))
f(s(x)) → d(f(p(s(x))))
p(s(x)) → x
The set Q consists of the following terms:
d(0(x0))
d(s(x0))
f(s(x0))
p(s(x0))
We have to consider all minimal (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:
d(0(x0))
d(s(x0))
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].
d(0(x0))
d(s(x0))
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) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(s(x)) → F(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(x)) → x
(24) Obligation:
Q DP problem:
P is empty.
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.
(25) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(26) YES