Termination w.r.t. Q of the following Term Rewriting System could be proven:
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.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS Reverse
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.
The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
f(0) → 0
Used ordering:
Polynomial interpretation [25]:
POL(0) = 1
POL(f(x1)) = 1 + x1
POL(p(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS Reverse
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.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
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))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
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.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QTRS Reverse
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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ UsableRulesProof
↳ QTRS Reverse
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
f(s(x0))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QTRS Reverse
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.
By using the rule removal processor [15] with the following polynomial ordering [25], 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: POLO with Polynomial interpretation [25]:
POL(F(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QTRS Reverse
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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QTRS Reverse
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
f(s(x0))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QTRS Reverse
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.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QTRS Reverse
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.
We have reversed the following QTRS:
The set of rules R is
f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x
The set Q is empty.
We have obtained the following QTRS:
s(f(x)) → s(p(f(s(s(x)))))
0'(f(x)) → 0'(x)
s(p(x)) → x
The set Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS Reverse
↳ QTRS
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.