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(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
Q is empty.
We have reversed the following QTRS:
The set of rules R is
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
0(f(x)) → 0(s(x))
0(d(x)) → 0(x)
s(d(x)) → s(p(d(s(s(x)))))
s(f(x)) → s(p(f(d(x))))
s(p(x)) → x
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
0(f(x)) → 0(s(x))
0(d(x)) → 0(x)
s(d(x)) → s(p(d(s(s(x)))))
s(f(x)) → s(p(f(d(x))))
s(p(x)) → x
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
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(x1)) → s(0(x1))
Used ordering:
Polynomial interpretation [25]:
POL(0(x1)) = 2·x1
POL(d(x1)) = x1
POL(f(x1)) = 1 + x1
POL(p(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
The set Q consists of the following terms:
d(0(x0))
d(s(x0))
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(x1)) → D(f(p(s(x1))))
F(s(x1)) → F(p(s(x1)))
F(s(x1)) → P(s(x1))
D(s(x1)) → P(s(x1))
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
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.
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → D(f(p(s(x1))))
F(s(x1)) → F(p(s(x1)))
F(s(x1)) → P(s(x1))
D(s(x1)) → P(s(x1))
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
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.
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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
d(0(x0))
d(s(x0))
f(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
Q is empty.
We have to consider all (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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
d(0(x0))
d(s(x0))
f(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D(s(x1)) → D(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
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(x1)) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(D(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 1 + x1
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D(s(x1)) → D(p(s(x1)))
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.
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → F(p(s(x1)))
The TRS R consists of the following rules:
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(p(s(x1)))))
f(s(x1)) → d(f(p(s(x1))))
p(s(x1)) → x1
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.
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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → F(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
d(0(x0))
d(s(x0))
f(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → F(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → F(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
Q is empty.
We have to consider all (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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → F(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
d(0(x0))
d(s(x0))
f(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → F(p(s(x1)))
The TRS R consists of the following rules:
p(s(x1)) → x1
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(x1)) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(F(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 1 + x1
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x1)) → F(p(s(x1)))
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.