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:
q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q is empty.
We have obtained the following QTRS:
0(q(x)) → s(s(s(s(0(s(s(p(p(x)))))))))
s(q(x)) → s(s(p(p(r(s(s(s(s(s(s(p(p(x)))))))))))))
0(r(x)) → s(s(s(p(p(p(0(s(p(s(p(x)))))))))))
s(r(x)) → s(p(s(p(q(s(s(p(s(p(x))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(0(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
0(q(x)) → s(s(s(s(0(s(s(p(p(x)))))))))
s(q(x)) → s(s(p(p(r(s(s(s(s(s(s(p(p(x)))))))))))))
0(r(x)) → s(s(s(p(p(p(0(s(p(s(p(x)))))))))))
s(r(x)) → s(p(s(p(q(s(s(p(s(p(x))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(0(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q is empty.
We have obtained the following QTRS:
0(q(x)) → s(s(s(s(0(s(s(p(p(x)))))))))
s(q(x)) → s(s(p(p(r(s(s(s(s(s(s(p(p(x)))))))))))))
0(r(x)) → s(s(s(p(p(p(0(s(p(s(p(x)))))))))))
s(r(x)) → s(p(s(p(q(s(s(p(s(p(x))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(0(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
0(q(x)) → s(s(s(s(0(s(s(p(p(x)))))))))
s(q(x)) → s(s(p(p(r(s(s(s(s(s(s(p(p(x)))))))))))))
0(r(x)) → s(s(s(p(p(p(0(s(p(s(p(x)))))))))))
s(r(x)) → s(p(s(p(q(s(s(p(s(p(x))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(0(x))))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(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:
q(0(x1)) → p(p(s(s(0(s(s(s(s(x1)))))))))
r(0(x1)) → p(s(p(s(0(p(p(p(s(s(s(x1)))))))))))
Used ordering:
Polynomial interpretation [25]:
POL(0(x1)) = 1 + 2·x1
POL(p(x1)) = x1
POL(q(x1)) = 2·x1
POL(r(x1)) = 2·x1
POL(s(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → P(s(s(x1)))
P(p(s(x1))) → P(x1)
R(s(x1)) → P(s(p(s(s(q(p(s(p(s(x1))))))))))
Q(s(x1)) → P(p(s(s(x1))))
R(s(x1)) → P(s(x1))
R(s(x1)) → P(s(p(s(x1))))
Q(s(x1)) → P(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
R(s(x1)) → Q(p(s(p(s(x1)))))
R(s(x1)) → P(s(s(q(p(s(p(s(x1))))))))
Q(s(x1)) → R(p(p(s(s(x1)))))
Q(s(x1)) → P(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))
The TRS R consists of the following rules:
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → P(s(s(x1)))
P(p(s(x1))) → P(x1)
R(s(x1)) → P(s(p(s(s(q(p(s(p(s(x1))))))))))
Q(s(x1)) → P(p(s(s(x1))))
R(s(x1)) → P(s(x1))
R(s(x1)) → P(s(p(s(x1))))
Q(s(x1)) → P(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
R(s(x1)) → Q(p(s(p(s(x1)))))
R(s(x1)) → P(s(s(q(p(s(p(s(x1))))))))
Q(s(x1)) → R(p(p(s(s(x1)))))
Q(s(x1)) → P(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))
The TRS R consists of the following rules:
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 8 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x1))) → P(x1)
The TRS R consists of the following rules:
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x1))) → P(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x1))) → P(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
P(p(s(x1))) → P(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(P(x1)) = 2·x1
POL(p(x1)) = 2·x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
R(s(x1)) → Q(p(s(p(s(x1)))))
Q(s(x1)) → R(p(p(s(s(x1)))))
The TRS R consists of the following rules:
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))
The TRS R consists of the following rules:
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q consists of the following terms:
q(s(x0))
r(s(x0))
p(s(x0))
p(0(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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q consists of the following terms:
q(s(x0))
r(s(x0))
p(s(x0))
p(0(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.
q(s(x0))
r(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
R(s(x1)) → Q(p(s(p(s(x1)))))
Q(s(x1)) → R(p(p(s(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q consists of the following terms:
p(s(x0))
p(0(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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
Q is empty.
We have to consider all (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))
The TRS R consists of the following rules:
q(s(x1)) → p(p(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))))
r(s(x1)) → p(s(p(s(s(q(p(s(p(s(x1))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q consists of the following terms:
q(s(x0))
r(s(x0))
p(s(x0))
p(0(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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(p(p(s(s(x1)))))
R(s(x1)) → Q(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q consists of the following terms:
q(s(x0))
r(s(x0))
p(s(x0))
p(0(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.
q(s(x0))
r(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
R(s(x1)) → Q(p(s(p(s(x1)))))
Q(s(x1)) → R(p(p(s(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule Q(s(x1)) → R(p(p(s(s(x1))))) at position [0,0] we obtained the following new rules:
Q(s(x1)) → R(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(p(s(x1)))
R(s(x1)) → Q(p(s(p(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(x1))))
The set Q consists of the following terms:
p(s(x0))
p(0(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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(p(s(x1)))
R(s(x1)) → Q(p(s(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))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule R(s(x1)) → Q(p(s(p(s(x1))))) at position [0] we obtained the following new rules:
R(s(x1)) → Q(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
R(s(x1)) → Q(p(s(x1)))
Q(s(x1)) → R(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))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule Q(s(x1)) → R(p(s(x1))) at position [0] we obtained the following new rules:
Q(s(x1)) → R(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
R(s(x1)) → Q(p(s(x1)))
Q(s(x1)) → R(x1)
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
p(0(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule R(s(x1)) → Q(p(s(x1))) at position [0] we obtained the following new rules:
R(s(x1)) → Q(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(x1)
R(s(x1)) → Q(x1)
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
p(s(x0))
p(0(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
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(x1)
R(s(x1)) → Q(x1)
R is empty.
The set Q consists of the following terms:
p(s(x0))
p(0(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.
p(s(x0))
p(0(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
Q(s(x1)) → R(x1)
R(s(x1)) → Q(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
Q(s(x1)) → R(x1)
R(s(x1)) → Q(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(Q(x1)) = 2 + x1
POL(R(x1)) = 1 + 2·x1
POL(s(x1)) = 2 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.