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:
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
s(a(x)) → s(s(p(p(b(s(p(s(s(s(x))))))))))
s(b(x)) → s(p(s(p(c(s(s(p(p(s(s(s(x))))))))))))
s(c(x)) → s(p(s(p(a(s(p(s(p(x)))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
s(a(x)) → s(s(p(p(b(s(p(s(s(s(x))))))))))
s(b(x)) → s(p(s(p(c(s(s(p(p(s(s(s(x))))))))))))
s(c(x)) → s(p(s(p(a(s(p(s(p(x)))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
s(a(x)) → s(s(p(p(b(s(p(s(s(s(x))))))))))
s(b(x)) → s(p(s(p(c(s(s(p(p(s(s(s(x))))))))))))
s(c(x)) → s(p(s(p(a(s(p(s(p(x)))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
s(a(x)) → s(s(p(p(b(s(p(s(s(s(x))))))))))
s(b(x)) → s(p(s(p(c(s(s(p(p(s(s(s(x))))))))))))
s(c(x)) → s(p(s(p(a(s(p(s(p(x)))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
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:
P(p(s(x1))) → P(x1)
B(s(x1)) → P(s(p(s(x1))))
B(s(x1)) → P(s(s(c(p(s(p(s(x1))))))))
C(s(x1)) → P(s(x1))
A(s(x1)) → P(s(s(x1)))
B(s(x1)) → P(s(x1))
B(s(x1)) → P(p(s(s(c(p(s(p(s(x1)))))))))
C(s(x1)) → P(s(a(p(s(p(s(x1)))))))
A(s(x1)) → B(p(p(s(s(x1)))))
A(s(x1)) → P(p(s(s(x1))))
A(s(x1)) → P(s(b(p(p(s(s(x1)))))))
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
C(s(x1)) → P(s(p(s(x1))))
C(s(x1)) → P(s(p(s(a(p(s(p(s(x1)))))))))
The TRS R consists of the following rules:
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x1))) → P(x1)
B(s(x1)) → P(s(p(s(x1))))
B(s(x1)) → P(s(s(c(p(s(p(s(x1))))))))
C(s(x1)) → P(s(x1))
A(s(x1)) → P(s(s(x1)))
B(s(x1)) → P(s(x1))
B(s(x1)) → P(p(s(s(c(p(s(p(s(x1)))))))))
C(s(x1)) → P(s(a(p(s(p(s(x1)))))))
A(s(x1)) → B(p(p(s(s(x1)))))
A(s(x1)) → P(p(s(s(x1))))
A(s(x1)) → P(s(b(p(p(s(s(x1)))))))
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
C(s(x1)) → P(s(p(s(x1))))
C(s(x1)) → P(s(p(s(a(p(s(p(s(x1)))))))))
The TRS R consists of the following rules:
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → 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 11 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ 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:
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ 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.
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ 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.
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ 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.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))
The TRS R consists of the following rules:
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))
The TRS R consists of the following rules:
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
The set Q consists of the following terms:
a(s(x0))
b(s(x0))
c(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
a(s(x0))
b(s(x0))
c(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.
a(s(x0))
b(s(x0))
c(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(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 rewriting [15] the rule B(s(x1)) → C(p(s(p(s(x1))))) at position [0] we obtained the following new rules:
B(s(x1)) → C(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))
B(s(x1)) → C(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 rewriting [15] the rule C(s(x1)) → A(p(s(p(s(x1))))) at position [0] we obtained the following new rules:
C(s(x1)) → A(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
A(s(x1)) → B(p(p(s(s(x1)))))
C(s(x1)) → A(p(s(x1)))
B(s(x1)) → C(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 rewriting [15] the rule A(s(x1)) → B(p(p(s(s(x1))))) at position [0,0] we obtained the following new rules:
A(s(x1)) → B(p(s(x1)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
A(s(x1)) → B(p(s(x1)))
B(s(x1)) → C(p(s(x1)))
C(s(x1)) → A(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 rewriting [15] the rule B(s(x1)) → C(p(s(x1))) at position [0] we obtained the following new rules:
B(s(x1)) → C(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(x1)
A(s(x1)) → B(p(s(x1)))
C(s(x1)) → A(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 rewriting [15] the rule C(s(x1)) → A(p(s(x1))) at position [0] we obtained the following new rules:
C(s(x1)) → A(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(x1)
C(s(x1)) → A(x1)
A(s(x1)) → B(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 rewriting [15] the rule A(s(x1)) → B(p(s(x1))) at position [0] we obtained the following new rules:
A(s(x1)) → B(x1)
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(x1)
A(s(x1)) → B(x1)
C(s(x1)) → A(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.
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(x1)
A(s(x1)) → B(x1)
C(s(x1)) → A(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.
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))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(x1)
A(s(x1)) → B(x1)
C(s(x1)) → A(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:
B(s(x1)) → C(x1)
A(s(x1)) → B(x1)
C(s(x1)) → A(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 1 + x1
POL(B(x1)) = x1
POL(C(x1)) = 2 + 2·x1
POL(s(x1)) = 2 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ MNOCProof
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.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))
The TRS R consists of the following rules:
a(s(x1)) → s(s(s(p(s(b(p(p(s(s(x1))))))))))
b(s(x1)) → s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
c(s(x1)) → p(s(p(s(a(p(s(p(s(x1)))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
The set Q consists of the following terms:
a(s(x0))
b(s(x0))
c(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(s(x1)))))
The TRS R consists of the following rules:
p(s(x1)) → x1
The set Q consists of the following terms:
a(s(x0))
b(s(x0))
c(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.
a(s(x0))
b(s(x0))
c(s(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ MNOCProof
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(s(x1)) → C(p(s(p(s(x1)))))
C(s(x1)) → A(p(s(p(s(x1)))))
A(s(x1)) → B(p(p(s(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.