Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(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:
C(b(x1)) → B(c(c(x1)))
C(b(x1)) → A(b(c(c(x1))))
B(c(x1)) → A(x1)
C(b(x1)) → C(c(x1))
C(b(x1)) → C(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → B(c(c(x1)))
C(b(x1)) → A(b(c(c(x1))))
B(c(x1)) → A(x1)
C(b(x1)) → C(c(x1))
C(b(x1)) → C(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(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 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(x1)) → A(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(x1)) → A(x1)
A(a(x1)) → B(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(x1)) → A(x1)
A(a(x1)) → B(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(c(x1)) → A(x1)
A(a(x1)) → B(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(B(x1)) = x1
POL(a(x1)) = 2·x1
POL(c(x1)) = 1 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QTRS Reverse
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → C(c(x1))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(x1)) → C(c(x1)) at position [0] we obtained the following new rules:
C(b(b(x0))) → C(a(b(c(c(x0)))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x0))) → C(a(b(c(c(x0)))))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
C(b(b(x0))) → C(a(b(c(c(x0)))))
C(b(x1)) → C(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
C(b(b(x0))) → C(a(b(c(c(x0)))))
C(b(x1)) → C(x1)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(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:
B(c(x)) → A(x)
B(b(C(x))) → B(a(C(x)))
B(b(C(x))) → A(C(x))
B(c(x)) → C1(c(b(a(x))))
B(c(x)) → C1(b(a(x)))
B(b(C(x))) → C1(b(a(C(x))))
B(b(C(x))) → C1(c(b(a(C(x)))))
B(c(x)) → B(a(x))
C1(b(x)) → A(x)
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(x)) → A(x)
B(b(C(x))) → B(a(C(x)))
B(b(C(x))) → A(C(x))
B(c(x)) → C1(c(b(a(x))))
B(c(x)) → C1(b(a(x)))
B(b(C(x))) → C1(b(a(C(x))))
B(b(C(x))) → C1(c(b(a(C(x)))))
B(c(x)) → B(a(x))
C1(b(x)) → A(x)
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(x)) → A(x)
B(c(x)) → C1(c(b(a(x))))
B(c(x)) → C1(b(a(x)))
B(b(C(x))) → C1(b(a(C(x))))
B(b(C(x))) → C1(c(b(a(C(x)))))
B(c(x)) → B(a(x))
C1(b(x)) → A(x)
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(x)) → C1(c(b(a(x)))) at position [0] we obtained the following new rules:
B(c(a(x0))) → C1(c(b(b(x0))))
B(c(y0)) → C1(a(a(y0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(x)) → A(x)
B(c(a(x0))) → C1(c(b(b(x0))))
B(c(y0)) → C1(a(a(y0)))
B(c(x)) → C1(b(a(x)))
B(b(C(x))) → C1(b(a(C(x))))
B(c(x)) → B(a(x))
B(b(C(x))) → C1(c(b(a(C(x)))))
C1(b(x)) → A(x)
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(x)) → B(a(x)) at position [0] we obtained the following new rules:
B(c(a(x0))) → B(b(x0))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(a(x0))) → B(b(x0))
B(c(x)) → A(x)
B(c(a(x0))) → C1(c(b(b(x0))))
B(c(y0)) → C1(a(a(y0)))
B(c(x)) → C1(b(a(x)))
B(b(C(x))) → C1(b(a(C(x))))
B(b(C(x))) → C1(c(b(a(C(x)))))
C1(b(x)) → A(x)
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(C(x))) → C1(c(b(a(C(x))))) at position [0] we obtained the following new rules:
B(b(C(y0))) → C1(a(a(C(y0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(x)) → A(x)
B(c(a(x0))) → B(b(x0))
B(c(a(x0))) → C1(c(b(b(x0))))
B(b(C(y0))) → C1(a(a(C(y0))))
B(c(y0)) → C1(a(a(y0)))
B(c(x)) → C1(b(a(x)))
B(b(C(x))) → C1(b(a(C(x))))
C1(b(x)) → A(x)
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
b(c(x)) → a(x)
c(b(x)) → a(b(c(c(x))))
C(b(b(x))) → C(a(b(c(c(x)))))
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(c(x)) → a(x)
c(b(x)) → a(b(c(c(x))))
C(b(b(x))) → C(a(b(c(c(x)))))
C(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
b(b(C(x))) → c(c(b(a(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
b(c(x)) → a(x)
c(b(x)) → a(b(c(c(x))))
C(b(b(x))) → C(a(b(c(c(x)))))
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(c(x)) → a(x)
c(b(x)) → a(b(c(c(x))))
C(b(b(x))) → C(a(b(c(c(x)))))
C(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(x1)
b(c(x1)) → a(x1)
c(b(x1)) → a(b(c(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(x)
c(b(x)) → a(x)
b(c(x)) → c(c(b(a(x))))
Q is empty.