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(x1) → b(x1)
a(c(x1)) → c(c(a(b(x1))))
b(b(x1)) → a(x1)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → c(c(a(b(x1))))
b(b(x1)) → a(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:
B(b(x1)) → A(x1)
A(c(x1)) → A(b(x1))
A(c(x1)) → B(x1)
A(x1) → B(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → c(c(a(b(x1))))
b(b(x1)) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(x1)) → A(x1)
A(c(x1)) → A(b(x1))
A(c(x1)) → B(x1)
A(x1) → B(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → c(c(a(b(x1))))
b(b(x1)) → a(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
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
a(c(x1)) → c(c(a(b(x1))))
b(b(x1)) → a(x1)
B(b(x1)) → A(x1)
A(c(x1)) → A(b(x1))
A(c(x1)) → B(x1)
A(x1) → B(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(c(x1)) → c(c(a(b(x1))))
b(b(x1)) → a(x1)
B(b(x1)) → A(x1)
A(c(x1)) → A(b(x1))
A(c(x1)) → B(x1)
A(x1) → B(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(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(a(x)) → C(c(x))
B1(B(x)) → A2(x)
C(a(x)) → C(x)
A1(x) → B1(x)
C(a(x)) → A1(c(c(x)))
B1(b(x)) → A1(x)
C(a(x)) → B1(a(c(c(x))))
C(A(x)) → B1(A(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(c(x))
B1(B(x)) → A2(x)
C(a(x)) → C(x)
A1(x) → B1(x)
C(a(x)) → A1(c(c(x)))
B1(b(x)) → A1(x)
C(a(x)) → B1(a(c(c(x))))
C(A(x)) → B1(A(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(x)
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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B1(x)
B1(b(x)) → A1(x)
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(x)
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
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B1(x)
B1(b(x)) → A1(x)
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:
B1(b(x)) → A1(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A1(x1)) = 2 + x1
POL(B1(x1)) = 2 + x1
POL(b(x1)) = 2 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B1(x)
R is empty.
Q is empty.
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.
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
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B1(x)
B1(b(x)) → A1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(c(x))
C(a(x)) → C(x)
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(x)) → C(c(x)) at position [0] we obtained the following new rules:
C(a(A(x0))) → C(B(x0))
C(a(A(x0))) → C(b(A(x0)))
C(a(a(x0))) → C(b(a(c(c(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(A(x0))) → C(B(x0))
C(a(x)) → C(x)
C(a(a(x0))) → C(b(a(c(c(x0)))))
C(a(A(x0))) → C(b(A(x0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(x)
C(a(a(x0))) → C(b(a(c(c(x0)))))
C(a(A(x0))) → C(b(A(x0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(A(x0))) → C(b(A(x0))) at position [0] we obtained the following new rules:
C(a(A(x0))) → C(b(B(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x)) → C(x)
C(a(a(x0))) → C(b(a(c(c(x0)))))
C(a(A(x0))) → C(b(B(x0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(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(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(c(x)) → c(c(a(b(x))))
b(b(x)) → a(x)
B(b(x)) → A(x)
A(c(x)) → A(b(x))
A(c(x)) → B(x)
A(x) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
a(c(x)) → c(c(a(b(x))))
b(b(x)) → a(x)
B(b(x)) → A(x)
A(c(x)) → A(b(x))
A(c(x)) → B(x)
A(x) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
b(B(x)) → A(x)
c(A(x)) → b(A(x))
c(A(x)) → B(x)
A(x) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(c(x)) → c(c(a(b(x))))
b(b(x)) → a(x)
B(b(x)) → A(x)
A(c(x)) → A(b(x))
A(c(x)) → B(x)
A(x) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
a(c(x)) → c(c(a(b(x))))
b(b(x)) → a(x)
B(b(x)) → A(x)
A(c(x)) → A(b(x))
A(c(x)) → B(x)
A(x) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(c(x1)) → c(c(a(b(x1))))
b(b(x1)) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(c(x1)) → c(c(a(b(x1))))
b(b(x1)) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(a(x)) → b(a(c(c(x))))
b(b(x)) → a(x)
Q is empty.