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(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(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(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(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:
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(x1)) → A(a(x1)) at position [0] we obtained the following new rules:
A(c(b(b(x0)))) → A(c(x0))
A(c(x0)) → A(b(x0))
A(c(c(x0))) → A(c(b(a(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(x0)) → A(b(x0))
A(c(c(x0))) → A(c(b(a(a(x0)))))
A(c(x1)) → A(x1)
A(c(b(b(x0)))) → A(c(x0))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(a(x1))))
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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(c(x0))) → A(c(b(a(a(x0)))))
A(c(x1)) → A(x1)
A(c(b(b(x0)))) → A(c(x0))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(a(x1))))
A(c(c(x0))) → A(c(b(a(a(x0)))))
A(c(x1)) → A(x1)
A(c(b(b(x0)))) → A(c(x0))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(a(x1))))
A(c(c(x0))) → A(c(b(a(a(x0)))))
A(c(x1)) → A(x1)
A(c(b(b(x0)))) → A(c(x0))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(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:
B(b(a(x))) → C(x)
C(c(A(x))) → B(c(A(x)))
C(a(x)) → B(c(x))
C(a(x)) → C(x)
A1(x) → B(x)
C(c(A(x))) → A1(b(c(A(x))))
C(c(A(x))) → A1(a(b(c(A(x)))))
C(a(x)) → A1(b(c(x)))
C(a(x)) → A1(a(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x))) → C(x)
C(c(A(x))) → B(c(A(x)))
C(a(x)) → B(c(x))
C(a(x)) → C(x)
A1(x) → B(x)
C(c(A(x))) → A1(b(c(A(x))))
C(c(A(x))) → A1(a(b(c(A(x)))))
C(a(x)) → A1(b(c(x)))
C(a(x)) → A1(a(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(A(x))) → B(c(A(x))) at position [0] we obtained the following new rules:
C(c(A(x0))) → B(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x))) → C(x)
C(a(x)) → B(c(x))
A1(x) → B(x)
C(a(x)) → C(x)
C(c(A(x))) → A1(b(c(A(x))))
C(c(A(x))) → A1(a(b(c(A(x)))))
C(a(x)) → A1(b(c(x)))
C(a(x)) → A1(a(b(c(x))))
C(c(A(x0))) → B(A(x0))
The TRS R consists of the following rules:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(b(a(x))) → C(x)
C(a(x)) → B(c(x))
C(a(x)) → C(x)
A1(x) → B(x)
C(c(A(x))) → A1(b(c(A(x))))
C(c(A(x))) → A1(a(b(c(A(x)))))
C(a(x)) → A1(b(c(x)))
C(a(x)) → A1(a(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(x)) → B(c(x)) at position [0] we obtained the following new rules:
C(a(a(x0))) → B(a(a(b(c(x0)))))
C(a(c(A(x0)))) → B(a(a(b(c(A(x0))))))
C(a(A(x0))) → B(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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))) → B(a(a(b(c(x0)))))
B(b(a(x))) → C(x)
A1(x) → B(x)
C(a(x)) → C(x)
C(c(A(x))) → A1(b(c(A(x))))
C(a(A(x0))) → B(A(x0))
C(c(A(x))) → A1(a(b(c(A(x)))))
C(a(x)) → A1(b(c(x)))
C(a(c(A(x0)))) → B(a(a(b(c(A(x0))))))
C(a(x)) → A1(a(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(x0))) → B(a(a(b(c(x0)))))
B(b(a(x))) → C(x)
C(a(x)) → C(x)
A1(x) → B(x)
C(c(A(x))) → A1(b(c(A(x))))
C(c(A(x))) → A1(a(b(c(A(x)))))
C(a(x)) → A1(b(c(x)))
C(a(c(A(x0)))) → B(a(a(b(c(A(x0))))))
C(a(x)) → A1(a(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(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)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(b(b(x))) → c(x)
a(c(x)) → c(b(a(a(x))))
A(c(c(x))) → A(c(b(a(a(x)))))
A(c(x)) → A(x)
A(c(b(b(x)))) → A(c(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(b(b(x))) → c(x)
a(c(x)) → c(b(a(a(x))))
A(c(c(x))) → A(c(b(a(a(x)))))
A(c(x)) → A(x)
A(c(b(b(x)))) → A(c(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
c(c(A(x))) → a(a(b(c(A(x)))))
c(A(x)) → A(x)
b(b(c(A(x)))) → c(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(b(b(x))) → c(x)
a(c(x)) → c(b(a(a(x))))
A(c(c(x))) → A(c(b(a(a(x)))))
A(c(x)) → A(x)
A(c(b(b(x)))) → A(c(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(b(b(x))) → c(x)
a(c(x)) → c(b(a(a(x))))
A(c(c(x))) → A(c(b(a(a(x)))))
A(c(x)) → A(x)
A(c(b(b(x)))) → A(c(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(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)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(b(x1))) → c(x1)
a(c(x1)) → c(b(a(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(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)
b(b(a(x))) → c(x)
c(a(x)) → a(a(b(c(x))))
Q is empty.