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(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(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)) → B(c(c(a(x1))))
B(c(x1)) → B(x1)
B(c(x1)) → A(b(x1))
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(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)) → B(c(c(a(x1))))
B(c(x1)) → B(x1)
B(c(x1)) → A(b(x1))
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(x1)) → A(b(x1)) at position [0] we obtained the following new rules:
B(c(c(x0))) → A(a(b(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(c(x0))) → A(a(b(x0)))
A(c(x1)) → A(x1)
A(c(x1)) → B(c(c(a(x1))))
B(c(x1)) → B(x1)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))
B(c(c(x0))) → A(a(b(x0)))
A(c(x1)) → A(x1)
A(c(x1)) → B(c(c(a(x1))))
B(c(x1)) → B(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))
B(c(c(x0))) → A(a(b(x0)))
A(c(x1)) → A(x1)
A(c(x1)) → B(c(c(a(x1))))
B(c(x1)) → B(x1)
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ 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:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(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(B(x)))
C(A(x)) → C(B(x))
C(b(x)) → B1(a(x))
C(c(B(x))) → B1(a(A(x)))
C(a(x)) → C(b(x))
C(a(x)) → B1(x)
C(a(x)) → C(c(b(x)))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ 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(B(x)))
C(A(x)) → C(B(x))
C(b(x)) → B1(a(x))
C(c(B(x))) → B1(a(A(x)))
C(a(x)) → C(b(x))
C(a(x)) → B1(x)
C(a(x)) → C(c(b(x)))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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(c(B(x)))
C(a(x)) → C(b(x))
C(a(x)) → C(c(b(x)))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(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(b(x)) at position [0] we obtained the following new rules:
C(a(a(x0))) → C(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(B(x)))
C(a(a(x0))) → C(x0)
C(a(x)) → C(c(b(x)))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(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(B(x))) at position [0] we obtained the following new rules:
C(A(x0)) → C(B(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(x0)) → C(B(x0))
C(a(a(x0))) → C(x0)
C(a(x)) → C(c(b(x)))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(x0))) → C(x0)
C(a(x)) → C(c(b(x)))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(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(b(x))) at position [0] we obtained the following new rules:
C(a(x0)) → C(b(a(x0)))
C(a(a(x0))) → C(c(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x0)) → C(b(a(x0)))
C(a(a(x0))) → C(c(x0))
C(a(a(x0))) → C(x0)
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(x0)) → C(b(a(x0))) at position [0] we obtained the following new rules:
C(a(x0)) → C(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(x0)) → C(x0)
C(a(a(x0))) → C(c(x0))
C(a(a(x0))) → C(x0)
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(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
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(x)) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → x
a(c(x)) → b(c(c(a(x))))
b(c(x)) → a(b(x))
B(c(c(x))) → A(a(b(x)))
A(c(x)) → A(x)
A(c(x)) → B(c(c(a(x))))
B(c(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ 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(b(x)) → x
a(c(x)) → b(c(c(a(x))))
b(c(x)) → a(b(x))
B(c(c(x))) → A(a(b(x)))
A(c(x)) → A(x)
A(c(x)) → B(c(c(a(x))))
B(c(x)) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(A(x)) → a(c(c(B(x))))
c(B(x)) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → x
a(c(x)) → b(c(c(a(x))))
b(c(x)) → a(b(x))
B(c(c(x))) → A(a(b(x)))
A(c(x)) → A(x)
A(c(x)) → B(c(c(a(x))))
B(c(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ 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(b(x)) → x
a(c(x)) → b(c(c(a(x))))
b(c(x)) → a(b(x))
B(c(c(x))) → A(a(b(x)))
A(c(x)) → A(x)
A(c(x)) → B(c(c(a(x))))
B(c(x)) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(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:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(c(x1)) → b(c(c(a(x1))))
b(c(x1)) → a(b(x1))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(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:
b(a(x)) → x
c(a(x)) → a(c(c(b(x))))
c(b(x)) → b(a(x))
Q is empty.