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(b(x1))
c(b(x1)) → a(a(c(c(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(b(x1))
c(b(x1)) → a(a(c(c(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:
C(b(x1)) → A(a(c(c(x1))))
C(b(x1)) → A(c(c(x1)))
C(b(x1)) → C(c(x1))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(b(x1))
c(b(x1)) → a(a(c(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → A(a(c(c(x1))))
C(b(x1)) → A(c(c(x1)))
C(b(x1)) → C(c(x1))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(b(x1))
c(b(x1)) → a(a(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 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ 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(b(x1)) → x1
a(c(x1)) → b(b(x1))
c(b(x1)) → a(a(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(a(c(c(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x0))) → C(a(a(c(c(x0)))))
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(c(x1)) → b(b(x1))
c(b(x1)) → a(a(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(x1))
c(b(x1)) → a(a(c(c(x1))))
C(b(b(x0))) → C(a(a(c(c(x0)))))
C(b(x1)) → C(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(c(x1)) → b(b(x1))
c(b(x1)) → a(a(c(c(x1))))
C(b(b(x0))) → C(a(a(c(c(x0)))))
C(b(x1)) → C(x1)
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
b(b(C(x))) → c(c(a(a(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
b(b(C(x))) → c(c(a(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)) → C1(c(a(a(x))))
C1(a(x)) → B(x)
B(b(C(x))) → C1(a(a(C(x))))
B(c(x)) → C1(a(a(x)))
B(b(C(x))) → C1(c(a(a(C(x)))))
C1(a(x)) → B(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
b(b(C(x))) → c(c(a(a(C(x)))))
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(c(x)) → C1(c(a(a(x))))
C1(a(x)) → B(x)
B(b(C(x))) → C1(a(a(C(x))))
B(c(x)) → C1(a(a(x)))
B(b(C(x))) → C1(c(a(a(C(x)))))
C1(a(x)) → B(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
b(b(C(x))) → c(c(a(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(a(a(x)))) at position [0] we obtained the following new rules:
B(c(y0)) → C1(b(b(a(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(y0)) → C1(b(b(a(y0))))
C1(a(x)) → B(x)
B(b(C(x))) → C1(a(a(C(x))))
B(c(x)) → C1(a(a(x)))
B(b(C(x))) → C1(c(a(a(C(x)))))
C1(a(x)) → B(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
b(b(C(x))) → c(c(a(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(a(a(C(x))))) at position [0] we obtained the following new rules:
B(b(C(y0))) → C1(b(b(a(C(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(y0)) → C1(b(b(a(y0))))
C1(a(x)) → B(x)
B(b(C(x))) → C1(a(a(C(x))))
B(c(x)) → C1(a(a(x)))
B(b(C(y0))) → C1(b(b(a(C(y0)))))
C1(a(x)) → B(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
b(b(C(x))) → c(c(a(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
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
b(b(C(x))) → c(c(a(a(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → x
a(c(x)) → b(b(x))
c(b(x)) → a(a(c(c(x))))
C(b(b(x))) → C(a(a(c(c(x)))))
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(x))
c(b(x)) → a(a(c(c(x))))
C(b(b(x))) → C(a(a(c(c(x)))))
C(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
b(b(C(x))) → c(c(a(a(C(x)))))
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → x
a(c(x)) → b(b(x))
c(b(x)) → a(a(c(c(x))))
C(b(b(x))) → C(a(a(c(c(x)))))
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(x))
c(b(x)) → a(a(c(c(x))))
C(b(b(x))) → C(a(a(c(c(x)))))
C(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(c(x1)) → b(b(x1))
c(b(x1)) → a(a(c(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(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)) → b(b(x))
b(c(x)) → c(c(a(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(b(x1))
c(b(x1)) → a(a(c(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(a(x)) → b(b(x))
b(c(x)) → c(c(a(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)) → b(b(x))
b(c(x)) → c(c(a(a(x))))
Q is empty.