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(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → 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(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → 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(x1) → B(b(x1))
B(b(c(x1))) → C(a(x1))
A(x1) → B(x1)
B(b(c(x1))) → A(x1)
B(b(c(x1))) → C(b(c(a(x1))))
B(b(c(x1))) → B(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → b(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → 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:
A(x1) → B(b(x1))
B(b(c(x1))) → C(a(x1))
A(x1) → B(x1)
B(b(c(x1))) → A(x1)
B(b(c(x1))) → C(b(c(a(x1))))
B(b(c(x1))) → B(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → b(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → 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:
A(x1) → B(b(x1))
A(x1) → B(x1)
B(b(c(x1))) → A(x1)
B(b(c(x1))) → B(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → b(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(c(x1))) → B(c(a(x1))) at position [0] we obtained the following new rules:
B(b(c(x0))) → B(c(b(b(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(x1) → B(b(x1))
B(b(c(x0))) → B(c(b(b(x0))))
A(x1) → B(x1)
B(b(c(x1))) → A(x1)
The TRS R consists of the following rules:
a(x1) → b(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → 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(x1) → b(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → x1
A(x1) → B(b(x1))
B(b(c(x0))) → B(c(b(b(x0))))
A(x1) → B(x1)
B(b(c(x1))) → A(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → x1
A(x1) → B(b(x1))
B(b(c(x0))) → B(c(b(b(x0))))
A(x1) → B(x1)
B(b(c(x1))) → A(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → A(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:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → 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(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
C(b(B(x))) → A2(x)
C(b(B(x))) → C(B(x))
C(b(b(x))) → A1(c(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → A(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
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
C(b(B(x))) → A2(x)
C(b(B(x))) → C(B(x))
C(b(b(x))) → A1(c(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → 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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
The TRS R consists of the following rules:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(b(x))) → C(b(c(x))) at position [0,0] we obtained the following new rules:
C(b(b(c(x0)))) → C(b(x0))
C(b(b(b(B(x0))))) → C(b(b(b(c(B(x0))))))
C(b(b(b(b(x0))))) → C(b(a(c(b(c(x0))))))
C(b(b(b(B(x0))))) → C(b(A(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(b(c(x0)))) → C(b(x0))
C(b(b(b(B(x0))))) → C(b(b(b(c(B(x0))))))
C(b(b(x))) → C(x)
C(b(b(b(b(x0))))) → C(b(a(c(b(c(x0))))))
C(b(b(b(B(x0))))) → C(b(A(x0)))
The TRS R consists of the following rules:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(b(b(B(x0))))) → C(b(A(x0))) at position [0,0] we obtained the following new rules:
C(b(b(b(B(x0))))) → C(b(B(x0)))
C(b(b(b(B(x0))))) → C(b(b(B(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(b(c(x0)))) → C(b(x0))
C(b(b(x))) → C(x)
C(b(b(b(B(x0))))) → C(b(b(b(c(B(x0))))))
C(b(b(b(B(x0))))) → C(b(b(B(x0))))
C(b(b(b(B(x0))))) → C(b(B(x0)))
C(b(b(b(b(x0))))) → C(b(a(c(b(c(x0))))))
The TRS R consists of the following rules:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → 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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(b(b(c(x0)))) → C(b(x0))
C(b(b(b(B(x0))))) → C(b(b(b(c(B(x0))))))
C(b(b(x))) → C(x)
C(b(b(b(B(x0))))) → C(b(b(B(x0))))
C(b(b(b(b(x0))))) → C(b(a(c(b(c(x0))))))
The TRS R consists of the following rules:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → 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(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(b(x))
b(b(c(x))) → c(b(c(a(x))))
c(c(x)) → x
A(x) → B(b(x))
B(b(c(x))) → B(c(b(b(x))))
A(x) → B(x)
B(b(c(x))) → A(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(x) → b(b(x))
b(b(c(x))) → c(b(c(a(x))))
c(c(x)) → x
A(x) → B(b(x))
B(b(c(x))) → B(c(b(b(x))))
A(x) → B(x)
B(b(c(x))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
A(x) → b(B(x))
c(b(B(x))) → b(b(c(B(x))))
A(x) → B(x)
c(b(B(x))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(b(x))
b(b(c(x))) → c(b(c(a(x))))
c(c(x)) → x
A(x) → B(b(x))
B(b(c(x))) → B(c(b(b(x))))
A(x) → B(x)
B(b(c(x))) → A(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(x) → b(b(x))
b(b(c(x))) → c(b(c(a(x))))
c(c(x)) → x
A(x) → B(b(x))
B(b(c(x))) → B(c(b(b(x))))
A(x) → B(x)
B(b(c(x))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → 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(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(x1))
b(b(c(x1))) → c(b(c(a(x1))))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → 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(b(x))
c(b(b(x))) → a(c(b(c(x))))
c(c(x)) → x
Q is empty.