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