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) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → 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) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → 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(b(c(x1))) → A(b(b(x1)))
A(b(c(x1))) → B(b(x1))
A(b(c(x1))) → B(x1)
B(x1) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → 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(b(c(x1))) → A(b(b(x1)))
A(b(c(x1))) → B(b(x1))
A(b(c(x1))) → B(x1)
B(x1) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(c(x1))) → A(b(b(x1))) at position [0] we obtained the following new rules:
A(b(c(x0))) → A(b(a(x0)))
A(b(c(y0))) → A(a(b(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(c(x1))) → B(b(x1))
A(b(c(x1))) → B(x1)
A(b(c(y0))) → A(a(b(y0)))
A(b(c(x0))) → A(b(a(x0)))
B(x1) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → 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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → a(x1)
A(b(c(x1))) → B(b(x1))
A(b(c(x1))) → B(x1)
A(b(c(y0))) → A(a(b(y0)))
A(b(c(x0))) → A(b(a(x0)))
B(x1) → A(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → a(x1)
A(b(c(x1))) → B(b(x1))
A(b(c(x1))) → B(x1)
A(b(c(y0))) → A(a(b(y0)))
A(b(c(x0))) → A(b(a(x0)))
B(x1) → A(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → a(x)
c(b(A(x))) → b(B(x))
c(b(A(x))) → B(x)
c(b(A(x))) → b(a(A(x)))
c(b(A(x))) → a(b(A(x)))
B(x) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ 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) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → a(x)
c(b(A(x))) → b(B(x))
c(b(A(x))) → B(x)
c(b(A(x))) → b(a(A(x)))
c(b(A(x))) → a(b(A(x)))
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(a(x))) → B1(a(c(c(x))))
C(b(A(x))) → B1(B(x))
C(b(A(x))) → B1(a(A(x)))
C(b(A(x))) → B2(x)
C(b(a(x))) → C(x)
B1(x) → A1(x)
C(b(a(x))) → B1(b(a(c(c(x)))))
C(b(A(x))) → A1(A(x))
C(b(a(x))) → C(c(x))
C(b(a(x))) → A1(c(c(x)))
C(b(A(x))) → A1(b(A(x)))
The TRS R consists of the following rules:
a(x) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → a(x)
c(b(A(x))) → b(B(x))
c(b(A(x))) → B(x)
c(b(A(x))) → b(a(A(x)))
c(b(A(x))) → a(b(A(x)))
B(x) → A(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
Q DP problem:
The TRS P consists of the following rules:
C(b(a(x))) → B1(a(c(c(x))))
C(b(A(x))) → B1(B(x))
C(b(A(x))) → B1(a(A(x)))
C(b(A(x))) → B2(x)
C(b(a(x))) → C(x)
B1(x) → A1(x)
C(b(a(x))) → B1(b(a(c(c(x)))))
C(b(A(x))) → A1(A(x))
C(b(a(x))) → C(c(x))
C(b(a(x))) → A1(c(c(x)))
C(b(A(x))) → A1(b(A(x)))
The TRS R consists of the following rules:
a(x) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → a(x)
c(b(A(x))) → b(B(x))
c(b(A(x))) → B(x)
c(b(A(x))) → b(a(A(x)))
c(b(A(x))) → a(b(A(x)))
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 9 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(a(x))) → C(x)
C(b(a(x))) → C(c(x))
The TRS R consists of the following rules:
a(x) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → a(x)
c(b(A(x))) → b(B(x))
c(b(A(x))) → B(x)
c(b(A(x))) → b(a(A(x)))
c(b(A(x))) → a(b(A(x)))
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) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → a(x)
c(b(A(x))) → b(B(x))
c(b(A(x))) → B(x)
c(b(A(x))) → b(a(A(x)))
c(b(A(x))) → a(b(A(x)))
B(x) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(c(x))) → c(c(a(b(b(x)))))
b(x) → a(x)
A(b(c(x))) → B(b(x))
A(b(c(x))) → B(x)
A(b(c(x))) → A(a(b(x)))
A(b(c(x))) → A(b(a(x)))
B(x) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ 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) → x
a(b(c(x))) → c(c(a(b(b(x)))))
b(x) → a(x)
A(b(c(x))) → B(b(x))
A(b(c(x))) → B(x)
A(b(c(x))) → A(a(b(x)))
A(b(c(x))) → A(b(a(x)))
B(x) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → 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:
a(x) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → a(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(c(x1))) → c(c(a(b(b(x1)))))
b(x1) → a(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → 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:
a(x) → x
c(b(a(x))) → b(b(a(c(c(x)))))
b(x) → a(x)
Q is empty.