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(b(x1))) → b(b(b(c(x1))))
c(b(x1)) → a(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(b(x1))) → b(b(b(c(x1))))
c(b(x1)) → a(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:
C(b(x1)) → A(x1)
A(b(b(x1))) → C(x1)
C(b(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(c(x1))))
c(b(x1)) → a(a(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → A(x1)
A(b(b(x1))) → C(x1)
C(b(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(c(x1))))
c(b(x1)) → a(a(x1))
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(x1) → x1
a(b(b(x1))) → b(b(b(c(x1))))
c(b(x1)) → a(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(b(a(x))) → c(b(b(b(x))))
b(c(x)) → 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:
a(x) → x
b(b(a(x))) → c(b(b(b(x))))
b(c(x)) → a(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(b(x1))) → b(b(b(c(x1))))
c(b(x1)) → a(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(b(a(x))) → c(b(b(b(x))))
b(c(x)) → 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:
a(x) → x
b(b(a(x))) → c(b(b(b(x))))
b(c(x)) → a(a(x))
Q is empty.