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(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(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(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(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:
A(c(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(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:
A(c(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(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(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(a(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(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(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(a(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(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(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
Q is empty.