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))) → a(b(b(a(x1))))
b(b(b(x1))) → a(x1)

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a(x1) → x1
a(a(b(x1))) → a(b(b(a(x1))))
b(b(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(a(b(x1))) → A(x1)
A(a(b(x1))) → B(b(a(x1)))
B(b(b(x1))) → A(x1)
A(a(b(x1))) → A(b(b(a(x1))))
A(a(b(x1))) → B(a(x1))

The TRS R consists of the following rules:

a(x1) → x1
a(a(b(x1))) → a(b(b(a(x1))))
b(b(b(x1))) → a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP

Q DP problem:
The TRS P consists of the following rules:

A(a(b(x1))) → A(x1)
A(a(b(x1))) → B(b(a(x1)))
B(b(b(x1))) → A(x1)
A(a(b(x1))) → A(b(b(a(x1))))
A(a(b(x1))) → B(a(x1))

The TRS R consists of the following rules:

a(x1) → x1
a(a(b(x1))) → a(b(b(a(x1))))
b(b(b(x1))) → a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.