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.