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
  ↳ 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.
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.