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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(x1) → B(c(x1)) at position [0] we obtained the following new rules:

A(c(c(x0))) → B(b(x0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(b(x1)) → A(x1)
C(c(c(x1))) → B(x1)
A(x1) → C(x1)
A(c(c(x0))) → B(b(x0))
B(b(x1)) → A(a(x1))

The TRS R consists of the following rules:

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

The set Q is empty.
We have obtained the following QTRS:

a(x) → x
a(x) → c(b(x))
b(b(x)) → a(a(x))
c(c(c(x))) → b(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
a(x) → c(b(x))
b(b(x)) → a(a(x))
c(c(c(x))) → b(x)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

a(x) → x
a(x) → c(b(x))
b(b(x)) → a(a(x))
c(c(c(x))) → b(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
a(x) → c(b(x))
b(b(x)) → a(a(x))
c(c(c(x))) → b(x)

Q is empty.