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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

Q is empty.