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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

A(a(b(b(x0)))) → A(b(a(x0)))



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

Q is empty.