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:

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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q restricted rewrite system:
The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

b(b(x1)) → c(c(c(c(x1))))
c(x1) → x1
b(c(b(x1))) → b(b(b(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 4 less nodes.

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

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

B(c(b(x1))) → B(b(b(x1)))
B(c(b(x1))) → B(b(x1))

The TRS R consists of the following rules:

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

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

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

b(b(x)) → c(c(c(c(x))))
c(x) → x
b(c(b(x))) → b(b(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:

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

Q is empty.

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

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

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

b(b(x)) → c(c(c(c(x))))
c(x) → x
b(c(b(x))) → b(b(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:

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

Q is empty.