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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
QTRS
          ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
QTRS
              ↳ DependencyPairsProof
              ↳ QTRS Reverse
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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(c(C(x))) → A1(A(x))
B(c(c(x))) → A1(x)
A1(x) → B(x)
B(c(C(x))) → A2(x)
B(c(c(x))) → A1(a(x))
A1(x) → B(b(x))

The TRS R consists of the following rules:

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

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

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

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

B(c(C(x))) → A1(A(x))
B(c(c(x))) → A1(x)
A1(x) → B(x)
B(c(C(x))) → A2(x)
B(c(c(x))) → A1(a(x))
A1(x) → B(b(x))

The TRS R consists of the following rules:

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

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

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

B(c(C(x))) → A1(A(x))
A1(x) → B(x)
B(c(c(x))) → A1(x)
B(c(c(x))) → A1(a(x))
A1(x) → B(b(x))

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
              ↳ QTRS Reverse
QTRS
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPToSRSProof
        ↳ QTRS
          ↳ QTRS Reverse
            ↳ QTRS
              ↳ DependencyPairsProof
              ↳ QTRS Reverse
              ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.