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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

a(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → a(b(c(c(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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

a(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → a(b(c(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPToSRSProof
QTRS
              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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(x))
b(b(x)) → x
b(c(c(x))) → c(c(b(a(x))))
b(c(C(x))) → c(C(x))
b(c(C(x))) → c(c(b(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(x)
A1(x) → B(x)
B(c(C(x))) → A2(x)
B(c(C(x))) → B(A(x))
B(c(c(x))) → B(a(x))

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(c(b(a(x))))
b(c(C(x))) → c(C(x))
b(c(C(x))) → c(c(b(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
QDP
                          ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(c(b(a(x))))
b(c(C(x))) → c(C(x))
b(c(C(x))) → c(c(b(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.
By narrowing [15] the rule B(c(C(x))) → B(A(x)) at position [0] we obtained the following new rules:

B(c(C(x0))) → B(C(x0))



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

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

B(c(C(x0))) → B(C(x0))
B(c(c(x))) → A1(x)
A1(x) → B(x)
B(c(c(x))) → B(a(x))

The TRS R consists of the following rules:

a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(c(b(a(x))))
b(c(C(x))) → c(C(x))
b(c(C(x))) → c(c(b(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
QDP
                                  ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(x) → B(x)
B(c(c(x))) → A1(x)
B(c(c(x))) → B(a(x))

The TRS R consists of the following rules:

a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(c(b(a(x))))
b(c(C(x))) → c(C(x))
b(c(C(x))) → c(c(b(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.
By narrowing [15] the rule B(c(c(x))) → B(a(x)) at position [0] we obtained the following new rules:

B(c(c(x0))) → B(c(b(x0)))



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

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

B(c(c(x0))) → B(c(b(x0)))
B(c(c(x))) → A1(x)
A1(x) → B(x)

The TRS R consists of the following rules:

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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(c(x))
b(b(x)) → x
c(c(b(x))) → a(b(c(c(x))))
C(c(b(x))) → C(c(x))
C(c(b(x))) → A(b(c(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(x))
b(b(x)) → x
b(c(c(x))) → c(c(b(a(x))))
b(c(C(x))) → c(C(x))
b(c(C(x))) → c(c(b(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(c(x))
b(b(x)) → x
c(c(b(x))) → a(b(c(c(x))))
C(c(b(x))) → C(c(x))
C(c(b(x))) → A(b(c(c(x))))
A(x) → C(x)
C(c(b(x))) → C(x)

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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(c(x))
b(b(x)) → x
c(c(b(x))) → a(b(c(c(x))))
C(c(b(x))) → C(c(x))
C(c(b(x))) → A(b(c(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(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → a(b(c(c(x1))))

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

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

Q is empty.

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

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

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

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

Q is empty.