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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

B(b(c(x0))) → B(c(b(b(x0))))



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

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

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

The TRS R consists of the following rules:

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

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

B(b(c(x0))) → B(b(x0))



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

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

B(b(c(x0))) → B(c(b(b(x0))))
B(b(c(x0))) → B(b(x0))
A(x1) → B(x1)
B(b(c(x1))) → A(x1)

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

a(x) → b(x)
b(b(c(x))) → c(b(c(b(a(x)))))
c(c(x)) → x
B(b(c(x))) → B(c(b(b(x))))
B(b(c(x))) → B(b(x))
A(x) → B(x)
B(b(c(x))) → A(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:

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(x) → b(x)
c(b(b(x))) → a(b(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → b(b(c(B(x))))
c(b(B(x))) → b(B(x))
A(x) → B(x)
c(b(B(x))) → A(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 3 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

C(b(b(c(x0)))) → C(b(x0))
C(b(b(b(b(x0))))) → C(b(a(b(c(b(c(x0)))))))
C(b(b(b(B(x0))))) → C(b(b(b(c(B(x0))))))
C(b(b(x))) → C(x)
C(b(b(b(B(x0))))) → C(b(b(B(x0))))
C(b(b(b(B(x0))))) → C(b(A(x0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(b(b(B(x0))))) → C(b(A(x0))) at position [0,0] we obtained the following new rules:

C(b(b(b(B(x0))))) → C(b(B(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.