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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(a(x1)) → x1
b(b(x1)) → a(b(c(x1)))
c(c(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 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

B(b(c(x0))) → B(b(b(b(x0))))



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ 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(a(x)) → x
b(b(x)) → c(b(a(x)))
c(c(x)) → b(b(b(x)))
c(b(B(x))) → b(b(b(B(x))))
b(B(x)) → C(x)
c(C(x)) → B(x)
c(C(x)) → b(B(x))
c(C(x)) → b(b(B(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:

C1(c(x)) → B1(b(b(x)))
B1(b(x)) → C1(b(a(x)))
C1(C(x)) → B1(B(x))
C1(b(B(x))) → B1(b(B(x)))
C1(b(B(x))) → B1(b(b(B(x))))
B1(b(x)) → A(x)
C1(C(x)) → B1(b(B(x)))
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))
B1(b(x)) → B1(a(x))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ 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:

C1(c(x)) → B1(b(b(x)))
B1(b(x)) → C1(b(a(x)))
C1(C(x)) → B1(B(x))
C1(b(B(x))) → B1(b(B(x)))
C1(b(B(x))) → B1(b(b(B(x))))
B1(b(x)) → A(x)
C1(C(x)) → B1(b(B(x)))
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))
B1(b(x)) → B1(a(x))

The TRS R consists of the following rules:

a(a(x)) → x
b(b(x)) → c(b(a(x)))
c(c(x)) → b(b(b(x)))
c(b(B(x))) → b(b(b(B(x))))
b(B(x)) → C(x)
c(C(x)) → B(x)
c(C(x)) → b(B(x))
c(C(x)) → b(b(B(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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ 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:

B1(b(x)) → C1(b(a(x)))
C1(c(x)) → B1(b(b(x)))
C1(b(B(x))) → B1(b(B(x)))
C1(b(B(x))) → B1(b(b(B(x))))
C1(C(x)) → B1(b(B(x)))
C1(c(x)) → B1(x)
B1(b(x)) → B1(a(x))
C1(c(x)) → B1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(x)) → C1(b(a(x))) at position [0] we obtained the following new rules:

B1(b(a(x0))) → C1(b(x0))



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

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

C1(c(x)) → B1(b(b(x)))
B1(b(a(x0))) → C1(b(x0))
C1(b(B(x))) → B1(b(B(x)))
C1(b(B(x))) → B1(b(b(B(x))))
C1(c(x)) → B1(b(x))
B1(b(x)) → B1(a(x))
C1(c(x)) → B1(x)
C1(C(x)) → B1(b(B(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(x)) → B1(a(x)) at position [0] we obtained the following new rules:

B1(b(a(x0))) → B1(x0)



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

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

C1(c(x)) → B1(b(b(x)))
C1(b(B(x))) → B1(b(B(x)))
B1(b(a(x0))) → C1(b(x0))
C1(b(B(x))) → B1(b(b(B(x))))
B1(b(a(x0))) → B1(x0)
C1(C(x)) → B1(b(B(x)))
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(b(B(x))) → B1(b(b(B(x)))) at position [0] we obtained the following new rules:

C1(b(B(x0))) → B1(b(C(x0)))
C1(b(B(y0))) → B1(c(b(a(B(y0)))))



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

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

C1(c(x)) → B1(b(b(x)))
C1(b(B(x0))) → B1(b(C(x0)))
C1(b(B(y0))) → B1(c(b(a(B(y0)))))
B1(b(a(x0))) → C1(b(x0))
C1(b(B(x))) → B1(b(B(x)))
B1(b(a(x0))) → B1(x0)
C1(c(x)) → B1(b(x))
C1(c(x)) → B1(x)
C1(C(x)) → B1(b(B(x)))

The TRS R consists of the following rules:

a(a(x)) → x
b(b(x)) → c(b(a(x)))
c(c(x)) → b(b(b(x)))
c(b(B(x))) → b(b(b(B(x))))
b(B(x)) → C(x)
c(C(x)) → B(x)
c(C(x)) → b(B(x))
c(C(x)) → b(b(B(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 2 less nodes.

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

C1(c(x)) → B1(b(b(x)))
C1(b(B(x))) → B1(b(B(x)))
B1(b(a(x0))) → C1(b(x0))
B1(b(a(x0))) → B1(x0)
C1(C(x)) → B1(b(B(x)))
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(b(B(x))) → B1(b(B(x))) at position [0] we obtained the following new rules:

C1(b(B(x0))) → B1(C(x0))



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

C1(c(x)) → B1(b(b(x)))
C1(b(B(x0))) → B1(C(x0))
B1(b(a(x0))) → C1(b(x0))
B1(b(a(x0))) → B1(x0)
C1(c(x)) → B1(b(x))
C1(c(x)) → B1(x)
C1(C(x)) → B1(b(B(x)))

The TRS R consists of the following rules:

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

C1(c(x)) → B1(b(b(x)))
B1(b(a(x0))) → C1(b(x0))
B1(b(a(x0))) → B1(x0)
C1(C(x)) → B1(b(B(x)))
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(C(x)) → B1(b(B(x))) at position [0] we obtained the following new rules:

C1(C(x0)) → B1(C(x0))



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

C1(c(x)) → B1(b(b(x)))
B1(b(a(x0))) → C1(b(x0))
C1(C(x0)) → B1(C(x0))
B1(b(a(x0))) → B1(x0)
C1(c(x)) → B1(b(x))
C1(c(x)) → B1(x)

The TRS R consists of the following rules:

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

C1(c(x)) → B1(b(b(x)))
B1(b(a(x0))) → C1(b(x0))
B1(b(a(x0))) → B1(x0)
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(x)) → B1(b(b(x))) at position [0] we obtained the following new rules:

C1(c(B(x0))) → B1(b(C(x0)))
C1(c(b(x0))) → B1(b(c(b(a(x0)))))
C1(c(x0)) → B1(c(b(a(x0))))



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

B1(b(a(x0))) → C1(b(x0))
B1(b(a(x0))) → B1(x0)
C1(c(B(x0))) → B1(b(C(x0)))
C1(c(x)) → B1(b(x))
C1(c(x)) → B1(x)
C1(c(x0)) → B1(c(b(a(x0))))
C1(c(b(x0))) → B1(b(c(b(a(x0)))))

The TRS R consists of the following rules:

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

B1(b(a(x0))) → C1(b(x0))
B1(b(a(x0))) → B1(x0)
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))
C1(c(x0)) → B1(c(b(a(x0))))
C1(c(b(x0))) → B1(b(c(b(a(x0)))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(b(x0))) → B1(b(c(b(a(x0))))) at position [0] we obtained the following new rules:

C1(c(b(a(x0)))) → B1(b(c(b(x0))))



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

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

C1(c(b(a(x0)))) → B1(b(c(b(x0))))
B1(b(a(x0))) → C1(b(x0))
B1(b(a(x0))) → B1(x0)
C1(c(x)) → B1(b(x))
C1(c(x)) → B1(x)
C1(c(x0)) → B1(c(b(a(x0))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(x0)) → B1(c(b(a(x0)))) at position [0] we obtained the following new rules:

C1(c(a(x0))) → B1(c(b(x0)))



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

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

B1(b(a(x0))) → C1(b(x0))
C1(c(b(a(x0)))) → B1(b(c(b(x0))))
B1(b(a(x0))) → B1(x0)
C1(c(a(x0))) → B1(c(b(x0)))
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


C1(c(a(x0))) → B1(c(b(x0)))
The remaining pairs can at least be oriented weakly.

B1(b(a(x0))) → C1(b(x0))
C1(c(b(a(x0)))) → B1(b(c(b(x0))))
B1(b(a(x0))) → B1(x0)
C1(c(x)) → B1(x)
C1(c(x)) → B1(b(x))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( C(x1) ) = 0


POL( c(x1) ) = x1


POL( B(x1) ) = 0


POL( a(x1) ) = 1


POL( B1(x1) ) = 0


POL( b(x1) ) = 0


POL( C1(x1) ) = x1



The following usable rules [17] were oriented:

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



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

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

C1(c(b(a(x0)))) → B1(b(c(b(x0))))
B1(b(a(x0))) → C1(b(x0))
B1(b(a(x0))) → B1(x0)
C1(c(x)) → B1(b(x))
C1(c(x)) → B1(x)

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

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

Q is empty.

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

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

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

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

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

Q is empty.