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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

B(c(c(x0))) → A(a(b(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ 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:

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ 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:

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

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(b(c(c(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
c(A(x)) → A(x)
c(c(A(x))) → a(b(c(c(B(x)))))
b(c(A(x))) → B(x)
c(B(x)) → 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 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ 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:

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

The TRS R consists of the following rules:

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

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

C(c(A(x0))) → C(B(x0))



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

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

The TRS R consists of the following rules:

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

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

C(a(x)) → C(c(x))
C(a(x)) → C(x)

The TRS R consists of the following rules:

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


C(a(x)) → C(x)
The remaining pairs can at least be oriented weakly.

C(a(x)) → C(c(x))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = 1


POL( C(x1) ) = x1


POL( c(x1) ) = x1 + 1


POL( b(x1) ) = max{0, x1 - 1}


POL( B(x1) ) = max{0, -1}


POL( a(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



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

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

C(a(x)) → C(c(x))

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

C(a(b(x0))) → C(x0)



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

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

The TRS R consists of the following rules:

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

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

C(a(c(B(y0)))) → C(A(y0))



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

C(a(c(B(y0)))) → C(A(y0))
C(a(a(x0))) → C(a(b(c(c(x0)))))
C(a(c(A(x0)))) → C(a(b(c(c(B(x0))))))
C(a(b(x0))) → C(x0)

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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


C(a(c(A(x0)))) → C(a(b(c(c(B(x0))))))
The remaining pairs can at least be oriented weakly.

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

POL( A(x1) ) = 1


POL( C(x1) ) = x1


POL( c(x1) ) = x1 + 1


POL( b(x1) ) = max{0, x1 - 1}


POL( B(x1) ) = max{0, -1}


POL( a(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

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

Q is empty.

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

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

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

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

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

Q is empty.