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(b(c(x1)))) → a(c(c(b(a(a(x1))))))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ MNOCProof
          ↳ Rewriting
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
QDP
          ↳ Rewriting
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

B(b(b(c(x1)))) → A(c(c(b(b(a(x1))))))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
QDP
              ↳ Rewriting
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

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

B(b(b(c(x1)))) → B(b(a(x1)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
QDP
                  ↳ Rewriting
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

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

B(b(b(c(x1)))) → A(b(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
QDP
                      ↳ Rewriting
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

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

B(b(b(c(x1)))) → A(c(c(b(b(b(x1))))))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
QDP
                          ↳ Rewriting
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

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

B(b(b(c(x1)))) → B(b(b(x1)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
QDP
                              ↳ ForwardInstantiation
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A(x1) → B(x1) we obtained the following new rules:

A(b(b(c(y_0)))) → B(b(b(c(y_0))))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
QDP
                                  ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(b(b(c(x1)))) → A(x1)
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(c(c(b(b(b(x1))))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
QDP
                                      ↳ ForwardInstantiation
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(b(b(c(x1)))) → A(x1)
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(b(b(c(x1)))) → A(x1) we obtained the following new rules:

B(b(b(c(b(b(c(y_0))))))) → A(b(b(c(y_0))))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
QDP
                                          ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(b(b(c(b(b(c(y_0))))))) → A(b(b(c(y_0))))
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ QDP
                                          ↳ QDPToSRSProof
QTRS
                                              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(x1) → b(x1)
b(b(b(c(x1)))) → a(c(c(b(a(a(x1))))))
B(b(b(c(b(b(c(y_0))))))) → A(b(b(c(y_0))))
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))

The set Q consists of the following terms:

a(x0)
b(b(b(c(x0))))


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

a(x1) → b(x1)
b(b(b(c(x1)))) → a(c(c(b(a(a(x1))))))
B(b(b(c(b(b(c(y_0))))))) → A(b(b(c(y_0))))
A(b(b(c(y_0)))) → B(b(b(c(y_0))))
B(b(b(c(x1)))) → A(b(x1))
B(b(b(c(x1)))) → B(b(b(x1)))

The set Q is {a(x0), b(b(b(c(x0))))}.
We have obtained the following QTRS:

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ 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) → b(x)
c(b(b(b(x)))) → a(a(b(c(c(a(x))))))
c(b(b(c(b(b(B(x))))))) → c(b(b(A(x))))
c(b(b(A(x)))) → c(b(b(B(x))))
c(b(b(B(x)))) → b(A(x))
c(b(b(B(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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ 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(b(b(c(b(b(B(x))))))) → C(b(b(A(x))))
C(b(b(A(x)))) → C(b(b(B(x))))
C(b(b(b(x)))) → A1(x)
C(b(b(b(x)))) → A1(b(c(c(a(x)))))
C(b(b(b(x)))) → C(c(a(x)))
C(b(b(b(x)))) → A1(a(b(c(c(a(x))))))
C(b(b(b(x)))) → C(a(x))

The TRS R consists of the following rules:

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ 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(b(b(b(x)))) → C(c(a(x)))
C(b(b(b(x)))) → C(a(x))

The TRS R consists of the following rules:

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

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

C(b(b(b(x0)))) → C(b(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ 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:

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

The TRS R consists of the following rules:

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

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

C(b(b(b(x0)))) → C(c(b(x0)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ QDP
                                          ↳ QDPToSRSProof
                                            ↳ QTRS
                                              ↳ QTRS Reverse
                                                ↳ QTRS
                                                  ↳ DependencyPairsProof
                                                    ↳ QDP
                                                      ↳ DependencyGraphProof
                                                        ↳ QDP
                                                          ↳ Narrowing
                                                            ↳ QDP
                                                              ↳ Narrowing
QDP
                                                  ↳ QTRS Reverse
                                                  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ 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(x)
b(b(b(c(x)))) → a(c(c(b(a(a(x))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(x))))
A(b(b(c(x)))) → B(b(b(c(x))))
B(b(b(c(x)))) → A(b(x))
B(b(b(c(x)))) → B(b(b(x)))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Rewriting
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ Rewriting
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ 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(x)
b(b(b(c(x)))) → a(c(c(b(a(a(x))))))
B(b(b(c(b(b(c(x))))))) → A(b(b(c(x))))
A(b(b(c(x)))) → B(b(b(c(x))))
B(b(b(c(x)))) → A(b(x))
B(b(b(c(x)))) → B(b(b(x)))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ Overlay + Local Confluence
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.