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

Q is empty.


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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q consists of the following terms:

a(a(b(x0)))
a(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:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(a(b(x0)))
a(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
QDP
          ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(a(b(x0)))
a(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Narrowing
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ DependencyGraphProof
QDP
                      ↳ ForwardInstantiation

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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



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

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

A(a(b(a(b(y_0))))) → A(a(b(y_0)))
A(a(b(b(x0)))) → A(c(c(a(a(a(x0))))))
A(a(b(a(b(b(y_0)))))) → A(a(b(b(y_0))))
A(c(x1)) → A(x1)
A(a(b(c(y_0)))) → A(c(y_0))
A(a(b(x1))) → A(a(x1))
A(a(b(a(b(c(y_0)))))) → A(a(b(c(y_0))))
A(a(b(c(x0)))) → A(a(b(a(x0))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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



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

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

A(a(b(a(b(y_0))))) → A(a(b(y_0)))
A(c(a(b(a(b(c(y_0))))))) → A(a(b(a(b(c(y_0))))))
A(c(c(y_0))) → A(c(y_0))
A(a(b(x1))) → A(a(x1))
A(a(b(a(b(c(y_0)))))) → A(a(b(c(y_0))))
A(c(a(b(y_0)))) → A(a(b(y_0)))
A(a(b(b(x0)))) → A(c(c(a(a(a(x0))))))
A(c(a(b(b(y_0))))) → A(a(b(b(y_0))))
A(a(b(a(b(b(y_0)))))) → A(a(b(b(y_0))))
A(c(a(b(c(y_0))))) → A(a(b(c(y_0))))
A(a(b(c(y_0)))) → A(c(y_0))
A(c(a(b(a(b(b(y_0))))))) → A(a(b(a(b(b(y_0))))))
A(c(a(b(a(b(y_0)))))) → A(a(b(a(b(y_0)))))
A(a(b(c(x0)))) → A(a(b(a(x0))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(a(b(x0)))
a(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Narrowing
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP
                          ↳ ForwardInstantiation
                            ↳ QDP
                              ↳ QDPToSRSProof
QTRS
                                  ↳ QTRS Reverse

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

a(a(b(x1))) → c(c(a(a(a(x1)))))
a(c(x1)) → b(a(x1))
A(a(b(a(b(y_0))))) → A(a(b(y_0)))
A(c(a(b(a(b(c(y_0))))))) → A(a(b(a(b(c(y_0))))))
A(c(c(y_0))) → A(c(y_0))
A(a(b(x1))) → A(a(x1))
A(a(b(a(b(c(y_0)))))) → A(a(b(c(y_0))))
A(c(a(b(y_0)))) → A(a(b(y_0)))
A(a(b(b(x0)))) → A(c(c(a(a(a(x0))))))
A(c(a(b(b(y_0))))) → A(a(b(b(y_0))))
A(a(b(a(b(b(y_0)))))) → A(a(b(b(y_0))))
A(c(a(b(c(y_0))))) → A(a(b(c(y_0))))
A(a(b(c(y_0)))) → A(c(y_0))
A(c(a(b(a(b(b(y_0))))))) → A(a(b(a(b(b(y_0))))))
A(c(a(b(a(b(y_0)))))) → A(a(b(a(b(y_0)))))
A(a(b(c(x0)))) → A(a(b(a(x0))))

The set Q consists of the following terms:

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


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

a(a(b(x1))) → c(c(a(a(a(x1)))))
a(c(x1)) → b(a(x1))
A(a(b(a(b(y_0))))) → A(a(b(y_0)))
A(c(a(b(a(b(c(y_0))))))) → A(a(b(a(b(c(y_0))))))
A(c(c(y_0))) → A(c(y_0))
A(a(b(x1))) → A(a(x1))
A(a(b(a(b(c(y_0)))))) → A(a(b(c(y_0))))
A(c(a(b(y_0)))) → A(a(b(y_0)))
A(a(b(b(x0)))) → A(c(c(a(a(a(x0))))))
A(c(a(b(b(y_0))))) → A(a(b(b(y_0))))
A(a(b(a(b(b(y_0)))))) → A(a(b(b(y_0))))
A(c(a(b(c(y_0))))) → A(a(b(c(y_0))))
A(a(b(c(y_0)))) → A(c(y_0))
A(c(a(b(a(b(b(y_0))))))) → A(a(b(a(b(b(y_0))))))
A(c(a(b(a(b(y_0)))))) → A(a(b(a(b(y_0)))))
A(a(b(c(x0)))) → A(a(b(a(x0))))

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

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

The set Q is empty.

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

b(a(a(x))) → a(a(a(c(c(x)))))
c(a(x)) → a(b(x))
b(a(b(a(A(x))))) → b(a(A(x)))
c(b(a(b(a(c(A(x))))))) → c(b(a(b(a(A(x))))))
c(c(A(x))) → c(A(x))
b(a(A(x))) → a(A(x))
c(b(a(b(a(A(x)))))) → c(b(a(A(x))))
b(a(c(A(x)))) → b(a(A(x)))
b(b(a(A(x)))) → a(a(a(c(c(A(x))))))
b(b(a(c(A(x))))) → b(b(a(A(x))))
b(b(a(b(a(A(x)))))) → b(b(a(A(x))))
c(b(a(c(A(x))))) → c(b(a(A(x))))
c(b(a(A(x)))) → c(A(x))
b(b(a(b(a(c(A(x))))))) → b(b(a(b(a(A(x))))))
b(a(b(a(c(A(x)))))) → b(a(b(a(A(x)))))
c(b(a(A(x)))) → a(b(a(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 9 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

B(b(a(c(A(x0))))) → B(a(A(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

B(b(a(b(a(A(x0)))))) → B(a(A(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

b(a(a(x))) → a(a(a(c(c(x)))))
c(a(x)) → a(b(x))
b(a(b(a(A(x))))) → b(a(A(x)))
c(b(a(b(a(c(A(x))))))) → c(b(a(b(a(A(x))))))
c(c(A(x))) → c(A(x))
b(a(A(x))) → a(A(x))
c(b(a(b(a(A(x)))))) → c(b(a(A(x))))
b(a(c(A(x)))) → b(a(A(x)))
b(b(a(A(x)))) → a(a(a(c(c(A(x))))))
b(b(a(c(A(x))))) → b(b(a(A(x))))
b(b(a(b(a(A(x)))))) → b(b(a(A(x))))
c(b(a(c(A(x))))) → c(b(a(A(x))))
c(b(a(A(x)))) → c(A(x))
b(b(a(b(a(c(A(x))))))) → b(b(a(b(a(A(x))))))
b(a(b(a(c(A(x)))))) → b(a(b(a(A(x)))))
c(b(a(A(x)))) → a(b(a(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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

B(a(b(a(c(A(x0)))))) → B(a(a(A(x0))))



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

C(b(a(b(a(A(x0)))))) → C(a(A(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

C(b(a(c(A(x0))))) → C(a(A(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

B(b(a(b(a(c(A(x0))))))) → B(a(A(x0)))



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

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

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

The TRS R consists of the following rules:

b(a(a(x))) → a(a(a(c(c(x)))))
c(a(x)) → a(b(x))
b(a(b(a(A(x))))) → b(a(A(x)))
c(b(a(b(a(c(A(x))))))) → c(b(a(b(a(A(x))))))
c(c(A(x))) → c(A(x))
b(a(A(x))) → a(A(x))
c(b(a(b(a(A(x)))))) → c(b(a(A(x))))
b(a(c(A(x)))) → b(a(A(x)))
b(b(a(A(x)))) → a(a(a(c(c(A(x))))))
b(b(a(c(A(x))))) → b(b(a(A(x))))
b(b(a(b(a(A(x)))))) → b(b(a(A(x))))
c(b(a(c(A(x))))) → c(b(a(A(x))))
c(b(a(A(x)))) → c(A(x))
b(b(a(b(a(c(A(x))))))) → b(b(a(b(a(A(x))))))
b(a(b(a(c(A(x)))))) → b(a(b(a(A(x)))))
c(b(a(A(x)))) → a(b(a(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
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Narrowing
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP
                          ↳ ForwardInstantiation
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
                                                            ↳ QDP
                                                              ↳ Narrowing
                                                                ↳ QDP
                                                                  ↳ DependencyGraphProof
                                                                    ↳ QDP
                                                                      ↳ Narrowing
                                                                        ↳ QDP
                                                                          ↳ Narrowing
                                                                            ↳ QDP
                                                                              ↳ Narrowing
                                                                                ↳ QDP
                                                                                  ↳ Narrowing
                                                                                    ↳ QDP
                                                                                      ↳ Narrowing
                                                                                        ↳ QDP
                                                                                          ↳ Narrowing
                                                                                            ↳ QDP
                                                                                              ↳ Narrowing
                                                                                                ↳ QDP
                                                                                                  ↳ DependencyGraphProof
QDP
                                                                                                      ↳ SemLabProof
                                                                                                      ↳ SemLabProof2
                                      ↳ QTRS Reverse
                                      ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.C: 0
c: 0
B: 0
a: 0
A: 1
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

C.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → B.0(a.0(a.1(A.0(x0))))
B.0(a.0(a.0(a.1(x0)))) → C.0(a.0(b.1(x0)))
C.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → B.0(a.0(a.1(A.1(x0))))
B.0(a.0(a.0(b.0(a.1(A.0(x0)))))) → C.0(a.0(b.0(a.1(A.0(x0)))))
B.0(a.0(b.0(a.0(c.1(A.0(x0)))))) → B.0(a.0(a.1(A.0(x0))))
B.0(a.0(a.0(b.0(a.1(A.1(x0)))))) → C.0(a.0(b.0(a.1(A.1(x0)))))
B.0(a.0(a.1(x))) → C.1(x)
C.0(a.0(x)) → B.0(x)
C.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → C.0(b.0(a.0(b.0(a.1(A.0(x))))))
B.0(a.0(a.0(b.0(a.0(b.0(a.1(A.1(x0)))))))) → C.0(c.0(b.0(a.1(A.1(x0)))))
B.0(a.0(b.0(a.0(c.1(A.1(x0)))))) → B.0(a.0(a.1(A.1(x0))))
B.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → B.0(b.0(a.0(a.1(A.1(x0)))))
B.0(a.0(a.0(a.0(x0)))) → C.0(a.0(b.0(x0)))
B.0(a.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → C.0(c.0(b.0(a.1(A.1(x0)))))
C.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → C.0(b.0(a.0(b.0(a.1(A.1(x))))))
C.0(a.1(x)) → B.1(x)
C.0(b.0(a.0(b.0(a.1(A.1(x0)))))) → C.0(a.1(A.1(x0)))
B.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → B.0(a.0(a.1(A.1(x0))))
B.0(a.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → C.0(c.0(b.0(a.1(A.0(x0)))))
C.0(b.0(a.0(b.0(a.1(A.0(x0)))))) → C.0(a.1(A.0(x0)))
C.0(b.0(a.0(c.1(A.0(x0))))) → C.0(a.1(A.0(x0)))
B.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → B.0(b.0(a.0(a.1(A.0(x0)))))
B.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → B.0(a.0(a.1(A.0(x0))))
B.0(a.0(a.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))))) → C.0(c.0(b.0(a.0(b.0(a.1(A.0(x0)))))))
B.0(a.0(a.0(x))) → C.0(x)
B.0(a.0(a.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))))) → C.0(c.0(b.0(a.0(b.0(a.1(A.1(x0)))))))
B.0(a.0(a.0(b.0(a.0(b.0(a.1(A.0(x0)))))))) → C.0(c.0(b.0(a.1(A.0(x0)))))
C.0(b.0(a.0(c.1(A.1(x0))))) → C.0(a.1(A.1(x0)))

The TRS R consists of the following rules:

c.0(b.0(a.0(b.0(a.1(A.1(x)))))) → c.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → c.0(b.0(a.0(b.0(a.1(A.0(x))))))
c.0(a.0(x)) → a.0(b.0(x))
b.0(a.0(b.0(a.0(c.1(A.1(x)))))) → b.0(a.0(b.0(a.1(A.1(x)))))
b.0(a.0(b.0(a.0(c.1(A.0(x)))))) → b.0(a.0(b.0(a.1(A.0(x)))))
c.0(b.0(a.0(c.1(A.0(x))))) → c.0(b.0(a.1(A.0(x))))
c.0(c.1(A.1(x))) → c.1(A.1(x))
c.0(c.1(A.0(x))) → c.1(A.0(x))
b.0(a.1(A.0(x))) → a.1(A.0(x))
b.0(a.0(a.1(x))) → a.0(a.0(a.0(c.0(c.1(x)))))
b.0(a.1(A.1(x))) → a.1(A.1(x))
c.0(a.1(x)) → a.0(b.1(x))
b.0(a.0(b.0(a.1(A.1(x))))) → b.0(a.1(A.1(x)))
c.0(b.0(a.1(A.0(x)))) → a.0(b.0(a.1(A.0(x))))
c.0(b.0(a.1(A.0(x)))) → c.1(A.0(x))
b.0(b.0(a.0(c.1(A.1(x))))) → b.0(b.0(a.1(A.1(x))))
c.0(b.0(a.1(A.1(x)))) → c.1(A.1(x))
b.0(b.0(a.1(A.0(x)))) → a.0(a.0(a.0(c.0(c.1(A.0(x))))))
b.0(a.0(b.0(a.1(A.0(x))))) → b.0(a.1(A.0(x)))
c.0(b.0(a.0(b.0(a.1(A.0(x)))))) → c.0(b.0(a.1(A.0(x))))
b.0(b.0(a.0(b.0(a.1(A.0(x)))))) → b.0(b.0(a.1(A.0(x))))
b.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → b.0(b.0(a.0(b.0(a.1(A.0(x))))))
b.0(a.0(c.1(A.0(x)))) → b.0(a.1(A.0(x)))
b.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → b.0(b.0(a.0(b.0(a.1(A.1(x))))))
b.0(a.0(c.1(A.1(x)))) → b.0(a.1(A.1(x)))
b.0(b.0(a.1(A.1(x)))) → a.0(a.0(a.0(c.0(c.1(A.1(x))))))
b.0(a.0(a.0(x))) → a.0(a.0(a.0(c.0(c.0(x)))))
b.0(b.0(a.0(b.0(a.1(A.1(x)))))) → b.0(b.0(a.1(A.1(x))))
c.0(b.0(a.1(A.1(x)))) → a.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(c.1(A.1(x))))) → c.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → c.0(b.0(a.0(b.0(a.1(A.1(x))))))
b.0(b.0(a.0(c.1(A.0(x))))) → b.0(b.0(a.1(A.0(x))))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Narrowing
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP
                          ↳ ForwardInstantiation
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
                                                            ↳ QDP
                                                              ↳ Narrowing
                                                                ↳ QDP
                                                                  ↳ DependencyGraphProof
                                                                    ↳ QDP
                                                                      ↳ Narrowing
                                                                        ↳ QDP
                                                                          ↳ Narrowing
                                                                            ↳ QDP
                                                                              ↳ Narrowing
                                                                                ↳ QDP
                                                                                  ↳ Narrowing
                                                                                    ↳ QDP
                                                                                      ↳ Narrowing
                                                                                        ↳ QDP
                                                                                          ↳ Narrowing
                                                                                            ↳ QDP
                                                                                              ↳ Narrowing
                                                                                                ↳ QDP
                                                                                                  ↳ DependencyGraphProof
                                                                                                    ↳ QDP
                                                                                                      ↳ SemLabProof
QDP
                                                                                                          ↳ DependencyGraphProof
                                                                                                      ↳ SemLabProof2
                                      ↳ QTRS Reverse
                                      ↳ QTRS Reverse

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

C.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → B.0(a.0(a.1(A.0(x0))))
B.0(a.0(a.0(a.1(x0)))) → C.0(a.0(b.1(x0)))
C.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → B.0(a.0(a.1(A.1(x0))))
B.0(a.0(a.0(b.0(a.1(A.0(x0)))))) → C.0(a.0(b.0(a.1(A.0(x0)))))
B.0(a.0(b.0(a.0(c.1(A.0(x0)))))) → B.0(a.0(a.1(A.0(x0))))
B.0(a.0(a.0(b.0(a.1(A.1(x0)))))) → C.0(a.0(b.0(a.1(A.1(x0)))))
B.0(a.0(a.1(x))) → C.1(x)
C.0(a.0(x)) → B.0(x)
C.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → C.0(b.0(a.0(b.0(a.1(A.0(x))))))
B.0(a.0(a.0(b.0(a.0(b.0(a.1(A.1(x0)))))))) → C.0(c.0(b.0(a.1(A.1(x0)))))
B.0(a.0(b.0(a.0(c.1(A.1(x0)))))) → B.0(a.0(a.1(A.1(x0))))
B.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → B.0(b.0(a.0(a.1(A.1(x0)))))
B.0(a.0(a.0(a.0(x0)))) → C.0(a.0(b.0(x0)))
B.0(a.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → C.0(c.0(b.0(a.1(A.1(x0)))))
C.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → C.0(b.0(a.0(b.0(a.1(A.1(x))))))
C.0(a.1(x)) → B.1(x)
C.0(b.0(a.0(b.0(a.1(A.1(x0)))))) → C.0(a.1(A.1(x0)))
B.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → B.0(a.0(a.1(A.1(x0))))
B.0(a.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → C.0(c.0(b.0(a.1(A.0(x0)))))
C.0(b.0(a.0(b.0(a.1(A.0(x0)))))) → C.0(a.1(A.0(x0)))
C.0(b.0(a.0(c.1(A.0(x0))))) → C.0(a.1(A.0(x0)))
B.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → B.0(b.0(a.0(a.1(A.0(x0)))))
B.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → B.0(a.0(a.1(A.0(x0))))
B.0(a.0(a.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))))) → C.0(c.0(b.0(a.0(b.0(a.1(A.0(x0)))))))
B.0(a.0(a.0(x))) → C.0(x)
B.0(a.0(a.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))))) → C.0(c.0(b.0(a.0(b.0(a.1(A.1(x0)))))))
B.0(a.0(a.0(b.0(a.0(b.0(a.1(A.0(x0)))))))) → C.0(c.0(b.0(a.1(A.0(x0)))))
C.0(b.0(a.0(c.1(A.1(x0))))) → C.0(a.1(A.1(x0)))

The TRS R consists of the following rules:

c.0(b.0(a.0(b.0(a.1(A.1(x)))))) → c.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → c.0(b.0(a.0(b.0(a.1(A.0(x))))))
c.0(a.0(x)) → a.0(b.0(x))
b.0(a.0(b.0(a.0(c.1(A.1(x)))))) → b.0(a.0(b.0(a.1(A.1(x)))))
b.0(a.0(b.0(a.0(c.1(A.0(x)))))) → b.0(a.0(b.0(a.1(A.0(x)))))
c.0(b.0(a.0(c.1(A.0(x))))) → c.0(b.0(a.1(A.0(x))))
c.0(c.1(A.1(x))) → c.1(A.1(x))
c.0(c.1(A.0(x))) → c.1(A.0(x))
b.0(a.1(A.0(x))) → a.1(A.0(x))
b.0(a.0(a.1(x))) → a.0(a.0(a.0(c.0(c.1(x)))))
b.0(a.1(A.1(x))) → a.1(A.1(x))
c.0(a.1(x)) → a.0(b.1(x))
b.0(a.0(b.0(a.1(A.1(x))))) → b.0(a.1(A.1(x)))
c.0(b.0(a.1(A.0(x)))) → a.0(b.0(a.1(A.0(x))))
c.0(b.0(a.1(A.0(x)))) → c.1(A.0(x))
b.0(b.0(a.0(c.1(A.1(x))))) → b.0(b.0(a.1(A.1(x))))
c.0(b.0(a.1(A.1(x)))) → c.1(A.1(x))
b.0(b.0(a.1(A.0(x)))) → a.0(a.0(a.0(c.0(c.1(A.0(x))))))
b.0(a.0(b.0(a.1(A.0(x))))) → b.0(a.1(A.0(x)))
c.0(b.0(a.0(b.0(a.1(A.0(x)))))) → c.0(b.0(a.1(A.0(x))))
b.0(b.0(a.0(b.0(a.1(A.0(x)))))) → b.0(b.0(a.1(A.0(x))))
b.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → b.0(b.0(a.0(b.0(a.1(A.0(x))))))
b.0(a.0(c.1(A.0(x)))) → b.0(a.1(A.0(x)))
b.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → b.0(b.0(a.0(b.0(a.1(A.1(x))))))
b.0(a.0(c.1(A.1(x)))) → b.0(a.1(A.1(x)))
b.0(b.0(a.1(A.1(x)))) → a.0(a.0(a.0(c.0(c.1(A.1(x))))))
b.0(a.0(a.0(x))) → a.0(a.0(a.0(c.0(c.0(x)))))
b.0(b.0(a.0(b.0(a.1(A.1(x)))))) → b.0(b.0(a.1(A.1(x))))
c.0(b.0(a.1(A.1(x)))) → a.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(c.1(A.1(x))))) → c.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → c.0(b.0(a.0(b.0(a.1(A.1(x))))))
b.0(b.0(a.0(c.1(A.0(x))))) → b.0(b.0(a.1(A.0(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 12 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Narrowing
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP
                          ↳ ForwardInstantiation
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
                                                            ↳ QDP
                                                              ↳ Narrowing
                                                                ↳ QDP
                                                                  ↳ DependencyGraphProof
                                                                    ↳ QDP
                                                                      ↳ Narrowing
                                                                        ↳ QDP
                                                                          ↳ Narrowing
                                                                            ↳ QDP
                                                                              ↳ Narrowing
                                                                                ↳ QDP
                                                                                  ↳ Narrowing
                                                                                    ↳ QDP
                                                                                      ↳ Narrowing
                                                                                        ↳ QDP
                                                                                          ↳ Narrowing
                                                                                            ↳ QDP
                                                                                              ↳ Narrowing
                                                                                                ↳ QDP
                                                                                                  ↳ DependencyGraphProof
                                                                                                    ↳ QDP
                                                                                                      ↳ SemLabProof
                                                                                                        ↳ QDP
                                                                                                          ↳ DependencyGraphProof
QDP
                                                                                                              ↳ RuleRemovalProof
                                                                                                      ↳ SemLabProof2
                                      ↳ QTRS Reverse
                                      ↳ QTRS Reverse

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

B.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → B.0(b.0(a.0(a.1(A.1(x0)))))
B.0(a.0(a.0(a.0(x0)))) → C.0(a.0(b.0(x0)))
B.0(a.0(a.0(a.1(x0)))) → C.0(a.0(b.1(x0)))
C.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → C.0(b.0(a.0(b.0(a.1(A.1(x))))))
B.0(a.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → C.0(c.0(b.0(a.1(A.1(x0)))))
B.0(a.0(a.0(b.0(a.1(A.0(x0)))))) → C.0(a.0(b.0(a.1(A.0(x0)))))
B.0(a.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → C.0(c.0(b.0(a.1(A.0(x0)))))
B.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → B.0(b.0(a.0(a.1(A.0(x0)))))
B.0(a.0(a.0(b.0(a.1(A.1(x0)))))) → C.0(a.0(b.0(a.1(A.1(x0)))))
B.0(a.0(a.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))))) → C.0(c.0(b.0(a.0(b.0(a.1(A.0(x0)))))))
C.0(a.0(x)) → B.0(x)
C.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → C.0(b.0(a.0(b.0(a.1(A.0(x))))))
B.0(a.0(a.0(x))) → C.0(x)
B.0(a.0(a.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))))) → C.0(c.0(b.0(a.0(b.0(a.1(A.1(x0)))))))
B.0(a.0(a.0(b.0(a.0(b.0(a.1(A.1(x0)))))))) → C.0(c.0(b.0(a.1(A.1(x0)))))
B.0(a.0(a.0(b.0(a.0(b.0(a.1(A.0(x0)))))))) → C.0(c.0(b.0(a.1(A.0(x0)))))

The TRS R consists of the following rules:

c.0(b.0(a.0(b.0(a.1(A.1(x)))))) → c.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → c.0(b.0(a.0(b.0(a.1(A.0(x))))))
c.0(a.0(x)) → a.0(b.0(x))
b.0(a.0(b.0(a.0(c.1(A.1(x)))))) → b.0(a.0(b.0(a.1(A.1(x)))))
b.0(a.0(b.0(a.0(c.1(A.0(x)))))) → b.0(a.0(b.0(a.1(A.0(x)))))
c.0(b.0(a.0(c.1(A.0(x))))) → c.0(b.0(a.1(A.0(x))))
c.0(c.1(A.1(x))) → c.1(A.1(x))
c.0(c.1(A.0(x))) → c.1(A.0(x))
b.0(a.1(A.0(x))) → a.1(A.0(x))
b.0(a.0(a.1(x))) → a.0(a.0(a.0(c.0(c.1(x)))))
b.0(a.1(A.1(x))) → a.1(A.1(x))
c.0(a.1(x)) → a.0(b.1(x))
b.0(a.0(b.0(a.1(A.1(x))))) → b.0(a.1(A.1(x)))
c.0(b.0(a.1(A.0(x)))) → a.0(b.0(a.1(A.0(x))))
c.0(b.0(a.1(A.0(x)))) → c.1(A.0(x))
b.0(b.0(a.0(c.1(A.1(x))))) → b.0(b.0(a.1(A.1(x))))
c.0(b.0(a.1(A.1(x)))) → c.1(A.1(x))
b.0(b.0(a.1(A.0(x)))) → a.0(a.0(a.0(c.0(c.1(A.0(x))))))
b.0(a.0(b.0(a.1(A.0(x))))) → b.0(a.1(A.0(x)))
c.0(b.0(a.0(b.0(a.1(A.0(x)))))) → c.0(b.0(a.1(A.0(x))))
b.0(b.0(a.0(b.0(a.1(A.0(x)))))) → b.0(b.0(a.1(A.0(x))))
b.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → b.0(b.0(a.0(b.0(a.1(A.0(x))))))
b.0(a.0(c.1(A.0(x)))) → b.0(a.1(A.0(x)))
b.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → b.0(b.0(a.0(b.0(a.1(A.1(x))))))
b.0(a.0(c.1(A.1(x)))) → b.0(a.1(A.1(x)))
b.0(b.0(a.1(A.1(x)))) → a.0(a.0(a.0(c.0(c.1(A.1(x))))))
b.0(a.0(a.0(x))) → a.0(a.0(a.0(c.0(c.0(x)))))
b.0(b.0(a.0(b.0(a.1(A.1(x)))))) → b.0(b.0(a.1(A.1(x))))
c.0(b.0(a.1(A.1(x)))) → a.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(c.1(A.1(x))))) → c.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → c.0(b.0(a.0(b.0(a.1(A.1(x))))))
b.0(b.0(a.0(c.1(A.0(x))))) → b.0(b.0(a.1(A.0(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

B.0(a.0(a.0(a.1(x0)))) → C.0(a.0(b.1(x0)))

Strictly oriented rules of the TRS R:

c.0(a.1(x)) → a.0(b.1(x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(A.0(x1)) = x1   
POL(A.1(x1)) = x1   
POL(B.0(x1)) = x1   
POL(C.0(x1)) = x1   
POL(a.0(x1)) = x1   
POL(a.1(x1)) = 1 + x1   
POL(b.0(x1)) = x1   
POL(b.1(x1)) = x1   
POL(c.0(x1)) = x1   
POL(c.1(x1)) = 1 + x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Narrowing
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP
                          ↳ ForwardInstantiation
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
                                                            ↳ QDP
                                                              ↳ Narrowing
                                                                ↳ QDP
                                                                  ↳ DependencyGraphProof
                                                                    ↳ QDP
                                                                      ↳ Narrowing
                                                                        ↳ QDP
                                                                          ↳ Narrowing
                                                                            ↳ QDP
                                                                              ↳ Narrowing
                                                                                ↳ QDP
                                                                                  ↳ Narrowing
                                                                                    ↳ QDP
                                                                                      ↳ Narrowing
                                                                                        ↳ QDP
                                                                                          ↳ Narrowing
                                                                                            ↳ QDP
                                                                                              ↳ Narrowing
                                                                                                ↳ QDP
                                                                                                  ↳ DependencyGraphProof
                                                                                                    ↳ QDP
                                                                                                      ↳ SemLabProof
                                                                                                        ↳ QDP
                                                                                                          ↳ DependencyGraphProof
                                                                                                            ↳ QDP
                                                                                                              ↳ RuleRemovalProof
QDP
                                                                                                      ↳ SemLabProof2
                                      ↳ QTRS Reverse
                                      ↳ QTRS Reverse

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

B.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → B.0(b.0(a.0(a.1(A.1(x0)))))
B.0(a.0(a.0(a.0(x0)))) → C.0(a.0(b.0(x0)))
C.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → C.0(b.0(a.0(b.0(a.1(A.1(x))))))
B.0(a.0(a.0(b.0(a.0(c.1(A.1(x0))))))) → C.0(c.0(b.0(a.1(A.1(x0)))))
B.0(a.0(a.0(b.0(a.1(A.0(x0)))))) → C.0(a.0(b.0(a.1(A.0(x0)))))
B.0(a.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → C.0(c.0(b.0(a.1(A.0(x0)))))
B.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))) → B.0(b.0(a.0(a.1(A.0(x0)))))
B.0(a.0(a.0(b.0(a.1(A.1(x0)))))) → C.0(a.0(b.0(a.1(A.1(x0)))))
B.0(a.0(a.0(b.0(a.0(b.0(a.0(c.1(A.0(x0))))))))) → C.0(c.0(b.0(a.0(b.0(a.1(A.0(x0)))))))
C.0(a.0(x)) → B.0(x)
C.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → C.0(b.0(a.0(b.0(a.1(A.0(x))))))
B.0(a.0(a.0(x))) → C.0(x)
B.0(a.0(a.0(b.0(a.0(b.0(a.0(c.1(A.1(x0))))))))) → C.0(c.0(b.0(a.0(b.0(a.1(A.1(x0)))))))
B.0(a.0(a.0(b.0(a.0(b.0(a.1(A.1(x0)))))))) → C.0(c.0(b.0(a.1(A.1(x0)))))
B.0(a.0(a.0(b.0(a.0(b.0(a.1(A.0(x0)))))))) → C.0(c.0(b.0(a.1(A.0(x0)))))

The TRS R consists of the following rules:

c.0(b.0(a.0(b.0(a.1(A.1(x)))))) → c.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → c.0(b.0(a.0(b.0(a.1(A.0(x))))))
c.0(a.0(x)) → a.0(b.0(x))
b.0(a.0(b.0(a.0(c.1(A.1(x)))))) → b.0(a.0(b.0(a.1(A.1(x)))))
b.0(a.0(b.0(a.0(c.1(A.0(x)))))) → b.0(a.0(b.0(a.1(A.0(x)))))
c.0(b.0(a.0(c.1(A.0(x))))) → c.0(b.0(a.1(A.0(x))))
c.0(c.1(A.1(x))) → c.1(A.1(x))
c.0(c.1(A.0(x))) → c.1(A.0(x))
b.0(a.1(A.0(x))) → a.1(A.0(x))
b.0(a.0(a.1(x))) → a.0(a.0(a.0(c.0(c.1(x)))))
b.0(a.1(A.1(x))) → a.1(A.1(x))
b.0(a.0(b.0(a.1(A.1(x))))) → b.0(a.1(A.1(x)))
c.0(b.0(a.1(A.0(x)))) → a.0(b.0(a.1(A.0(x))))
c.0(b.0(a.1(A.0(x)))) → c.1(A.0(x))
b.0(b.0(a.0(c.1(A.1(x))))) → b.0(b.0(a.1(A.1(x))))
c.0(b.0(a.1(A.1(x)))) → c.1(A.1(x))
b.0(b.0(a.1(A.0(x)))) → a.0(a.0(a.0(c.0(c.1(A.0(x))))))
b.0(a.0(b.0(a.1(A.0(x))))) → b.0(a.1(A.0(x)))
c.0(b.0(a.0(b.0(a.1(A.0(x)))))) → c.0(b.0(a.1(A.0(x))))
b.0(b.0(a.0(b.0(a.1(A.0(x)))))) → b.0(b.0(a.1(A.0(x))))
b.0(b.0(a.0(b.0(a.0(c.1(A.0(x))))))) → b.0(b.0(a.0(b.0(a.1(A.0(x))))))
b.0(a.0(c.1(A.0(x)))) → b.0(a.1(A.0(x)))
b.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → b.0(b.0(a.0(b.0(a.1(A.1(x))))))
b.0(a.0(c.1(A.1(x)))) → b.0(a.1(A.1(x)))
b.0(b.0(a.1(A.1(x)))) → a.0(a.0(a.0(c.0(c.1(A.1(x))))))
b.0(a.0(a.0(x))) → a.0(a.0(a.0(c.0(c.0(x)))))
b.0(b.0(a.0(b.0(a.1(A.1(x)))))) → b.0(b.0(a.1(A.1(x))))
c.0(b.0(a.1(A.1(x)))) → a.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(c.1(A.1(x))))) → c.0(b.0(a.1(A.1(x))))
c.0(b.0(a.0(b.0(a.0(c.1(A.1(x))))))) → c.0(b.0(a.0(b.0(a.1(A.1(x))))))
b.0(b.0(a.0(c.1(A.0(x))))) → b.0(b.0(a.1(A.0(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used. Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Narrowing
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP
                          ↳ ForwardInstantiation
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
                                                            ↳ QDP
                                                              ↳ Narrowing
                                                                ↳ QDP
                                                                  ↳ DependencyGraphProof
                                                                    ↳ QDP
                                                                      ↳ Narrowing
                                                                        ↳ QDP
                                                                          ↳ Narrowing
                                                                            ↳ QDP
                                                                              ↳ Narrowing
                                                                                ↳ QDP
                                                                                  ↳ Narrowing
                                                                                    ↳ QDP
                                                                                      ↳ Narrowing
                                                                                        ↳ QDP
                                                                                          ↳ Narrowing
                                                                                            ↳ QDP
                                                                                              ↳ Narrowing
                                                                                                ↳ QDP
                                                                                                  ↳ DependencyGraphProof
                                                                                                    ↳ QDP
                                                                                                      ↳ SemLabProof
                                                                                                      ↳ SemLabProof2
QDP
                                                                                                          ↳ Narrowing
                                      ↳ QTRS Reverse
                                      ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
          ↳ Narrowing
            ↳ QDP
              ↳ Rewriting
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP
                          ↳ ForwardInstantiation
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
                                                            ↳ QDP
                                                              ↳ Narrowing
                                                                ↳ QDP
                                                                  ↳ DependencyGraphProof
                                                                    ↳ QDP
                                                                      ↳ Narrowing
                                                                        ↳ QDP
                                                                          ↳ Narrowing
                                                                            ↳ QDP
                                                                              ↳ Narrowing
                                                                                ↳ QDP
                                                                                  ↳ Narrowing
                                                                                    ↳ QDP
                                                                                      ↳ Narrowing
                                                                                        ↳ QDP
                                                                                          ↳ Narrowing
                                                                                            ↳ QDP
                                                                                              ↳ Narrowing
                                                                                                ↳ QDP
                                                                                                  ↳ DependencyGraphProof
                                                                                                    ↳ QDP
                                                                                                      ↳ SemLabProof
                                                                                                      ↳ SemLabProof2
                                                                                                        ↳ QDP
                                                                                                          ↳ Narrowing
QDP
                                      ↳ QTRS Reverse
                                      ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.