Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

A(b(a(c(x0)))) → A(a(a(b(x0))))
A(b(a(c(y0)))) → A(b(a(a(y0))))
A(b(a(c(x1)))) → A(a(a(x1)))
A(b(a(c(y0)))) → A(a(b(a(y0))))
A(b(a(c(b(a(c(x0))))))) → A(a(a(c(c(a(a(a(a(x0)))))))))
A(b(a(c(x1)))) → A(a(x1))
A(b(a(c(x1)))) → A(x1)

The TRS R consists of the following rules:

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

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

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



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

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

A(b(a(c(x0)))) → A(a(a(b(x0))))
A(b(a(c(y0)))) → A(b(a(a(y0))))
A(b(a(c(y0)))) → A(b(a(y0)))
A(b(a(c(y0)))) → A(a(b(a(y0))))
A(b(a(c(b(a(c(x0))))))) → A(a(a(c(c(a(a(a(a(x0)))))))))
A(b(a(c(x0)))) → A(a(b(x0)))
A(b(a(c(b(a(c(x0))))))) → A(a(c(c(a(a(a(a(x0))))))))
A(b(a(c(x1)))) → A(x1)
A(b(a(c(x1)))) → A(a(x1))

The TRS R consists of the following rules:

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

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

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



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

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

A(b(a(c(x0)))) → A(b(x0))
A(b(a(c(b(a(c(x0))))))) → A(c(c(a(a(a(a(x0)))))))
A(b(a(c(x0)))) → A(a(a(b(x0))))
A(b(a(c(y0)))) → A(b(a(a(y0))))
A(b(a(c(y0)))) → A(a(b(a(y0))))
A(b(a(c(y0)))) → A(b(a(y0)))
A(b(a(c(b(a(c(x0))))))) → A(a(a(c(c(a(a(a(a(x0)))))))))
A(b(a(c(x0)))) → A(a(b(x0)))
A(b(a(c(x1)))) → A(x1)
A(b(a(c(b(a(c(x0))))))) → A(a(c(c(a(a(a(a(x0))))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

A(b(a(c(x0)))) → A(b(x0))
A(b(a(c(x0)))) → A(a(a(b(x0))))
A(b(a(c(y0)))) → A(b(a(a(y0))))
A(b(a(c(y0)))) → A(b(a(y0)))
A(b(a(c(y0)))) → A(a(b(a(y0))))
A(b(a(c(b(a(c(x0))))))) → A(a(a(c(c(a(a(a(a(x0)))))))))
A(b(a(c(x0)))) → A(a(b(x0)))
A(b(a(c(b(a(c(x0))))))) → A(a(c(c(a(a(a(a(x0))))))))
A(b(a(c(x1)))) → A(x1)

The TRS R consists of the following rules:

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

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

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



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

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

A(b(a(c(x0)))) → A(b(x0))
A(b(a(c(x0)))) → A(a(a(b(x0))))
A(b(a(c(y0)))) → A(b(a(a(y0))))
A(b(a(c(y0)))) → A(b(b(y0)))
A(b(a(c(y0)))) → A(a(b(a(y0))))
A(b(a(c(y0)))) → A(b(a(y0)))
A(b(a(c(a(c(x0)))))) → A(c(c(a(a(a(a(x0)))))))
A(b(a(c(b(a(c(x0))))))) → A(a(a(c(c(a(a(a(a(x0)))))))))
A(b(a(c(x1)))) → A(x1)
A(b(a(c(b(a(c(x0))))))) → A(a(c(c(a(a(a(a(x0))))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

A(b(a(c(x0)))) → A(b(x0))
A(b(a(c(x0)))) → A(a(a(b(x0))))
A(b(a(c(y0)))) → A(b(a(a(y0))))
A(b(a(c(y0)))) → A(b(a(y0)))
A(b(a(c(y0)))) → A(a(b(a(y0))))
A(b(a(c(b(a(c(x0))))))) → A(a(a(c(c(a(a(a(a(x0)))))))))
A(b(a(c(b(a(c(x0))))))) → A(a(c(c(a(a(a(a(x0))))))))
A(b(a(c(x1)))) → A(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


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

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

a(x1) → b(x1)
a(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
A(b(a(c(x0)))) → A(b(x0))
A(b(a(c(x0)))) → A(a(a(b(x0))))
A(b(a(c(y0)))) → A(b(a(a(y0))))
A(b(a(c(y0)))) → A(b(a(y0)))
A(b(a(c(y0)))) → A(a(b(a(y0))))
A(b(a(c(b(a(c(x0))))))) → A(a(a(c(c(a(a(a(a(x0)))))))))
A(b(a(c(b(a(c(x0))))))) → A(a(c(c(a(a(a(a(x0))))))))
A(b(a(c(x1)))) → A(x1)

Q is empty.

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

a(x1) → b(x1)
a(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
A(b(a(c(x0)))) → A(b(x0))
A(b(a(c(x0)))) → A(a(a(b(x0))))
A(b(a(c(y0)))) → A(b(a(a(y0))))
A(b(a(c(y0)))) → A(b(a(y0)))
A(b(a(c(y0)))) → A(a(b(a(y0))))
A(b(a(c(b(a(c(x0))))))) → A(a(a(c(c(a(a(a(a(x0)))))))))
A(b(a(c(b(a(c(x0))))))) → A(a(c(c(a(a(a(a(x0))))))))
A(b(a(c(x1)))) → A(x1)

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

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

The set Q is empty.

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

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ 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(a(b(a(x)))) → A1(a(a(c(c(x)))))
C(a(b(c(a(b(A(x))))))) → C(a(A(x)))
C(a(b(c(a(b(A(x))))))) → A1(a(a(a(c(c(a(A(x))))))))
C(a(b(A(x)))) → A1(a(b(A(x))))
C(a(b(c(a(b(A(x))))))) → C(c(a(a(A(x)))))
C(a(b(c(a(b(A(x))))))) → A1(a(A(x)))
C(a(b(c(a(b(A(x))))))) → A1(c(c(a(A(x)))))
C(a(b(c(a(b(A(x))))))) → A1(a(c(c(a(a(A(x)))))))
C(a(b(a(x)))) → A1(c(c(x)))
C(a(b(c(a(b(A(x))))))) → A1(c(c(a(a(A(x))))))
C(a(b(c(a(b(A(x))))))) → A1(a(a(c(c(a(a(A(x))))))))
C(a(b(c(a(b(A(x))))))) → A1(A(x))
C(a(b(A(x)))) → A1(A(x))
C(a(b(A(x)))) → A1(b(a(A(x))))
C(a(b(c(a(b(A(x))))))) → A1(a(a(c(c(a(A(x)))))))
C(a(b(a(x)))) → A1(a(a(a(c(c(x))))))
C(a(b(a(x)))) → A1(a(c(c(x))))
C(a(b(c(a(b(A(x))))))) → C(c(a(A(x))))
C(a(b(c(a(b(A(x))))))) → C(a(a(A(x))))
C(a(b(c(a(b(A(x))))))) → A1(a(c(c(a(A(x))))))
C(a(b(A(x)))) → A1(a(A(x)))
C(a(b(a(x)))) → C(x)
C(a(b(a(x)))) → C(c(x))
C(a(b(c(a(b(A(x))))))) → A1(a(a(a(c(c(a(a(A(x)))))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 18 less nodes.

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ 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(a(b(c(a(b(A(x))))))) → C(a(A(x)))
C(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(c(a(b(A(x))))))) → C(a(a(A(x))))
C(a(b(c(a(b(A(x))))))) → C(c(a(A(x))))
C(a(b(c(a(b(A(y0))))))) → C(c(b(a(A(y0)))))
C(a(b(a(x)))) → C(x)
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ 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(a(b(c(a(b(A(x))))))) → C(a(A(x)))
C(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(c(a(b(A(x))))))) → C(c(a(A(x))))
C(a(b(c(a(b(A(x))))))) → C(a(a(A(x))))
C(a(b(a(x)))) → C(x)
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ 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(a(b(c(a(b(A(y0))))))) → C(b(a(A(y0))))
C(a(b(c(a(b(A(x))))))) → C(a(A(x)))
C(a(b(c(a(b(A(x))))))) → C(c(a(A(x))))
C(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(a(x)))) → C(x)
C(a(b(c(a(b(A(y0))))))) → C(a(b(A(y0))))
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ 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(a(b(c(a(b(A(x))))))) → C(a(A(x)))
C(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(c(a(b(A(x))))))) → C(c(a(A(x))))
C(a(b(a(x)))) → C(x)
C(a(b(c(a(b(A(y0))))))) → C(a(b(A(y0))))
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ 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(a(b(c(a(b(A(x))))))) → C(a(A(x)))
C(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(a(x)))) → C(x)
C(a(b(c(a(b(A(y0))))))) → C(c(b(A(y0))))
C(a(b(c(a(b(A(y0))))))) → C(a(b(A(y0))))
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ 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(a(b(c(a(b(A(x))))))) → C(a(A(x)))
C(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(a(x)))) → C(x)
C(a(b(c(a(b(A(y0))))))) → C(a(b(A(y0))))
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

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

C(a(b(c(a(b(A(y0))))))) → C(b(A(y0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ 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(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(c(a(b(A(y0))))))) → C(b(A(y0)))
C(a(b(a(x)))) → C(x)
C(a(b(c(a(b(A(y0))))))) → C(a(b(A(y0))))
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ 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(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(a(x)))) → C(x)
C(a(b(c(a(b(A(y0))))))) → C(a(b(A(y0))))
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(b(c(a(b(A(y0))))))) → C(a(b(A(y0)))) at position [0] we obtained the following new rules:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ 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(a(b(c(a(b(A(y0))))))) → C(b(b(A(y0))))
C(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(a(x)))) → C(x)
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

C(a(b(c(a(b(A(y0))))))) → C(c(a(b(A(y0)))))
C(a(b(a(x)))) → C(x)
C(a(b(a(x)))) → C(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.