MAYBE Termination w.r.t. Q proof of ../tpdb/SRS/Waldmann/uni-7.srs
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(a(a(x1)))) → b(b(c(c(x1))))
b(b(b(b(x1)))) → a(a(c(c(x1))))
c(c(c(c(x1)))) → b(a(a(a(x1))))

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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


The following pairs can be oriented strictly and are deleted.


B(b(b(b(x1)))) → C(x1)
A(a(a(a(x1)))) → B(c(c(x1)))
C(c(c(c(x1)))) → A(x1)
A(a(a(a(x1)))) → C(c(x1))
B(b(b(b(x1)))) → C(c(x1))
A(a(a(a(x1)))) → C(x1)
B(b(b(b(x1)))) → A(c(c(x1)))
C(c(c(c(x1)))) → A(a(a(x1)))
C(c(c(c(x1)))) → A(a(x1))
The remaining pairs can at least be oriented weakly.

A(a(a(a(x1)))) → B(b(c(c(x1))))
C(c(c(c(x1)))) → B(a(a(a(x1))))
B(b(b(b(x1)))) → A(a(c(c(x1))))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( c(x1) ) = x1 + 1


POL( C(x1) ) = x1 + 1


POL( a(x1) ) = x1 + 1


POL( b(x1) ) = x1 + 1


POL( B(x1) ) = x1 + 1


POL( A(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(a(a(a(x1)))) → b(b(c(c(x1))))
b(b(b(b(x1)))) → a(a(c(c(x1))))
c(c(c(c(x1)))) → b(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

a(a(a(a(x)))) → b(b(c(c(x))))
b(b(b(b(x)))) → a(a(c(c(x))))
c(c(c(c(x)))) → b(a(a(a(x))))
A(a(a(a(c(c(c(x))))))) → B(b(c(b(a(a(a(x)))))))
B(b(b(b(c(c(c(x))))))) → A(a(c(b(a(a(a(x)))))))
B(b(b(b(c(c(x)))))) → A(a(b(a(a(a(x))))))
A(a(a(a(c(c(x)))))) → B(b(b(a(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(c(c(b(b(b(B(x))))))) → B1(c(a(A(x))))
A1(a(a(a(x)))) → B1(x)
C(c(c(c(x)))) → A1(a(b(x)))
C(c(b(b(b(B(x)))))) → A1(a(b(a(A(x)))))
C(c(b(b(b(B(x)))))) → A1(b(a(A(x))))
C(c(c(b(b(b(B(x))))))) → A1(b(c(a(A(x)))))
C(c(c(a(a(a(A(x))))))) → A1(a(a(b(c(b(B(x)))))))
B1(b(b(b(x)))) → C(c(a(a(x))))
C(c(c(a(a(a(A(x))))))) → B1(B(x))
C(c(c(b(b(b(B(x))))))) → A1(A(x))
C(c(b(b(b(B(x)))))) → A1(a(a(b(a(A(x))))))
A1(a(a(a(x)))) → B1(b(x))
C(c(c(a(a(a(A(x))))))) → C(b(B(x)))
A1(a(a(a(x)))) → C(b(b(x)))
C(c(c(a(a(a(A(x))))))) → A1(b(c(b(B(x)))))
C(c(c(c(x)))) → A1(b(x))
C(c(c(c(x)))) → A1(a(a(b(x))))
C(c(a(a(a(A(x)))))) → A1(a(a(b(b(B(x))))))
C(c(c(c(x)))) → B1(x)
B1(b(b(b(x)))) → A1(a(x))
C(c(c(b(b(b(B(x))))))) → A1(a(b(c(a(A(x))))))
B1(b(b(b(x)))) → A1(x)
A1(a(a(a(x)))) → C(c(b(b(x))))
C(c(b(b(b(B(x)))))) → A1(A(x))
C(c(a(a(a(A(x)))))) → B1(B(x))
C(c(c(a(a(a(A(x))))))) → B1(c(b(B(x))))
C(c(b(b(b(B(x)))))) → B1(a(A(x)))
C(c(a(a(a(A(x)))))) → B1(b(B(x)))
C(c(a(a(a(A(x)))))) → A1(a(b(b(B(x)))))
C(c(a(a(a(A(x)))))) → A1(b(b(B(x))))
C(c(c(b(b(b(B(x))))))) → C(a(A(x)))
B1(b(b(b(x)))) → C(a(a(x)))
C(c(c(b(b(b(B(x))))))) → A1(a(a(b(c(a(A(x)))))))
C(c(c(a(a(a(A(x))))))) → A1(a(b(c(b(B(x))))))

The TRS R consists of the following rules:

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

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

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

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

C(c(c(b(b(b(B(x))))))) → B1(c(a(A(x))))
A1(a(a(a(x)))) → B1(x)
C(c(c(c(x)))) → A1(a(b(x)))
C(c(b(b(b(B(x)))))) → A1(a(b(a(A(x)))))
C(c(b(b(b(B(x)))))) → A1(b(a(A(x))))
C(c(c(b(b(b(B(x))))))) → A1(b(c(a(A(x)))))
C(c(c(a(a(a(A(x))))))) → A1(a(a(b(c(b(B(x)))))))
B1(b(b(b(x)))) → C(c(a(a(x))))
C(c(c(a(a(a(A(x))))))) → B1(B(x))
C(c(c(b(b(b(B(x))))))) → A1(A(x))
C(c(b(b(b(B(x)))))) → A1(a(a(b(a(A(x))))))
A1(a(a(a(x)))) → B1(b(x))
C(c(c(a(a(a(A(x))))))) → C(b(B(x)))
A1(a(a(a(x)))) → C(b(b(x)))
C(c(c(a(a(a(A(x))))))) → A1(b(c(b(B(x)))))
C(c(c(c(x)))) → A1(b(x))
C(c(c(c(x)))) → A1(a(a(b(x))))
C(c(a(a(a(A(x)))))) → A1(a(a(b(b(B(x))))))
C(c(c(c(x)))) → B1(x)
B1(b(b(b(x)))) → A1(a(x))
C(c(c(b(b(b(B(x))))))) → A1(a(b(c(a(A(x))))))
B1(b(b(b(x)))) → A1(x)
A1(a(a(a(x)))) → C(c(b(b(x))))
C(c(b(b(b(B(x)))))) → A1(A(x))
C(c(a(a(a(A(x)))))) → B1(B(x))
C(c(c(a(a(a(A(x))))))) → B1(c(b(B(x))))
C(c(b(b(b(B(x)))))) → B1(a(A(x)))
C(c(a(a(a(A(x)))))) → B1(b(B(x)))
C(c(a(a(a(A(x)))))) → A1(a(b(b(B(x)))))
C(c(a(a(a(A(x)))))) → A1(b(b(B(x))))
C(c(c(b(b(b(B(x))))))) → C(a(A(x)))
B1(b(b(b(x)))) → C(a(a(x)))
C(c(c(b(b(b(B(x))))))) → A1(a(a(b(c(a(A(x)))))))
C(c(c(a(a(a(A(x))))))) → A1(a(b(c(b(B(x))))))

The TRS R consists of the following rules:

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

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

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

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

A1(a(a(a(x)))) → B1(x)
A1(a(a(a(x)))) → B1(b(x))
C(c(c(c(x)))) → A1(a(b(x)))
B1(b(b(b(x)))) → A1(x)
A1(a(a(a(x)))) → C(b(b(x)))
A1(a(a(a(x)))) → C(c(b(b(x))))
B1(b(b(b(x)))) → C(a(a(x)))
C(c(c(c(x)))) → A1(b(x))
B1(b(b(b(x)))) → C(c(a(a(x))))
C(c(c(c(x)))) → A1(a(a(b(x))))
B1(b(b(b(x)))) → A1(a(x))
C(c(c(c(x)))) → B1(x)

The TRS R consists of the following rules:

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

A1(a(a(a(x)))) → B1(x)
A1(a(a(a(x)))) → B1(b(x))
C(c(c(c(x)))) → A1(a(b(x)))
B1(b(b(b(x)))) → A1(x)
A1(a(a(a(x)))) → C(b(b(x)))
B1(b(b(b(x)))) → C(a(a(x)))
C(c(c(c(x)))) → A1(b(x))
B1(b(b(b(x)))) → C(c(a(a(x))))
B1(b(b(b(x)))) → A1(a(x))
C(c(c(c(x)))) → B1(x)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(A1(x1)) = x1   
POL(B(x1)) = 2·x1   
POL(B1(x1)) = 2·x1   
POL(C(x1)) = x1   
POL(a(x1)) = 1 + 2·x1   
POL(b(x1)) = 1 + 2·x1   
POL(c(x1)) = 1 + 2·x1   



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(a(x)))) → C(c(b(b(x)))) at position [0] we obtained the following new rules:

A1(a(a(a(b(b(b(x0))))))) → C(c(b(c(c(a(a(x0)))))))
A1(a(a(a(b(b(x0)))))) → C(c(c(c(a(a(x0))))))



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

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

A1(a(a(a(b(b(b(x0))))))) → C(c(b(c(c(a(a(x0)))))))
A1(a(a(a(b(b(x0)))))) → C(c(c(c(a(a(x0))))))
C(c(c(c(x)))) → A1(a(a(b(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(c(c(x)))) → A1(a(a(b(x)))) at position [0] we obtained the following new rules:

C(c(c(c(b(b(b(x0))))))) → A1(a(a(c(c(a(a(x0)))))))



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

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

A1(a(a(a(b(b(b(x0))))))) → C(c(b(c(c(a(a(x0)))))))
C(c(c(c(b(b(b(x0))))))) → A1(a(a(c(c(a(a(x0)))))))
A1(a(a(a(b(b(x0)))))) → C(c(c(c(a(a(x0))))))

The TRS R consists of the following rules:

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

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS

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

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

Q is empty.