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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
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(c(y_0))) → A(c(y_0))
A(c(c(c(y_0)))) → A(c(c(y_0)))



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

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

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

The TRS R consists of the following rules:

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

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

C(b(b(c(y_0)))) → A(c(y_0))
C(b(b(c(c(y_0))))) → A(c(c(y_0)))
C(b(b(c(c(c(y_0)))))) → A(c(c(c(y_0))))



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

C1(c(A(x))) → B(c(C(x)))
C1(c(A(x))) → B(C(x))
C1(a(x)) → B(x)
C1(A(x)) → B(C(x))
C1(c(A(x))) → C1(b(c(C(x))))
C1(c(A(x))) → A1(c(c(b(C(x)))))
C1(c(A(x))) → A1(c(c(b(c(C(x))))))
C1(c(c(b(b(C(x)))))) → C1(A(x))
C1(c(b(b(C(x))))) → C1(A(x))
C1(c(A(x))) → C1(c(b(c(C(x)))))
C1(a(x)) → C1(b(x))
C1(a(x)) → A1(c(c(b(x))))
C1(c(A(x))) → C1(C(x))
C1(b(b(C(x)))) → C1(A(x))
C1(c(A(x))) → C1(b(C(x)))
A1(x) → B(x)
C1(a(x)) → C1(c(b(x)))
C1(A(x)) → B(c(C(x)))
B(b(c(x))) → A1(x)
C1(A(x)) → C1(C(x))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(A(x))) → C1(c(b(C(x))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(c(b(b(C(x))))) → C1(c(A(x)))

The TRS R consists of the following rules:

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

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

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

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

C1(c(A(x))) → B(c(C(x)))
C1(c(A(x))) → B(C(x))
C1(a(x)) → B(x)
C1(A(x)) → B(C(x))
C1(c(A(x))) → C1(b(c(C(x))))
C1(c(A(x))) → A1(c(c(b(C(x)))))
C1(c(A(x))) → A1(c(c(b(c(C(x))))))
C1(c(c(b(b(C(x)))))) → C1(A(x))
C1(c(b(b(C(x))))) → C1(A(x))
C1(c(A(x))) → C1(c(b(c(C(x)))))
C1(a(x)) → C1(b(x))
C1(a(x)) → A1(c(c(b(x))))
C1(c(A(x))) → C1(C(x))
C1(b(b(C(x)))) → C1(A(x))
C1(c(A(x))) → C1(b(C(x)))
A1(x) → B(x)
C1(a(x)) → C1(c(b(x)))
C1(A(x)) → B(c(C(x)))
B(b(c(x))) → A1(x)
C1(A(x)) → C1(C(x))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(A(x))) → C1(c(b(C(x))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(c(b(b(C(x))))) → C1(c(A(x)))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ QDPToSRSProof
                        ↳ QTRS
                          ↳ QTRS Reverse
                            ↳ QTRS
                              ↳ QTRS Reverse
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
QDP
                                        ↳ UsableRulesProof
                                        ↳ UsableRulesProof
                                      ↳ QDP
                              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(x) → B(x)
B(b(c(x))) → A1(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ QDPToSRSProof
                        ↳ QTRS
                          ↳ QTRS Reverse
                            ↳ QTRS
                              ↳ QTRS Reverse
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                        ↳ UsableRulesProof
QDP
                                        ↳ UsableRulesProof
                                      ↳ QDP
                              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(x) → B(x)
B(b(c(x))) → A1(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ QDPToSRSProof
                        ↳ QTRS
                          ↳ QTRS Reverse
                            ↳ QTRS
                              ↳ QTRS Reverse
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                        ↳ UsableRulesProof
QDP
                                            ↳ UsableRulesReductionPairsProof
                                      ↳ QDP
                              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(x) → B(x)
B(b(c(x))) → A1(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

A1(x) → B(x)
B(b(c(x))) → A1(x)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A1(x1)) = 2 + x1   
POL(B(x1)) = 1 + x1   
POL(b(x1)) = 2 + x1   
POL(c(x1)) = 2 + 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ QDPToSRSProof
                        ↳ QTRS
                          ↳ QTRS Reverse
                            ↳ QTRS
                              ↳ QTRS Reverse
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ UsableRulesReductionPairsProof
QDP
                                                ↳ PisEmptyProof
                                      ↳ QDP
                              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(a(x)) → C1(b(x))
C1(c(b(b(C(x))))) → C1(c(A(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(b(b(C(x))))) → C1(c(A(x))) at position [0] we obtained the following new rules:

C1(c(b(b(C(x0))))) → C1(b(C(x0)))
C1(c(b(b(C(x0))))) → C1(b(c(C(x0))))



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

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

C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(b(b(C(x0))))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(c(b(b(C(x0))))) → C1(b(C(x0)))
C1(a(x)) → C1(b(x))

The TRS R consists of the following rules:

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

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

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

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

C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(c(c(b(b(C(x)))))) → C1(c(A(x)))
C1(a(x)) → C1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x)))))) → C1(c(A(x))) at position [0] we obtained the following new rules:

C1(c(c(b(b(C(x0)))))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(b(C(x0)))



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

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

C1(c(c(b(b(C(x0)))))) → C1(b(c(C(x0))))
C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(a(x)) → C1(b(x))
C1(c(c(b(b(C(x0)))))) → C1(b(C(x0)))

The TRS R consists of the following rules:

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

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

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

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

C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(a(x)) → C1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(x)) → C1(b(x)) at position [0] we obtained the following new rules:

C1(a(b(c(x0)))) → C1(a(x0))



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

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

C1(a(x)) → C1(c(b(x)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(a(b(c(x0)))) → C1(a(x0))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(x)) → C1(c(b(x))) at position [0] we obtained the following new rules:

C1(a(b(C(x0)))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(c(a(x0)))



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

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

C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x)))))) → C1(c(c(A(x))))
C1(a(b(C(x0)))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x)))))) → C1(c(c(A(x)))) at position [0] we obtained the following new rules:

C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(c(c(b(b(C(x0)))))) → C1(c(b(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(c(b(c(C(x0)))))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))



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

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

C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(a(b(C(x0)))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))
C1(c(c(b(b(C(x0)))))) → C1(c(b(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(c(b(c(C(x0)))))

The TRS R consists of the following rules:

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

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

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

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

C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(a(b(C(x0)))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(b(C(x0)))) → C1(c(A(x0))) at position [0] we obtained the following new rules:

C1(a(b(C(x0)))) → C1(b(c(C(x0))))
C1(a(b(C(x0)))) → C1(b(C(x0)))



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

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

C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(C(x0)))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))
C1(a(b(C(x0)))) → C1(b(C(x0)))

The TRS R consists of the following rules:

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

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

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

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

C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(c(c(b(b(C(x0)))))) → C1(c(A(x0)))
C1(a(b(c(x0)))) → C1(a(x0))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x0)))))) → C1(c(A(x0))) at position [0] we obtained the following new rules:

C1(c(c(b(b(C(x0)))))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(b(C(x0)))



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

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

C1(c(c(b(b(C(x0)))))) → C1(b(c(C(x0))))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(c(x0)))) → C1(a(x0))
C1(c(c(b(b(C(x0)))))) → C1(b(C(x0)))

The TRS R consists of the following rules:

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

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

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

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

C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0))))))
C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(c(x0)))) → C1(a(x0))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(C(x0)))))) at position [0] we obtained the following new rules:

C1(c(c(b(b(C(y0)))))) → C1(b(c(c(b(C(y0))))))



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

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

C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(c(x0)))) → C1(a(x0))
C1(c(c(b(b(C(y0)))))) → C1(b(c(c(b(C(y0))))))

The TRS R consists of the following rules:

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

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

C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0)))))))
C1(a(b(c(x0)))) → C1(a(x0))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(c(c(b(b(C(x0)))))) → C1(a(c(c(b(c(C(x0))))))) at position [0] we obtained the following new rules:

C1(c(c(b(b(C(y0)))))) → C1(b(c(c(b(c(C(y0)))))))



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

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

C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(c(c(b(b(C(y0)))))) → C1(b(c(c(b(c(C(y0)))))))
C1(a(b(c(x0)))) → C1(a(x0))

The TRS R consists of the following rules:

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

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

C1(a(b(c(x0)))) → C1(c(a(x0)))
C1(a(b(c(x0)))) → C1(a(x0))

The TRS R consists of the following rules:

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

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.