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

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

The TRS R consists of the following rules:

a(a(a(x1))) → b(x1)
b(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
      ↳ ForwardInstantiation
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

B(c(a(y_0))) → A(a(a(y_0)))



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

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

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

The TRS R consists of the following rules:

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

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

B(c(a(a(y_0)))) → A(a(a(y_0)))



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

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

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

The TRS R consists of the following rules:

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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(a(a(x))) → b(x)
b(c(x)) → c(c(a(a(a(a(x))))))
A(a(a(x))) → B(x)
B(c(x)) → A(a(a(x)))
B(c(a(x))) → A(a(a(x)))
B(c(a(a(x)))) → A(a(a(x)))
B(c(x)) → A(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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 11 less nodes.

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

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

C(b(x)) → C(c(x))
C(b(x)) → C(x)

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

C(b(B(x0))) → C(B(x0))



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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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(a(a(x))) → b(x)
b(c(x)) → c(c(a(a(a(a(x))))))
A(a(a(x))) → B(x)
B(c(x)) → A(a(a(x)))
B(c(a(x))) → A(a(a(x)))
B(c(a(a(x)))) → A(a(a(x)))
B(c(x)) → A(a(a(a(x))))

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.