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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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


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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

C(c(B(x0))) → C(B(x0))



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

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

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

The TRS R consists of the following rules:

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

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

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

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

C(a(x)) → C(c(x))
C(a(x)) → C(x)

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

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

Q is empty.

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

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

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

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

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

Q is empty.