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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

A(b(c(x0))) → B(x0)



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

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

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

The TRS R consists of the following rules:

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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:

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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:

B1(a(x)) → B1(b(x))
B1(a(x)) → B1(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(x)) → B1(b(x)) at position [0] we obtained the following new rules:

B1(a(b(A(x0)))) → B1(a(c(b(b(c(B(x0)))))))
B1(a(B(x0))) → B1(A(x0))
B1(a(A(x0))) → B1(a(c(b(B(x0)))))
B1(a(a(x0))) → B1(a(c(b(b(x0)))))
B1(a(b(x0))) → B1(a(x0))
B1(a(A(x0))) → B1(A(x0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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:

B1(a(B(x0))) → B1(A(x0))
B1(a(x)) → B1(x)
B1(a(b(x0))) → B1(a(x0))
B1(a(A(x0))) → B1(A(x0))
B1(a(b(A(x0)))) → B1(a(c(b(b(c(B(x0)))))))
B1(a(A(x0))) → B1(a(c(b(B(x0)))))
B1(a(a(x0))) → B1(a(c(b(b(x0)))))

The TRS R consists of the following rules:

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

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

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

B1(a(x)) → B1(x)
B1(a(b(x0))) → B1(a(x0))
B1(a(b(A(x0)))) → B1(a(c(b(b(c(B(x0)))))))
B1(a(A(x0))) → B1(a(c(b(B(x0)))))
B1(a(a(x0))) → B1(a(c(b(b(x0)))))

The TRS R consists of the following rules:

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

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.