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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.

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

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

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 18 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:

A1(b(b(x))) → A1(b(a(x)))
A1(b(b(x))) → A1(x)

The TRS R consists of the following rules:

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

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

A1(b(b(b(a(A(x0)))))) → A1(b(a(b(a(b(B(x0)))))))
A1(b(b(a(A(x0))))) → A1(b(a(b(a(B(x0))))))
A1(b(b(A(x0)))) → A1(b(B(x0)))
A1(b(b(a(b(B(x0)))))) → A1(b(b(b(b(b(A(x0)))))))
A1(b(b(A(x0)))) → A1(b(b(B(x0))))
A1(b(b(b(b(x0))))) → A1(b(a(b(a(x0)))))
A1(b(b(a(x0)))) → A1(b(b(b(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:

A1(b(b(b(a(A(x0)))))) → A1(b(a(b(a(b(B(x0)))))))
A1(b(b(a(A(x0))))) → A1(b(a(b(a(B(x0))))))
A1(b(b(A(x0)))) → A1(b(B(x0)))
A1(b(b(A(x0)))) → A1(b(b(B(x0))))
A1(b(b(a(b(B(x0)))))) → A1(b(b(b(b(b(A(x0)))))))
A1(b(b(b(b(x0))))) → A1(b(a(b(a(x0)))))
A1(b(b(a(x0)))) → A1(b(b(b(b(x0)))))
A1(b(b(x))) → A1(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

A1(b(b(A(x0)))) → A1(b(b(B(x0))))
A1(b(b(a(b(B(x0)))))) → A1(b(b(b(b(b(A(x0)))))))
A1(b(b(b(b(x0))))) → A1(b(a(b(a(x0)))))
A1(b(b(a(x0)))) → A1(b(b(b(b(x0)))))
A1(b(b(x))) → A1(x)

The TRS R consists of the following rules:

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

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.