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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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:

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

B(b(a(a(x0)))) → B(b(b(b(x0))))



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(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
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ QTRS Reverse
                  ↳ DependencyPairsProof
                    ↳ 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:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(A(x0)))) → B(b(a(A(x0)))) at position [0] we obtained the following new rules:

A1(b(a(A(y0)))) → B(b(b(b(b(A(y0))))))



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

A1(x) → B(x)
B(b(a(a(x)))) → A1(b(a(x)))
A1(a(A(x0))) → B(a(A(x0)))
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(b(a(A(y0)))) → B(b(b(b(b(A(y0))))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))

The TRS R consists of the following rules:

b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(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 1 less node.

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(A(x0))) → B(a(A(x0))) at position [0] we obtained the following new rules:

A1(a(A(y0))) → B(b(b(b(A(y0)))))



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

A1(a(A(y0))) → B(b(b(b(A(y0)))))
B(b(a(a(x)))) → A1(b(a(x)))
A1(x) → B(x)
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))

The TRS R consists of the following rules:

b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(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 1 less node.

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

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

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.