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(a(b(b(x1)))) → b(b(a(a(b(a(x1))))))

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

a(x) → x
b(b(a(a(x)))) → a(b(a(a(b(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:

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(a(b(b(x1)))) → b(b(a(a(b(a(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ Narrowing
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

A(a(b(b(a(b(b(x0))))))) → A(a(b(b(b(a(a(b(a(x0)))))))))
A(a(b(b(x0)))) → A(a(b(x0)))
A(a(b(b(y0)))) → A(b(a(y0)))



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

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

A(a(b(b(y0)))) → A(b(a(y0)))
A(a(b(b(a(b(b(x0))))))) → A(a(b(b(b(a(a(b(a(x0)))))))))
A(a(b(b(x0)))) → A(a(b(x0)))
A(a(b(b(x1)))) → A(x1)

The TRS R consists of the following rules:

a(x1) → x1
a(a(b(b(x1)))) → b(b(a(a(b(a(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
QDP
                  ↳ QDPToSRSProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(a(b(b(x1)))) → b(b(a(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
QTRS
                      ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(x) → x
b(b(a(a(x)))) → a(b(a(a(b(b(x))))))
b(b(a(b(b(a(A(x))))))) → a(b(a(a(b(b(b(a(A(x)))))))))
b(b(a(A(x)))) → b(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 6 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ 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)))) → B(x)
B(b(a(b(b(a(A(x))))))) → B(a(a(b(b(b(a(A(x))))))))
B(b(a(a(x)))) → B(a(a(b(b(x)))))
B(b(a(a(x)))) → B(b(x))
B(b(a(b(b(a(A(x))))))) → B(b(b(a(A(x)))))

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.