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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

a(a(a(x1))) → x1
b(a(b(b(x1)))) → a(b(b(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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(B(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + 2·x1   



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

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

B(a(b(b(x1)))) → B(b(b(a(x1))))

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ 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(b(b(b(a(B(x)))))) → B1(b(a(b(B(x)))))
B1(b(a(b(x)))) → B1(b(a(x)))
B1(b(a(b(x)))) → A(b(b(b(a(x)))))
A(a(b(b(a(B(x)))))) → B1(B(x))
B1(b(b(b(a(B(x)))))) → A(b(b(b(a(b(B(x)))))))
B1(b(a(b(x)))) → B1(b(b(a(x))))
B1(b(a(b(x)))) → B1(a(x))
B1(b(a(b(x)))) → A(x)
B1(b(b(b(a(B(x)))))) → A(b(B(x)))
A(a(b(b(a(B(x)))))) → B1(b(B(x)))
B1(b(b(b(a(B(x)))))) → B1(b(b(a(b(B(x))))))
B1(b(b(b(a(B(x)))))) → B1(a(b(B(x))))
B1(b(b(b(a(B(x)))))) → B1(B(x))

The TRS R consists of the following rules:

a(a(a(x))) → x
b(b(a(b(x)))) → a(b(b(b(a(x)))))
b(b(b(b(a(B(x)))))) → a(b(b(b(a(b(B(x)))))))
a(a(b(b(a(B(x)))))) → b(b(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 8 less nodes.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(B(x1)) = x1   
POL(B1(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   



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

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

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

The TRS R consists of the following rules:

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

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

B1(b(b(b(a(B(y0)))))) → B1(a(b(b(b(a(B(y0)))))))



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

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

B1(b(a(b(x)))) → B1(b(b(a(x))))
B1(b(b(b(a(B(y0)))))) → B1(a(b(b(b(a(B(y0)))))))

The TRS R consists of the following rules:

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

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

B1(b(a(b(x)))) → B1(b(b(a(x))))

The TRS R consists of the following rules:

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

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

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



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

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

The TRS R consists of the following rules:

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

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

B1(b(a(b(b(x0))))) → B1(a(b(b(b(a(x0))))))
B1(b(a(b(a(a(x0)))))) → B1(b(b(x0)))

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

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

Q is empty.

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

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

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

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

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

Q is empty.