Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

A(a(a(x1))) → B(a(x1))


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(a(a(x1))) → b(b(a(x1)))
a(b(a(x1))) → b(b(a(x1)))
b(a(b(x1))) → a(a(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
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ 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(a(a(x1))) → b(b(a(x1)))
a(b(a(x1))) → b(b(a(x1)))
b(a(b(x1))) → a(a(b(x1)))
B(a(b(a(b(x0))))) → A(a(a(a(b(x0)))))
A(b(a(b(a(x0))))) → B(b(b(b(a(x0)))))
A(b(a(b(x0)))) → B(a(a(b(x0))))
A(a(a(b(x0)))) → B(a(a(b(x0))))
B(a(b(a(x0)))) → A(b(b(a(x0))))
A(a(a(b(a(x0))))) → B(b(b(b(a(x0)))))
A(a(a(a(a(x0))))) → B(b(b(b(a(x0)))))
A(b(a(a(a(x0))))) → B(b(b(b(a(x0)))))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ 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:

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ 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(a(x))) → b(b(a(x)))
a(b(a(x))) → b(b(a(x)))
b(a(b(x))) → a(a(b(x)))
B(a(b(a(b(x))))) → A(a(a(a(b(x)))))
A(b(a(b(a(x))))) → B(b(b(b(a(x)))))
A(b(a(b(x)))) → B(a(a(b(x))))
A(a(a(b(x)))) → B(a(a(b(x))))
B(a(b(a(x)))) → A(b(b(a(x))))
A(a(a(b(a(x))))) → B(b(b(b(a(x)))))
A(a(a(a(a(x))))) → B(b(b(b(a(x)))))
A(b(a(a(a(x))))) → B(b(b(b(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(a(a(b(A(x))))) → A1(b(b(b(B(x)))))
B1(a(b(x))) → B1(a(a(x)))
A1(b(a(b(A(x))))) → B1(b(B(x)))
A1(b(a(b(A(x))))) → B1(B(x))
B1(a(b(x))) → A1(a(x))
B1(a(b(A(x)))) → A1(B(x))
B1(a(b(x))) → A1(x)
A1(b(a(b(A(x))))) → B1(b(b(B(x))))
A1(a(a(b(A(x))))) → B1(B(x))
A1(a(a(a(A(x))))) → B1(b(B(x)))
A1(a(a(a(A(x))))) → B1(B(x))
A1(b(a(B(x)))) → B1(b(A(x)))
B1(a(a(A(x)))) → B1(a(a(B(x))))
A1(b(a(x))) → A1(b(b(x)))
A1(b(a(a(A(x))))) → B1(B(x))
A1(b(a(B(x)))) → A1(b(b(A(x))))
B1(a(a(A(x)))) → A1(B(x))
B1(a(a(A(x)))) → A1(a(B(x)))
B1(a(b(a(B(x))))) → A1(a(A(x)))
A1(a(a(x))) → B1(x)
B1(a(b(A(x)))) → B1(a(a(B(x))))
A1(b(a(x))) → B1(b(x))
A1(b(a(a(A(x))))) → A1(b(b(b(B(x)))))
A1(a(a(x))) → B1(b(x))
A1(a(a(b(A(x))))) → B1(b(B(x)))
A1(a(a(b(A(x))))) → B1(b(b(B(x))))
A1(b(a(x))) → B1(x)
A1(b(a(a(A(x))))) → B1(b(b(B(x))))
A1(b(a(b(A(x))))) → A1(b(b(b(B(x)))))
A1(b(a(B(x)))) → B1(A(x))
A1(b(a(a(A(x))))) → B1(b(B(x)))
B1(a(b(a(B(x))))) → A1(a(a(A(x))))
B1(a(b(a(B(x))))) → B1(a(a(a(A(x)))))
A1(a(a(x))) → A1(b(b(x)))
A1(a(a(a(A(x))))) → B1(b(b(B(x))))
B1(a(b(a(B(x))))) → A1(A(x))
A1(a(a(a(A(x))))) → A1(b(b(b(B(x)))))
B1(a(b(A(x)))) → A1(a(B(x)))

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(a(a(x))) → a(b(b(x)))
a(b(a(x))) → a(b(b(x)))
b(a(b(x))) → b(a(a(x)))
b(a(b(a(B(x))))) → b(a(a(a(A(x)))))
a(b(a(b(A(x))))) → a(b(b(b(B(x)))))
b(a(b(A(x)))) → b(a(a(B(x))))
b(a(a(A(x)))) → b(a(a(B(x))))
a(b(a(B(x)))) → a(b(b(A(x))))
a(b(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(b(A(x))))) → a(b(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 27 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 1 + x1   
POL(A1(x1)) = 2 + 2·x1   
POL(B(x1)) = 1 + x1   
POL(B1(x1)) = 2·x1   
POL(a(x1)) = 2 + x1   
POL(b(x1)) = 2 + x1   



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

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

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

The TRS R consists of the following rules:

a(a(a(x))) → a(b(b(x)))
a(b(a(x))) → a(b(b(x)))
b(a(b(x))) → b(a(a(x)))
b(a(b(a(B(x))))) → b(a(a(a(A(x)))))
a(b(a(b(A(x))))) → a(b(b(b(B(x)))))
b(a(b(A(x)))) → b(a(a(B(x))))
b(a(a(A(x)))) → b(a(a(B(x))))
a(b(a(B(x)))) → a(b(b(A(x))))
a(b(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(b(A(x))))) → a(b(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 2 SCCs with 1 less node.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


A1(a(a(x))) → A1(b(b(x)))
The remaining pairs can at least be oriented weakly.

A1(b(a(x))) → A1(b(b(x)))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = x1 + 1


POL( b(x1) ) = max{0, -1}


POL( A1(x1) ) = max{0, x1 - 1}


POL( B(x1) ) = x1


POL( a(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



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

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

A1(b(a(x))) → A1(b(b(x)))

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(a(a(x))) → a(b(b(x)))
a(b(a(x))) → a(b(b(x)))
b(a(b(x))) → b(a(a(x)))
b(a(b(a(B(x))))) → b(a(a(a(A(x)))))
a(b(a(b(A(x))))) → a(b(b(b(B(x)))))
b(a(b(A(x)))) → b(a(a(B(x))))
b(a(a(A(x)))) → b(a(a(B(x))))
a(b(a(B(x)))) → a(b(b(A(x))))
a(b(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(b(A(x))))) → a(b(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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


A1(b(a(a(b(a(B(x0))))))) → A1(b(b(a(a(a(A(x0)))))))
A1(b(a(a(b(x0))))) → A1(b(b(a(a(x0)))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(A1(x1)) = (1/2)x_1   
POL(B(x1)) = 0   
POL(a(x1)) = 1/4   
POL(A(x1)) = 0   
POL(b(x1)) = (1/4)x_1   
The value of delta used in the strict ordering is 3/128.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ QDPToSRSProof
                        ↳ QTRS
                          ↳ QTRS Reverse
                            ↳ QTRS
                              ↳ QTRS Reverse
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ AND
                                              ↳ QDP
                                                ↳ QDPOrderProof
                                                  ↳ QDP
                                                    ↳ Narrowing
                                                      ↳ QDP
                                                        ↳ DependencyGraphProof
                                                          ↳ QDP
                                                            ↳ QDPOrderProof
QDP
                                                                ↳ PisEmptyProof
                                              ↳ QDP
                              ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

B1(a(b(x))) → B1(a(a(x)))

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(a(a(x))) → a(b(b(x)))
a(b(a(x))) → a(b(b(x)))
b(a(b(x))) → b(a(a(x)))
b(a(b(a(B(x))))) → b(a(a(a(A(x)))))
a(b(a(b(A(x))))) → a(b(b(b(B(x)))))
b(a(b(A(x)))) → b(a(a(B(x))))
b(a(a(A(x)))) → b(a(a(B(x))))
a(b(a(B(x)))) → a(b(b(A(x))))
a(b(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(b(A(x))))) → a(b(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 7 less nodes.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.B: 0
a: 0
A: 0
B1: 0
b: 1
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

B1.0(a.1(b.1(b.0(a.0(x0))))) → B1.0(a.0(a.1(b.1(b.0(x0)))))
B1.0(a.1(b.0(a.1(x0)))) → B1.0(a.1(b.1(b.1(x0))))
B1.0(a.1(b.0(a.0(a.0(x0))))) → B1.0(a.0(a.1(b.1(b.0(x0)))))
B1.0(a.1(b.1(b.0(a.1(x0))))) → B1.0(a.0(a.1(b.1(b.1(x0)))))
B1.0(a.1(b.0(a.0(x0)))) → B1.0(a.1(b.1(b.0(x0))))
B1.0(a.1(b.0(a.0(a.1(x0))))) → B1.0(a.0(a.1(b.1(b.1(x0)))))

The TRS R consists of the following rules:

a.0(a.0(a.1(b.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
b.0(a.1(b.0(A.1(x)))) → b.0(a.0(a.0(B.1(x))))
a.1(b.0(a.0(B.0(x)))) → a.1(b.1(b.0(A.0(x))))
a.0(a.0(a.0(x))) → a.1(b.1(b.0(x)))
b.0(a.0(a.0(A.1(x)))) → b.0(a.0(a.0(B.1(x))))
a.1(b.0(a.0(a.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
b.0(a.1(b.0(a.0(B.0(x))))) → b.0(a.0(a.0(a.0(A.0(x)))))
b.0(a.1(b.0(x))) → b.0(a.0(a.0(x)))
a.1(b.0(a.0(x))) → a.1(b.1(b.0(x)))
a.0(a.0(a.0(a.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.1(b.0(a.1(b.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
a.0(a.0(a.1(x))) → a.1(b.1(b.1(x)))
a.1(b.0(a.1(b.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.1(b.0(a.0(a.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.0(a.0(a.0(a.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
b.0(a.1(b.0(A.0(x)))) → b.0(a.0(a.0(B.0(x))))
b.0(a.1(b.1(x))) → b.0(a.0(a.1(x)))
a.1(b.0(a.1(x))) → a.1(b.1(b.1(x)))
a.1(b.0(a.0(B.1(x)))) → a.1(b.1(b.0(A.1(x))))
b.0(a.0(a.0(A.0(x)))) → b.0(a.0(a.0(B.0(x))))
b.0(a.1(b.0(a.0(B.1(x))))) → b.0(a.0(a.0(a.0(A.1(x)))))
a.0(a.0(a.1(b.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))

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

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

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

B1.0(a.1(b.1(b.0(a.0(x0))))) → B1.0(a.0(a.1(b.1(b.0(x0)))))
B1.0(a.1(b.0(a.1(x0)))) → B1.0(a.1(b.1(b.1(x0))))
B1.0(a.1(b.0(a.0(a.0(x0))))) → B1.0(a.0(a.1(b.1(b.0(x0)))))
B1.0(a.1(b.1(b.0(a.1(x0))))) → B1.0(a.0(a.1(b.1(b.1(x0)))))
B1.0(a.1(b.0(a.0(x0)))) → B1.0(a.1(b.1(b.0(x0))))
B1.0(a.1(b.0(a.0(a.1(x0))))) → B1.0(a.0(a.1(b.1(b.1(x0)))))

The TRS R consists of the following rules:

a.0(a.0(a.1(b.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
b.0(a.1(b.0(A.1(x)))) → b.0(a.0(a.0(B.1(x))))
a.1(b.0(a.0(B.0(x)))) → a.1(b.1(b.0(A.0(x))))
a.0(a.0(a.0(x))) → a.1(b.1(b.0(x)))
b.0(a.0(a.0(A.1(x)))) → b.0(a.0(a.0(B.1(x))))
a.1(b.0(a.0(a.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
b.0(a.1(b.0(a.0(B.0(x))))) → b.0(a.0(a.0(a.0(A.0(x)))))
b.0(a.1(b.0(x))) → b.0(a.0(a.0(x)))
a.1(b.0(a.0(x))) → a.1(b.1(b.0(x)))
a.0(a.0(a.0(a.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.1(b.0(a.1(b.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
a.0(a.0(a.1(x))) → a.1(b.1(b.1(x)))
a.1(b.0(a.1(b.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.1(b.0(a.0(a.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.0(a.0(a.0(a.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
b.0(a.1(b.0(A.0(x)))) → b.0(a.0(a.0(B.0(x))))
b.0(a.1(b.1(x))) → b.0(a.0(a.1(x)))
a.1(b.0(a.1(x))) → a.1(b.1(b.1(x)))
a.1(b.0(a.0(B.1(x)))) → a.1(b.1(b.0(A.1(x))))
b.0(a.0(a.0(A.0(x)))) → b.0(a.0(a.0(B.0(x))))
b.0(a.1(b.0(a.0(B.1(x))))) → b.0(a.0(a.0(a.0(A.1(x)))))
a.0(a.0(a.1(b.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 6 less nodes.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.