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

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

b(a(a(x))) → a(a(a(b(x))))
a(b(a(b(x)))) → b(b(a(x)))
b(a(b(a(A(x))))) → a(a(a(b(a(a(B(x)))))))
b(b(a(A(x)))) → a(a(a(b(a(B(x))))))
a(b(a(a(b(a(B(x))))))) → b(b(a(B(x))))
a(b(a(a(b(a(B(x))))))) → b(b(a(b(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 12 less nodes.

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

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

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

The TRS R consists of the following rules:

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


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(A(x))))) → A1(a(a(b(a(a(B(x))))))) at position [0] we obtained the following new rules:

B1(a(b(a(A(y0))))) → A1(a(a(a(a(a(b(B(y0))))))))



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

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

B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(x))) → A1(a(a(b(x))))
B1(a(b(a(A(x))))) → A1(b(a(a(B(x)))))
B1(a(a(x))) → A1(a(b(x)))
B1(a(a(x))) → B1(x)
B1(a(a(x))) → A1(b(x))
B1(a(b(a(A(y0))))) → A1(a(a(a(a(a(b(B(y0))))))))
A1(b(a(b(x)))) → B1(b(a(x)))
B1(a(b(a(A(x))))) → A1(a(b(a(a(B(x))))))

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(A(x))))) → A1(a(b(a(a(B(x)))))) at position [0] we obtained the following new rules:

B1(a(b(a(A(y0))))) → A1(a(a(a(a(b(B(y0)))))))



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

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

B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(x))) → A1(a(a(b(x))))
B1(a(b(a(A(x))))) → A1(b(a(a(B(x)))))
B1(a(b(a(A(y0))))) → A1(a(a(a(a(b(B(y0)))))))
B1(a(a(x))) → A1(a(b(x)))
B1(a(a(x))) → A1(b(x))
B1(a(a(x))) → B1(x)
A1(b(a(b(x)))) → B1(b(a(x)))

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(A(x))))) → A1(b(a(a(B(x))))) at position [0] we obtained the following new rules:

B1(a(b(a(A(y0))))) → A1(a(a(a(b(B(y0))))))



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

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

B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(x))) → A1(a(a(b(x))))
B1(a(a(x))) → A1(a(b(x)))
B1(a(a(x))) → A1(b(x))
B1(a(a(x))) → B1(x)
B1(a(b(a(A(y0))))) → A1(a(a(a(b(B(y0))))))
A1(b(a(b(x)))) → B1(b(a(x)))

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(x))) → A1(a(a(b(x)))) at position [0] we obtained the following new rules:

B1(a(a(a(a(b(a(B(x0)))))))) → A1(a(b(b(a(b(A(x0)))))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(a(b(a(a(B(x0))))))))))
B1(a(a(a(a(b(a(B(x0)))))))) → A1(a(b(b(a(B(x0))))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(b(a(A(x0)))))) → A1(a(a(a(a(a(b(a(B(x0)))))))))



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

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

B1(a(a(a(a(b(a(B(x0)))))))) → A1(a(b(b(a(b(A(x0)))))))
B1(a(a(a(a(b(a(B(x0)))))))) → A1(a(b(b(a(B(x0))))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(a(b(a(a(B(x0))))))))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(x))) → A1(a(b(x)))
B1(a(a(x))) → A1(b(x))
B1(a(a(x))) → B1(x)
A1(b(a(b(x)))) → B1(b(a(x)))
B1(a(a(b(a(A(x0)))))) → A1(a(a(a(a(a(b(a(B(x0)))))))))

The TRS R consists of the following rules:

b(a(a(x))) → a(a(a(b(x))))
a(b(a(b(x)))) → b(b(a(x)))
b(a(b(a(A(x))))) → a(a(a(b(a(a(B(x)))))))
b(b(a(A(x)))) → a(a(a(b(a(B(x))))))
a(b(a(a(b(a(B(x))))))) → b(b(a(B(x))))
a(b(a(a(b(a(B(x))))))) → b(b(a(b(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 3 less nodes.

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

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

B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(a(b(a(a(B(x0))))))))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(x))) → A1(a(b(x)))
B1(a(a(x))) → B1(x)
B1(a(a(x))) → A1(b(x))
A1(b(a(b(x)))) → B1(b(a(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(x))) → A1(a(b(x))) at position [0] we obtained the following new rules:

B1(a(a(b(a(A(x0)))))) → A1(a(a(a(a(b(a(B(x0))))))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(a(a(b(a(B(x0)))))))) → A1(b(b(a(B(x0)))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(b(a(a(B(x0)))))))))
B1(a(a(a(a(b(a(B(x0)))))))) → A1(b(b(a(b(A(x0))))))



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

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

B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(a(b(a(a(B(x0))))))))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(b(a(a(B(x0)))))))))
B1(a(a(a(a(b(a(B(x0)))))))) → A1(b(b(a(b(A(x0))))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(x))) → A1(b(x))
B1(a(a(b(a(A(x0)))))) → A1(a(a(a(a(b(a(B(x0))))))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(a(a(b(a(B(x0)))))))) → A1(b(b(a(B(x0)))))
B1(a(a(x))) → B1(x)
A1(b(a(b(x)))) → B1(b(a(x)))

The TRS R consists of the following rules:

b(a(a(x))) → a(a(a(b(x))))
a(b(a(b(x)))) → b(b(a(x)))
b(a(b(a(A(x))))) → a(a(a(b(a(a(B(x)))))))
b(b(a(A(x)))) → a(a(a(b(a(B(x))))))
a(b(a(a(b(a(B(x))))))) → b(b(a(B(x))))
a(b(a(a(b(a(B(x))))))) → b(b(a(b(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 3 less nodes.

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

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

B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(a(b(a(a(B(x0))))))))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(b(a(a(B(x0)))))))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(x))) → B1(x)
B1(a(a(x))) → A1(b(x))
A1(b(a(b(x)))) → B1(b(a(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(b(x)))) → B1(b(a(x))) at position [0] we obtained the following new rules:

A1(b(a(b(b(a(a(b(a(B(x0)))))))))) → B1(b(b(b(a(b(A(x0)))))))
A1(b(a(b(b(a(A(x0))))))) → B1(a(a(a(b(a(a(B(x0))))))))
A1(b(a(b(b(a(a(b(a(B(x0)))))))))) → B1(b(b(b(a(B(x0))))))
A1(b(a(b(a(x0))))) → B1(a(a(a(b(x0)))))
A1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(a(x0)))))



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

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

B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
A1(b(a(b(b(a(A(x0))))))) → B1(a(a(a(b(a(a(B(x0))))))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(a(b(a(a(B(x0))))))))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
A1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(a(x0)))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(b(a(a(B(x0)))))))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
A1(b(a(b(b(a(a(b(a(B(x0)))))))))) → B1(b(b(b(a(b(A(x0)))))))
B1(a(a(x))) → A1(b(x))
A1(b(a(b(b(a(a(b(a(B(x0)))))))))) → B1(b(b(b(a(B(x0))))))
A1(b(a(b(a(x0))))) → B1(a(a(a(b(x0)))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(x))) → B1(x)

The TRS R consists of the following rules:

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

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

B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
A1(b(a(b(b(a(A(x0))))))) → B1(a(a(a(b(a(a(B(x0))))))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(a(b(a(a(B(x0))))))))))
A1(b(a(b(a(x0))))) → B1(a(a(a(b(x0)))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(b(a(a(B(x0)))))))))
A1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(a(x0)))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(x))) → B1(x)
B1(a(a(x))) → A1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(a(b(a(a(B(x0)))))))))) at position [0] we obtained the following new rules:

B1(a(a(a(b(a(A(y0))))))) → A1(a(a(a(a(a(a(a(a(b(B(y0)))))))))))



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

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

A1(b(a(b(b(a(A(x0))))))) → B1(a(a(a(b(a(a(B(x0))))))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
A1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(a(x0)))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(b(a(a(B(x0)))))))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(x))) → A1(b(x))
B1(a(a(a(b(a(A(y0))))))) → A1(a(a(a(a(a(a(a(a(b(B(y0)))))))))))
A1(b(a(b(a(x0))))) → B1(a(a(a(b(x0)))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(x))) → B1(x)

The TRS R consists of the following rules:

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

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

B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
A1(b(a(b(b(a(A(x0))))))) → B1(a(a(a(b(a(a(B(x0))))))))
A1(b(a(b(a(x0))))) → B1(a(a(a(b(x0)))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(b(a(a(B(x0)))))))))
A1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(a(x0)))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(x))) → B1(x)
B1(a(a(x))) → A1(b(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(a(b(a(A(x0))))))) → A1(a(a(a(a(b(a(a(B(x0))))))))) at position [0] we obtained the following new rules:

B1(a(a(a(b(a(A(y0))))))) → A1(a(a(a(a(a(a(a(b(B(y0))))))))))



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

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

A1(b(a(b(b(a(A(x0))))))) → B1(a(a(a(b(a(a(B(x0))))))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
A1(b(a(b(a(x0))))) → B1(a(a(a(b(x0)))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
A1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(a(x0)))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(x))) → A1(b(x))
B1(a(a(x))) → B1(x)
B1(a(a(a(b(a(A(y0))))))) → A1(a(a(a(a(a(a(a(b(B(y0))))))))))

The TRS R consists of the following rules:

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

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

B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
A1(b(a(b(b(a(A(x0))))))) → B1(a(a(a(b(a(a(B(x0))))))))
A1(b(a(b(a(x0))))) → B1(a(a(a(b(x0)))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
A1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(a(x0)))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(x))) → B1(x)
B1(a(a(x))) → A1(b(x))

The TRS R consists of the following rules:

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


B1(a(a(a(a(x0))))) → A1(a(a(a(a(b(x0))))))
A1(b(a(b(b(a(A(x0))))))) → B1(a(a(a(b(a(a(B(x0))))))))
A1(b(a(b(a(x0))))) → B1(a(a(a(b(x0)))))
B1(a(a(a(b(x0))))) → A1(a(b(b(a(x0)))))
B1(a(a(a(a(x0))))) → A1(a(a(a(a(a(b(x0)))))))
B1(a(b(a(A(x))))) → B1(a(a(B(x))))
A1(b(a(b(b(a(b(x0))))))) → B1(b(b(b(a(x0)))))
B1(a(a(a(b(x0))))) → A1(b(b(a(x0))))
B1(a(a(x))) → B1(x)
B1(a(a(x))) → A1(b(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(A1(x1)) = 4 + (2)x_1   
POL(B(x1)) = 4 + (9/4)x_1   
POL(a(x1)) = 5/4 + x_1   
POL(B1(x1)) = 7/4 + (3)x_1   
POL(A(x1)) = 4 + (3/2)x_1   
POL(b(x1)) = (3/2)x_1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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



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

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

b(a(a(x))) → a(a(a(b(x))))
a(b(a(b(x)))) → b(b(a(x)))
b(a(b(a(A(x))))) → a(a(a(b(a(a(B(x)))))))
b(b(a(A(x)))) → a(a(a(b(a(B(x))))))
a(b(a(a(b(a(B(x))))))) → b(b(a(B(x))))
a(b(a(a(b(a(B(x))))))) → b(b(a(b(A(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.
We have reversed the following QTRS:
The set of rules R is

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

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

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

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

Q is empty.