Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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.


B(b(x1)) → B(a(x1))
B(a(a(a(b(x1))))) → A(x1)
B(b(x1)) → A(b(a(x1)))
B(b(x1)) → A(x1)
B(a(a(a(b(x1))))) → A(a(x1))
B(a(a(a(b(x1))))) → A(a(a(x1)))
B(a(a(a(b(x1))))) → A(a(b(a(a(a(x1))))))
B(a(a(a(b(x1))))) → A(b(a(a(a(x1)))))
B(a(a(a(b(x1))))) → B(a(a(a(x1))))
The remaining pairs can at least be oriented weakly.

B(a(a(a(b(x1))))) → A(a(a(b(a(a(a(x1)))))))
A(a(a(b(b(x1))))) → B(b(b(x1)))
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(B(x1)) = 12 + 8·x1   
POL(a(x1)) = 2 + 2·x1   
POL(b(x1)) = 14 + 8·x1   

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
QDP
              ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
QDP
                  ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ 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(b(b(x1))))) → b(b(b(x1)))
b(a(a(a(b(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(x1))) → a(a(x1))
b(b(x1)) → a(b(a(x1)))
A(a(a(b(b(a(a(a(b(x0))))))))) → B(b(a(a(a(b(a(a(a(x0)))))))))
A(a(a(b(b(x0))))) → B(a(b(a(x0))))
B(a(a(a(b(b(b(x0))))))) → A(a(a(b(b(b(b(x0)))))))
B(a(a(a(b(a(a(b(b(x0))))))))) → A(a(a(b(a(a(b(b(b(x0)))))))))
B(a(a(a(b(b(x0)))))) → A(a(a(a(a(a(b(a(a(a(x0))))))))))
B(a(a(a(b(a(a(x0))))))) → A(a(a(b(a(a(a(a(x0))))))))
B(a(a(a(b(x0))))) → A(a(a(b(a(a(x0))))))
B(a(a(a(b(a(b(b(x0)))))))) → A(a(a(b(a(b(b(b(x0))))))))
B(a(a(a(b(a(x0)))))) → A(a(a(b(a(a(a(x0)))))))
A(a(a(b(b(b(x0)))))) → B(b(a(b(a(x0)))))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
QTRS
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

b(b(a(a(a(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(x)))))))
a(a(a(x))) → a(a(x))
b(b(x)) → a(b(a(x)))
b(a(a(a(b(b(a(a(A(x))))))))) → a(a(a(b(a(a(a(b(B(x)))))))))
b(b(a(a(A(x))))) → a(b(a(B(x))))
b(b(b(a(a(a(B(x))))))) → b(b(b(b(a(a(A(x)))))))
b(b(a(a(b(a(a(a(B(x))))))))) → b(b(b(a(a(b(a(a(A(x)))))))))
b(b(a(a(a(B(x)))))) → a(a(a(b(a(a(a(a(a(A(x))))))))))
a(a(b(a(a(a(B(x))))))) → a(a(a(a(b(a(a(A(x))))))))
b(a(a(a(B(x))))) → a(a(b(a(a(A(x))))))
b(b(a(b(a(a(a(B(x)))))))) → b(b(b(a(b(a(a(A(x))))))))
a(b(a(a(a(B(x)))))) → a(a(a(b(a(a(A(x)))))))
b(b(b(a(a(A(x)))))) → a(b(a(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 61 less nodes.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

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

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

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

The TRS R consists of the following rules:

a(a(b(a(a(a(B(x))))))) → a(a(a(a(b(a(a(A(x))))))))
a(b(a(a(a(B(x)))))) → a(a(a(b(a(a(A(x)))))))
a(a(a(x))) → a(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 0 SCCs with 1 less node.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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


B1(a(a(a(b(b(a(a(A(x))))))))) → B1(a(a(a(b(B(x))))))
B1(b(a(a(a(x))))) → B1(b(x))
B1(a(a(a(b(x))))) → B1(a(a(a(x))))
B1(b(a(a(a(B(x)))))) → B1(a(a(a(a(a(A(x)))))))
B1(b(a(b(a(a(a(B(x)))))))) → B1(b(a(b(a(a(A(x)))))))
B1(b(b(a(a(a(B(x))))))) → B1(b(b(a(a(A(x))))))
B1(b(x)) → B1(a(x))
B1(b(a(a(a(x))))) → B1(x)
B1(b(a(a(b(a(a(a(B(x))))))))) → B1(b(a(a(b(a(a(A(x))))))))
B1(b(b(a(a(a(B(x))))))) → B1(b(a(a(A(x)))))
The remaining pairs can at least be oriented weakly.

B1(b(a(b(a(a(a(B(x)))))))) → B1(b(b(a(b(a(a(A(x))))))))
B1(b(a(a(a(x))))) → B1(b(b(x)))
B1(b(a(a(b(a(a(a(B(x))))))))) → B1(b(b(a(a(b(a(a(A(x)))))))))
B1(b(b(a(a(a(B(x))))))) → B1(b(b(b(a(a(A(x)))))))
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(B(x1)) = 8 + x1   
POL(B1(x1)) = 8·x1   
POL(a(x1)) = 4 + x1   
POL(b(x1)) = 12 + x1   

The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(b(a(a(a(B(y0))))))) → B1(b(a(b(a(a(a(A(y0)))))))) at position [0] we obtained the following new rules:

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

b(b(a(a(a(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(x)))))))
a(a(a(x))) → a(a(x))
b(b(x)) → a(b(a(x)))
b(a(a(a(b(b(a(a(A(x))))))))) → a(a(a(b(a(a(a(b(B(x)))))))))
b(b(a(a(A(x))))) → a(b(a(B(x))))
b(b(b(a(a(a(B(x))))))) → b(b(b(b(a(a(A(x)))))))
b(b(a(a(b(a(a(a(B(x))))))))) → b(b(b(a(a(b(a(a(A(x)))))))))
b(b(a(a(a(B(x)))))) → a(a(a(b(a(a(a(a(a(A(x))))))))))
a(a(b(a(a(a(B(x))))))) → a(a(a(a(b(a(a(A(x))))))))
b(a(a(a(B(x))))) → a(a(b(a(a(A(x))))))
b(b(a(b(a(a(a(B(x)))))))) → b(b(b(a(b(a(a(A(x))))))))
a(b(a(a(a(B(x)))))) → a(a(a(b(a(a(A(x)))))))
b(b(b(a(a(A(x)))))) → a(b(a(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 5 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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


B1(b(a(a(a(x0))))) → B1(a(b(a(x0))))
B1(b(a(a(b(a(a(a(B(y0))))))))) → B1(a(b(a(a(a(b(a(a(A(y0))))))))))
B1(b(a(a(a(a(a(a(B(x0))))))))) → B1(a(a(a(b(a(a(a(a(a(A(x0)))))))))))
The remaining pairs can at least be oriented weakly.

B1(b(a(a(a(b(a(a(a(B(x0)))))))))) → B1(b(b(b(b(a(a(A(x0))))))))
B1(b(a(a(a(b(a(b(a(a(a(B(x0)))))))))))) → B1(b(b(b(b(a(b(a(a(A(x0))))))))))
B1(b(a(a(a(b(a(a(b(a(a(a(B(x0))))))))))))) → B1(b(b(b(b(a(a(b(a(a(A(x0)))))))))))
B1(b(a(a(a(b(a(a(a(B(x0)))))))))) → B1(b(a(a(a(b(a(a(a(a(a(A(x0))))))))))))
B1(b(a(a(a(b(a(a(a(x0))))))))) → B1(b(b(b(b(x0)))))
B1(b(a(a(a(b(b(a(a(a(B(x0))))))))))) → B1(b(b(b(b(b(a(a(A(x0)))))))))
B1(b(a(a(a(b(x0)))))) → B1(b(a(b(a(x0)))))
B1(b(a(a(a(a(a(a(b(b(a(a(A(x0))))))))))))) → B1(b(a(a(a(b(a(a(a(b(B(x0)))))))))))
B1(b(a(a(a(a(a(b(a(a(a(B(x0)))))))))))) → B1(b(b(b(a(a(b(a(a(A(x0))))))))))
B1(b(a(a(a(a(a(a(x0)))))))) → B1(b(b(b(x0))))
B1(b(a(a(a(a(a(a(b(x0))))))))) → B1(b(a(a(a(b(a(a(a(x0)))))))))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( B1(x1) ) = x1


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


POL( b(x1) ) = 1


POL( B(x1) ) = 0


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



The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.