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

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

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

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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

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

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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

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

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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
                                                                      ↳ SemLabProof
                                                                      ↳ SemLabProof2

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.c: 1
B: 0
a: 0
A: 0
b: x0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
A.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → B.0(b.0(a.0(a.0(b.0(b.0(x1))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → B.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → A.0(a.0(a.0(a.1(b.1(b.1(x1))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.0(x0)))))))) → B.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(b.0(a.0(a.0(x0))))))))) → B.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))))
B.0(b.0(a.0(a.0(x1)))) → B.0(x1)
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → B.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
A.0(a.0(a.0(a.0(b.0(b.0(b.0(a.0(a.1(x0))))))))) → B.0(a.0(a.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → A.0(a.0(a.0(a.0(b.0(b.0(x1))))))
A.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → B.0(b.0(a.0(a.1(b.1(b.1(x1))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → B.1(x1)
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.1(x0))))))))) → A.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → B.0(x1)
B.0(b.0(a.0(a.0(x1)))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.1(x0))))))))) → A.0(a.0(a.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.1(x0)))))))) → B.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.0(b.0(a.0(a.1(x1)))) → B.1(x1)
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.1(x1)))) → B.1(b.1(x1))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.1(x0))))))))) → A.0(a.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → B.1(b.1(x1))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))

The TRS R consists of the following rules:

a.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → b.0(b.0(a.0(a.0(b.0(b.0(x1))))))
b.0(b.0(a.0(a.1(x1)))) → a.0(a.1(b.1(b.1(b.1(b.1(x1))))))
a.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → b.0(b.0(a.0(a.1(b.1(b.1(x1))))))
b.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.0(b.0(b.0(x1))))))))))
b.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.1(b.1(b.1(x1))))))))))
b.0(b.0(a.0(a.0(x1)))) → a.0(a.0(b.0(b.0(b.0(b.0(x1))))))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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
                                                                      ↳ SemLabProof
                                                                        ↳ QDP
                                                                          ↳ DependencyGraphProof
                                                                      ↳ SemLabProof2

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

B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
A.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → B.0(b.0(a.0(a.0(b.0(b.0(x1))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → B.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → A.0(a.0(a.0(a.1(b.1(b.1(x1))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.0(x0)))))))) → B.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(b.0(a.0(a.0(x0))))))))) → B.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))))
B.0(b.0(a.0(a.0(x1)))) → B.0(x1)
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → B.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
A.0(a.0(a.0(a.0(b.0(b.0(b.0(a.0(a.1(x0))))))))) → B.0(a.0(a.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → A.0(a.0(a.0(a.0(b.0(b.0(x1))))))
A.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → B.0(b.0(a.0(a.1(b.1(b.1(x1))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → B.1(x1)
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.1(x0))))))))) → A.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → B.0(x1)
B.0(b.0(a.0(a.0(x1)))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.1(x0))))))))) → A.0(a.0(a.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.1(x0)))))))) → B.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.0(b.0(a.0(a.1(x1)))) → B.1(x1)
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.1(x1)))) → B.1(b.1(x1))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.1(x0))))))))) → A.0(a.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → B.1(b.1(x1))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))

The TRS R consists of the following rules:

a.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → b.0(b.0(a.0(a.0(b.0(b.0(x1))))))
b.0(b.0(a.0(a.1(x1)))) → a.0(a.1(b.1(b.1(b.1(b.1(x1))))))
a.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → b.0(b.0(a.0(a.1(b.1(b.1(x1))))))
b.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.0(b.0(b.0(x1))))))))))
b.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.1(b.1(b.1(x1))))))))))
b.0(b.0(a.0(a.0(x1)))) → a.0(a.0(b.0(b.0(b.0(b.0(x1))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 6 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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
                                                                      ↳ SemLabProof
                                                                        ↳ QDP
                                                                          ↳ DependencyGraphProof
                                                                            ↳ QDP
                                                                              ↳ RuleRemovalProof
                                                                      ↳ SemLabProof2

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

B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
A.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → B.0(b.0(a.0(a.0(b.0(b.0(x1))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → A.0(a.0(a.0(a.1(b.1(b.1(x1))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.0(x0)))))))) → B.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(b.0(a.0(a.0(x0))))))))) → B.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))))
B.0(b.0(a.0(a.0(x1)))) → B.0(x1)
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → B.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → A.0(a.0(a.0(a.0(b.0(b.0(x1))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
A.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → B.0(b.0(a.0(a.1(b.1(b.1(x1))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → B.1(x1)
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → B.0(x1)
B.0(b.0(a.0(a.0(x1)))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.1(x0)))))))) → B.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.0(b.0(a.0(a.1(x1)))) → B.1(x1)
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.1(x1)))) → B.1(b.1(x1))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → B.1(b.1(x1))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))

The TRS R consists of the following rules:

a.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → b.0(b.0(a.0(a.0(b.0(b.0(x1))))))
b.0(b.0(a.0(a.1(x1)))) → a.0(a.1(b.1(b.1(b.1(b.1(x1))))))
a.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → b.0(b.0(a.0(a.1(b.1(b.1(x1))))))
b.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.0(b.0(b.0(x1))))))))))
b.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.1(b.1(b.1(x1))))))))))
b.0(b.0(a.0(a.0(x1)))) → a.0(a.0(b.0(b.0(b.0(b.0(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.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → B.1(x1)
B.0(b.0(a.0(a.1(x1)))) → B.1(x1)
B.0(b.0(a.0(a.1(x1)))) → B.1(b.1(x1))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → B.1(b.1(x1))


Used ordering: POLO with Polynomial interpretation [25]:

POL(A.0(x1)) = 1 + x1   
POL(B.0(x1)) = 1 + x1   
POL(B.1(x1)) = 1 + x1   
POL(a.0(x1)) = x1   
POL(a.1(x1)) = 1 + x1   
POL(b.0(x1)) = x1   
POL(b.1(x1)) = x1   
POL(c.0(x1)) = x1   
POL(c.1(x1)) = x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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
                                                                      ↳ SemLabProof
                                                                        ↳ QDP
                                                                          ↳ DependencyGraphProof
                                                                            ↳ QDP
                                                                              ↳ RuleRemovalProof
                                                                                ↳ QDP
                                                                                  ↳ DependencyGraphProof
                                                                      ↳ SemLabProof2

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

B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
A.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → B.0(b.0(a.0(a.0(b.0(b.0(x1))))))
B.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → A.0(a.0(a.0(a.1(b.1(b.1(x1))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.0(x0)))))))) → B.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(b.0(a.0(a.0(x0))))))))) → B.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(x1)))) → B.0(x1)
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → B.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → A.0(a.0(a.0(a.0(b.0(b.0(x1))))))
B.1(b.1(c.1(c.0(a.0(a.0(b.0(a.0(a.0(x0))))))))) → A.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
A.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → B.0(b.0(a.0(a.1(b.1(b.1(x1))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → B.0(x1)
B.0(b.0(a.0(a.0(x1)))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.1(x0)))))))) → B.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.1(x0)))))))) → A.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))
B.1(b.1(c.1(c.0(a.0(a.0(a.0(a.0(x0)))))))) → A.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))

The TRS R consists of the following rules:

a.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → b.0(b.0(a.0(a.0(b.0(b.0(x1))))))
b.0(b.0(a.0(a.1(x1)))) → a.0(a.1(b.1(b.1(b.1(b.1(x1))))))
a.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → b.0(b.0(a.0(a.1(b.1(b.1(x1))))))
b.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.0(b.0(b.0(x1))))))))))
b.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.1(b.1(b.1(x1))))))))))
b.0(b.0(a.0(a.0(x1)))) → a.0(a.0(b.0(b.0(b.0(b.0(x1))))))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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
                                                                      ↳ SemLabProof
                                                                        ↳ QDP
                                                                          ↳ DependencyGraphProof
                                                                            ↳ QDP
                                                                              ↳ RuleRemovalProof
                                                                                ↳ QDP
                                                                                  ↳ DependencyGraphProof
                                                                                    ↳ QDP
                                                                      ↳ SemLabProof2

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

B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → B.0(b.0(a.0(a.0(b.0(b.0(x1))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → B.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.0(x0)))))))) → B.0(a.0(a.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(x1)))) → B.0(b.0(x1))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(a.0(a.1(x0)))))))) → B.0(a.0(a.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))
A.0(a.0(a.0(a.0(b.0(b.0(b.0(a.0(a.0(x0))))))))) → B.0(a.0(a.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(x1)))) → B.0(x1)
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))))))
B.0(b.0(a.0(a.0(a.0(a.0(x0)))))) → B.0(a.0(a.0(b.0(b.0(b.0(b.0(x0)))))))
B.0(b.0(a.0(a.0(a.0(a.1(x0)))))) → A.0(a.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.1(x0))))))) → A.0(a.0(b.0(b.0(b.0(a.0(a.1(b.1(b.1(b.1(b.1(x0)))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → A.0(b.0(b.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))))
B.0(b.0(a.0(a.0(b.0(a.0(a.0(x0))))))) → B.0(b.0(a.0(a.0(b.0(b.0(b.0(b.0(x0))))))))

The TRS R consists of the following rules:

a.0(a.0(a.0(a.0(b.0(b.0(x1)))))) → b.0(b.0(a.0(a.0(b.0(b.0(x1))))))
b.0(b.0(a.0(a.1(x1)))) → a.0(a.1(b.1(b.1(b.1(b.1(x1))))))
a.0(a.0(a.0(a.1(b.1(b.1(x1)))))) → b.0(b.0(a.0(a.1(b.1(b.1(x1))))))
b.1(b.1(c.1(c.0(a.0(a.0(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.0(b.0(b.0(x1))))))))))
b.1(b.1(c.1(c.0(a.0(a.1(x1)))))) → c.1(c.1(c.1(c.0(a.0(a.0(a.0(a.1(b.1(b.1(x1))))))))))
b.0(b.0(a.0(a.0(x1)))) → a.0(a.0(b.0(b.0(b.0(b.0(x1))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used. Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models). 

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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
                                                                      ↳ SemLabProof
                                                                      ↳ SemLabProof2
                                                                        ↳ QDP
                                                                          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

a(a(a(a(b(b(x1)))))) → b(b(a(a(b(b(x1))))))
b(b(a(a(x1)))) → a(a(b(b(b(b(x1))))))
b(b(c(c(a(a(x1)))))) → c(c(c(c(a(a(a(a(b(b(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(a(a(a(a(x0)))))) → A(a(b(b(a(a(b(b(b(b(x0))))))))))
B(b(a(a(b(a(a(x0))))))) → B(b(a(a(b(b(b(b(x0))))))))
B(b(a(a(a(a(x0)))))) → B(b(a(a(b(b(b(b(x0))))))))
B(b(a(a(b(a(a(x0))))))) → A(a(b(b(b(a(a(b(b(b(b(x0)))))))))))
B(b(a(a(x1)))) → B(b(x1))
B(b(a(a(x1)))) → B(x1)
B(b(a(a(a(a(x0)))))) → A(b(b(a(a(b(b(b(b(x0)))))))))
B(b(a(a(b(a(a(x0))))))) → B(b(b(a(a(b(b(b(b(x0)))))))))
B(b(a(a(a(a(x0)))))) → B(a(a(b(b(b(b(x0)))))))
B(b(a(a(b(a(a(x0))))))) → A(b(b(b(a(a(b(b(b(b(x0))))))))))
The remaining pairs can at least be oriented weakly.

A(a(a(a(b(b(a(a(x0)))))))) → B(a(a(a(a(b(b(b(b(x0)))))))))
A(a(a(a(b(b(b(a(a(x0))))))))) → B(a(a(b(a(a(b(b(b(b(x0))))))))))
A(a(a(a(b(b(x1)))))) → B(b(a(a(b(b(x1))))))
Used ordering: Polynomial Order [21,25] with Interpretation:

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


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


POL( b(x1) ) = x1


POL( B(x1) ) = x1


POL( a(x1) ) = x1 + 1



The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ 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
                                                                      ↳ SemLabProof
                                                                      ↳ SemLabProof2
                                                                        ↳ QDP
                                                                          ↳ QDPOrderProof
                                                                            ↳ QDP
                                                                              ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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 3 less nodes.