Termination w.r.t. Q proof of /tmp/tmpD5gbhv/16.srs

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

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x₁) ) = x₁ + 1

POL( b(x₁) ) = max{0, -1}

POL( B(x₁) ) = 1

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                          ↳ SemLabProof2

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ RuleRemovalProof

                                          ↳ SemLabProof2

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Strictly oriented rules of the TRS R:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ RuleRemovalProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                          ↳ SemLabProof2

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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 25 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ RuleRemovalProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ AND

                                                          ↳ QDP

                                                            ↳ UsableRulesReductionPairsProof

                                                          ↳ QDP

                                          ↳ SemLabProof2

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

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

The following rules are removed from R:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ RuleRemovalProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ AND

                                                          ↳ QDP

                                                            ↳ UsableRulesReductionPairsProof

                                                              ↳ QDP

                                                                ↳ PisEmptyProof

                                                          ↳ QDP

                                          ↳ SemLabProof2

  ↳ QTRS Reverse

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ RuleRemovalProof

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ AND

                                                          ↳ QDP

                                                          ↳ QDP

                                          ↳ SemLabProof2

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ QDPOrderProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(a(x₁)) = 4 + x_1
POL(A(x₁)) = x_1
POL(b(x₁)) = (1/2)x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                                ↳ QDP

                                                  ↳ PisEmptyProof

  ↳ QTRS Reverse

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

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

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

Q is empty.