Termination w.r.t. Q proof of TRS/D3320.trs

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:

not₁(not₁(x)) -> x
not₁(or₂(x, y)) -> and₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))
not₁(and₂(x, y)) -> or₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

not₁(not₁(x)) -> x
not₁(or₂(x, y)) -> and₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))
not₁(and₂(x, y)) -> or₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

NOT₁(and₂(x, y)) -> NOT₁(x)
NOT₁(and₂(x, y)) -> NOT₁(not₁(x))
NOT₁(and₂(x, y)) -> NOT₁(y)
NOT₁(or₂(x, y)) -> NOT₁(x)
NOT₁(or₂(x, y)) -> NOT₁(not₁(not₁(x)))
NOT₁(or₂(x, y)) -> NOT₁(y)
NOT₁(and₂(x, y)) -> NOT₁(not₁(not₁(y)))
NOT₁(or₂(x, y)) -> NOT₁(not₁(x))
NOT₁(and₂(x, y)) -> NOT₁(not₁(y))
NOT₁(or₂(x, y)) -> NOT₁(not₁(y))
NOT₁(and₂(x, y)) -> NOT₁(not₁(not₁(x)))
NOT₁(or₂(x, y)) -> NOT₁(not₁(not₁(y)))

The TRS R consists of the following rules:

not₁(not₁(x)) -> x
not₁(or₂(x, y)) -> and₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))
not₁(and₂(x, y)) -> or₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

NOT₁(and₂(x, y)) -> NOT₁(x)
NOT₁(and₂(x, y)) -> NOT₁(not₁(x))
NOT₁(and₂(x, y)) -> NOT₁(y)
NOT₁(or₂(x, y)) -> NOT₁(x)
NOT₁(or₂(x, y)) -> NOT₁(not₁(not₁(x)))
NOT₁(or₂(x, y)) -> NOT₁(y)
NOT₁(and₂(x, y)) -> NOT₁(not₁(not₁(y)))
NOT₁(or₂(x, y)) -> NOT₁(not₁(x))
NOT₁(and₂(x, y)) -> NOT₁(not₁(y))
NOT₁(or₂(x, y)) -> NOT₁(not₁(y))
NOT₁(and₂(x, y)) -> NOT₁(not₁(not₁(x)))
NOT₁(or₂(x, y)) -> NOT₁(not₁(not₁(y)))

The TRS R consists of the following rules:

not₁(not₁(x)) -> x
not₁(or₂(x, y)) -> and₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))
not₁(and₂(x, y)) -> or₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

NOT₁(and₂(x, y)) -> NOT₁(x)
NOT₁(and₂(x, y)) -> NOT₁(not₁(x))
NOT₁(and₂(x, y)) -> NOT₁(y)
NOT₁(or₂(x, y)) -> NOT₁(x)
NOT₁(or₂(x, y)) -> NOT₁(not₁(not₁(x)))
NOT₁(or₂(x, y)) -> NOT₁(y)
NOT₁(and₂(x, y)) -> NOT₁(not₁(not₁(y)))
NOT₁(or₂(x, y)) -> NOT₁(not₁(x))
NOT₁(and₂(x, y)) -> NOT₁(not₁(y))
NOT₁(or₂(x, y)) -> NOT₁(not₁(y))
NOT₁(and₂(x, y)) -> NOT₁(not₁(not₁(x)))
NOT₁(or₂(x, y)) -> NOT₁(not₁(not₁(y)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(NOT₁(x₁)) = x₁
POL(and₂(x₁, x₂)) = 1 + x₁ + x₂
POL(not₁(x₁)) = x₁
POL(or₂(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [14] were oriented:

not₁(and₂(x, y)) -> or₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))
not₁(or₂(x, y)) -> and₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))
not₁(not₁(x)) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

not₁(not₁(x)) -> x
not₁(or₂(x, y)) -> and₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))
not₁(and₂(x, y)) -> or₂(not₁(not₁(not₁(x))), not₁(not₁(not₁(y))))

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.