Termination w.r.t. Q proof of TRS/D3312.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(x), not₁(y))
not₁(and₂(x, y)) -> or₂(not₁(x), not₁(y))
and₂(x, or₂(y, z)) -> or₂(and₂(x, y), and₂(x, z))
and₂(or₂(y, z), x) -> or₂(and₂(x, y), and₂(x, z))

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

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(y)
NOT₁(or₂(x, y)) -> NOT₁(x)
NOT₁(or₂(x, y)) -> NOT₁(y)
AND₂(x, or₂(y, z)) -> AND₂(x, z)
NOT₁(or₂(x, y)) -> AND₂(not₁(x), not₁(y))
AND₂(x, or₂(y, z)) -> AND₂(x, y)
AND₂(or₂(y, z), x) -> AND₂(x, z)
AND₂(or₂(y, z), x) -> AND₂(x, y)

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

NOT₁(and₂(x, y)) -> NOT₁(x)
NOT₁(and₂(x, y)) -> NOT₁(y)
NOT₁(or₂(x, y)) -> NOT₁(x)
NOT₁(or₂(x, y)) -> NOT₁(y)
AND₂(x, or₂(y, z)) -> AND₂(x, z)
NOT₁(or₂(x, y)) -> AND₂(not₁(x), not₁(y))
AND₂(x, or₂(y, z)) -> AND₂(x, y)
AND₂(or₂(y, z), x) -> AND₂(x, z)
AND₂(or₂(y, z), x) -> AND₂(x, y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

AND₂(x, or₂(y, z)) -> AND₂(x, z)
AND₂(x, or₂(y, z)) -> AND₂(x, y)
AND₂(or₂(y, z), x) -> AND₂(x, z)
AND₂(or₂(y, z), x) -> AND₂(x, y)

The TRS R consists of the following rules:

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

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.

AND₂(x, or₂(y, z)) -> AND₂(x, z)
AND₂(x, or₂(y, z)) -> AND₂(x, y)
AND₂(or₂(y, z), x) -> AND₂(x, z)
AND₂(or₂(y, z), x) -> AND₂(x, y)

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

POL(AND₂(x₁, x₂)) = 3·x₁ + 3·x₁·x₂ + 3·x₂
POL(or₂(x₁, x₂)) = 3 + 2·x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

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

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

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

POL(NOT₁(x₁)) = 3·x₁ + 3·x₁²
POL(and₂(x₁, x₂)) = 3 + x₁ + 2·x₂
POL(or₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

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

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.