Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/D33SLASH12.trs

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

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

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

          ↳ 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.

↳ 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 strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
NOT₁(x1) = x1
and₂(x1, x2) = and₂(x1, x2)
or₂(x1, x2) = or₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

trivial

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.