Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/D33SLASH20.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 strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
NOT₁(x1) = NOT₁(x1)
and₂(x1, x2) = and₂(x1, x2)
not₁(x1) = x1
or₂(x1, x2) = or₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

[NOT₁, and₂, or_2]

The following usable rules [14] were oriented:

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))))

↳ 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.