Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/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.

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

      ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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)))

Used argument filtering: NOT₁(x1) = x1
and₂(x1, x2) = and₂(x1, x2)
not₁(x1) = x1
or₂(x1, x2) = or₂(x1, x2)
Used ordering: Quasi Precedence: [and_2, or_2]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ 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.