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:

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(y))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(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:

NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(y)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(not(not(y)))

The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(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:

NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(y)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(not(not(y)))

The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(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.

NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(y)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(not(not(y)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
NOT(x1) = NOT(x1)
and(x1, x2) = and(x1, x2)
not(x1) = x1
or(x1, x2) = or(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:

NOT₁ > [and₂, or_2]

Status:

and₂: [1,2]
or₂: [1,2]
NOT₁: [1]

The following usable rules [14] were oriented:

not(and(x, y)) → or(not(not(not(x))), not(not(not(y))))
not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(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.