Termination w.r.t. Q proof of TRS/D3320.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:

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

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.