So, the question of the validity of \(¬(¬Q) ≡ Q\) always reduces to one of the two cases I already checked in the truth table. (Note that for this argument to be valid, the same Q must be substituted ...So, the question of the validity of \(¬(¬Q) ≡ Q\) always reduces to one of the two cases I already checked in the truth table. (Note that for this argument to be valid, the same Q must be substituted for p in every position where it occurs.) While this argument may be “obvious,” it is not exactly a proof, but for now we will just accept the validity of the following theorem:
