Then: not (p and q)= 1

That is, it is neither p nor q= 1.

If: p and q=0

Then: not (p and q)= 1

That is, non-p is not q= 1.

Then: (non-p non-q) and (non-p non-q)= 1.

Namely: (non-P, non-Q, non-P) and (non-P, non-Q, non-q)= 1.

Simplified as: (non-p, non-q) union (non-p, non-q)= 1.

De: (p and q)=0

Therefore, the proposition is reasonable.