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.