Convert the following expression into NOR-only 3

Question

I have the expression B'C + A'B and I must implement it using NORs only.

I have worked out:

I don't have enough breadboard room for this. How do I simplify this?

Answer 1

Your circuit looks mostly correct, save you have inputs of B' and A' each ouble entry into nor gates. You should not have negations as inputs, and further, since $B'\veebar B'\equiv B$ , this is redundant.

$\lnot P\equiv P\veebar P\\ \\P+Q\equiv(P\veebar Q)\veebar (P\veebar Q)\\PQ \equiv (P\veebar P)\veebar (Q\veebar Q)\\P'Q = P\veebar (Q\veebar Q)$

So $B'C+A'B \equiv ((B\veebar (C\veebar C))\veebar(A\veebar (B\veebar B)))\veebar((B\veebar (C\veebar C))\veebar(A\veebar (B\veebar B)))$

$$\raise{2ex}\boxed{\raise{2ex}\boxed{\raise{4ex}B\veebar \raise{2ex}\boxed{\raise{2ex}C\veebar \raise{2ex}C}}\veebar\raise{2ex}\boxed{\raise{4ex}A\veebar \raise{2ex}\boxed{\raise{2ex}B\veebar \raise{2ex}B}}}\veebar\raise{2ex}\boxed{\raise{2ex}\boxed{\raise{4ex}B\veebar\raise{2ex}\boxed{\raise{2ex}C\veebar\raise{2ex}C}}\veebar\raise{2ex}\boxed{\raise{4ex}A\veebar\raise{2ex}\boxed{\raise{2ex}B\veebar \raise{2ex}B}}}$$