(A + B·C)′ — De Morgan's laws
coretwo applications, outer then inner
Answer
A′ · (B′ + C′)
Why this example is worth doing
A nested case, where the law has to be applied twice: once at the top to split the OR, then again inside to handle the negated product. The result is Ā·(B̄ + C̄). The page's emphasis is on order — work outward in, never inside out — and on the bookkeeping trick of drawing the bar as a bracket so that its scope is unambiguous. Scope errors, not sign errors, are what actually break these problems once the expression is more than one level deep.
Try your own input in the De Morgan’s laws. Push a negation through any expression and see both forms side by side.
How the answer is reached
Negation pushed inwards
(A + B · C)′A′ · (B′ + C′)— De Morgan swaps the operator as the bar passes through it.
Truth table
| # | A | B | C | F |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 1 | 1 | 0 |
| 4 | 1 | 0 | 0 | 0 |
| 5 | 1 | 0 | 1 | 0 |
| 6 | 1 | 1 | 0 | 0 |
| 7 | 1 | 1 | 1 | 0 |