(A·B)′ — De Morgan's laws
introNAND expands to a sum
Answer
A′ + B′
Why this example is worth doing
The first law in its bare form. The bar over a product becomes a sum of barred variables, and the operator flips — that flip is the entire content of the theorem and the thing that gets dropped. The page animates the negation moving inward one node at a time rather than presenting before and after, because the mechanical rule students need is per-node: break the bar, change the sign, negate each operand. It also shows the resulting NAND-equals-inverted-input-OR symbol, which is the same statement in IEEE 91 notation.
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)′A′ + B′— De Morgan swaps the operator as the bar passes through it.
Truth table
| # | A | B | F |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 2 | 1 | 0 | 1 |
| 3 | 1 | 1 | 0 |