((A + B)·C)′ — De Morgan's laws
coreproduct of a sum, negated
Answer
A′ · B′ + C′
Why this example is worth doing
A negated product whose first factor is itself a sum, so the first application yields (A + B)′ + C̄ and the second turns the leading term into Ā·B̄. Expanding gives a two-term sum of products. The page uses this to show that De Morgan and distribution commute for the purposes of getting to a normal form — you may push all negations in first and then flatten, or flatten first — but that pushing negations in first is far less error-prone, which is why the tool's own normaliser does it in that order.
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 | 1 |
| 5 | 1 | 0 | 1 | 0 |
| 6 | 1 | 1 | 0 | 1 |
| 7 | 1 | 1 | 1 | 0 |