(Ā·B + A·B̄)′ — De Morgan's laws
edge casenegating a full SOP expression
Answer
(A + B′) · (A′ + B)
Why this example is worth doing
A complete sum of products under one bar, which is the shape that actually turns up when a student is asked for F′ having just minimised F. Two applications of the law give a product of sums, and expanding it recovers a sum of products for the complement. The page makes the strategic point alongside: if you want the complement, it is usually faster to re-minimise the OFF-set from the truth table than to push a bar through a finished expression, and it is much less error-prone.
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′)′(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 | 0 |
| 2 | 1 | 0 | 0 |
| 3 | 1 | 1 | 1 |