(Ā + B̄)′ — De Morgan's laws
exam standardDe Morgan then involution
Answer
A · B
Why this example is worth doing
Applying the law to an expression whose operands are already complemented, which produces a double negation on each and then cancels it: the answer is A·B. This is the reverse direction, turning a sum of complements back into a product, and it is the step that recovers an AND gate from a NOR-based implementation. The page pairs it with the bubble-pushing diagram in which two inversions meet on a wire and vanish, making the algebraic cancellation and the graphical one the same event.
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 | 0 |
| 1 | 0 | 1 | 0 |
| 2 | 1 | 0 | 0 |
| 3 | 1 | 1 | 1 |