(A + B)′ — Boolean simplification
coreDe Morgan; NOR expands to Ā·B̄
Answer
A′ · B′
Why this example is worth doing
The single most common error in first-year Boolean algebra is writing (A + B)′ = Ā + B̄. It is wrong on three of four rows, and the page shows exactly which. Negation does not distribute over OR; it swaps the operator as it passes through. The result Ā·B̄ is why a NOR gate is drawn as an AND gate with inverted inputs in the alternative IEEE symbol set, a connection the gate pages pick up and one that makes the law stick better than any amount of algebra.
Try your own input in the Boolean algebra simplifier. Simplify any expression to its minimal SOP and POS, one named law at a time.
How the answer is reached
Step by step
(A + B)′A′ · B′— De Morgan — (A + B)' = A'·B'
(A + B)′ simplifies to A′ · B′ in 1 step. That is a proven minimum (1 term, 2 literals), not a best effort. It is the only minimum-cost form.
Quine–McCluskey
| # | Term |
|---|---|
| 1 | A′ · B′ |
| # | Cover |
|---|---|
| 1 | A′ · B′ |
1 term(s), 2 literal(s); 1 essential prime implicant(s).
Truth table
| # | A | B | F |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 2 | 1 | 0 | 0 |
| 3 | 1 | 1 | 0 |