(A ⊕ B)′ — De Morgan's laws
exam standardXOR complement is XNOR
Answer
A′ ⊕ B
Why this example is worth doing
De Morgan does not apply directly to XOR, and that is the point of including it. You must first expand A ⊕ B to A·B̄ + Ā·B, then negate, then push the negations in, arriving at A·B + Ā·B̄ — which is XNOR. The page flags the shortcut students try instead, complementing either operand: Ā ⊕ B also equals XNOR, which is true and often useful, but complementing both operands leaves XOR unchanged. Those two facts together are worth more than the derivation.
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 | 0 |
| 2 | 1 | 0 | 0 |
| 3 | 1 | 1 | 1 |