XNOR expanded to SOP — XNOR gate
exam standardthe complement pair of XOR's SOP
Answer
A′ · B′ + A · B
Why this example is worth doing
As a sum of products XNOR is A·B + Ā·B̄, the two agreeing minterms, and like XOR it will not reduce further. Setting it beside the XOR expansion shows that the two functions partition the four minterms between them with nothing shared and nothing left over — which is what being complements means at the level of index sets, and is the cleanest way to see it.
Try your own input in the XNOR gate. Truth table, symbol and algebraic form for A ⊙ B, the equality detector.
How the answer is reached
Step by step
(A ⊕ B)′(A · B′ + A′ · B)′— Definition of XOR — A ⊕ B = A·B' + A'·B
(A · B′ + A′ · B)′(A · B′)′ · (A′ · B)′— De Morgan — (A + B)' = A'·B'
(A · B′)′ · (A′ · B)′(A′ + B′′) · (A′ · B)′— De Morgan — (A·B)' = A' + B'
(A′ + B′′) · (A′ · B)′(A′ + B) · (A′ · B)′— Involution (double negation) — (A')' = A
(A′ + B) · (A′ · B)′(A′ + B) · (A′′ + B′)— De Morgan — (A·B)' = A' + B'
(A′ + B) · (A′′ + B′)(A′ + B) · (A + B′)— Involution (double negation) — (A')' = A
(A′ + B) · (A + B′)A′ · (A + B′) + B · (A + B′)— Distributive — A · (B + C) = A · B + A · C
A′ · (A + B′) + B · (A + B′)A′ · A + A′ · B′ + B · (A + B′)— Distributive — A · (B + C) = A · B + A · C
A′ · A + A′ · B′ + B · (A + B′)0 + A′ · B′ + B · (A + B′)— Complement — A · A′ = 0
0 + A′ · B′ + B · (A + B′)A′ · B′ + B · (A + B′)— Identity — A + 0 = A
A′ · B′ + B · (A + B′)A′ · B′ + B · A + B · B′— Distributive — A · (B + C) = A · B + A · C
A′ · B′ + B · A + B · B′A′ · B′ + B · A + 0— Complement — A · A′ = 0
A′ · B′ + B · A + 0A′ · B′ + B · A— Identity — A + 0 = A
A′ · B′ + B · AA′ · B′ + A · B— Commutative — A · B = B · A; A + B = B + A
(A ⊕ B)′ simplifies to A′ · B′ + A · B in 14 steps. That is a proven minimum (2 terms, 4 literals), not a best effort. It is the only minimum-cost form.
Quine–McCluskey
| # | Term |
|---|---|
| 1 | A′ · B′ |
| 2 | A · B |
| # | Cover |
|---|---|
| 1 | A′ · B′ + A · B |
2 term(s), 4 literal(s); 2 essential prime implicant(s).
Truth table
| # | A | B | F |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 2 | 1 | 0 | 0 |
| 3 | 1 | 1 | 1 |