\((A + B)\cdot\overline{AB}\) is equivalent to:
2011
\((A + B)\cdot\overline{AB}\) is equivalent to:
Answer: A. \(A ⊕ B\) — Concept — Two standard results decide any question of this shape. First, De Morgan’s theorem: the complement of a product is the sum of the complements,…
- A.
\(A ⊕ B\)
- B.
\(A ⊙ B\)
- C.
\((A ⊕ B) ⊙ A\)
- D.
\((A ⊙ B) ⊕ A\)
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Correct answer: A
Concept — Two standard results decide any question of this shape. First, De Morgan’s theorem: the complement of a product is the sum of the complements, \(\overline{XY} = \bar{X} + \bar{Y}\). Second, the definition of the exclusive-OR function: \(X ⊕ Y\) is 1 exactly when its two inputs differ, and its canonical sum-of-products form is \(X\bar{Y} + \bar{X}Y\). So whenever a Boolean expression can be reduced to that particular sum of products, the expression is an exclusive-OR; if it reduces to \(XY + \bar{X}\bar{Y}\) it is an exclusive-NOR instead.
Application — reduce the given expression \((A + B)\cdot\overline{AB}\) step by step:
Apply De Morgan’s theorem to the complemented product: \(\overline{AB} = \bar{A} + \bar{B}\).
Substitute it back into the expression: \((A + B)(\bar{A} + \bar{B})\).
Expand the product term by term: \(A\bar{A} + A\bar{B} + B\bar{A} + B\bar{B}\).
Apply the complement law \(X\bar{X} = 0\) to the first and last terms: \(A\bar{A} = 0\) and \(B\bar{B} = 0\), so both vanish under the OR.
What remains is \(A\bar{B} + \bar{A}B\), which is precisely the canonical sum-of-products form of the exclusive-OR stated in the concept above. Hence \((A + B)\cdot\overline{AB} = A ⊕ B\).
Cross-check — the same conclusion from the full truth table. The output of the given expression is 1 only on the two rows where A and B differ, which is exactly the exclusive-OR column:
A | B | A + B | \(\overline{AB}\) | \((A+B)\overline{AB}\) | \(A ⊕ B\) |
|---|---|---|---|---|---|
0 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 |
Contrast — how the other candidate expressions behave, so the near misses are clear:
\(A ⊙ B\) (exclusive-NOR) has the output column 1, 0, 0, 1 over the rows above — it is the complement of the exclusive-OR, high when the inputs agree rather than when they differ.
\((A ⊕ B) ⊙ A\) collapses to \(\bar{B}\): with \(A = 0\) it is \((0 ⊕ B) ⊙ 0 = B ⊙ 0 = \bar{B}\), and with \(A = 1\) it is \((1 ⊕ B) ⊙ 1 = \bar{B} ⊙ 1 = \bar{B}\). Its output column is 1, 0, 1, 0 and it does not depend on A at all.
\((A ⊙ B) ⊕ A\) also collapses to \(\bar{B}\): with \(A = 0\) it is \((0 ⊙ B) ⊕ 0 = \bar{B}\), and with \(A = 1\) it is \((1 ⊙ B) ⊕ 1 = B ⊕ 1 = \bar{B}\). Its output column is likewise 1, 0, 1, 0.
Result — \((A + B)\cdot\overline{AB} = A\bar{B} + \bar{A}B = A ⊕ B\).