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Q28E

Expert-verifiedFound in: Page 818

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Find the duals of these Boolean expressions.**

\(\begin{array}{l}{\bf{a) x + y}}\\{\bf{b) \bar x\bar y}}\\{\bf{c) xyz + \bar x\bar y\bar z}}\\{\bf{d) x\bar z + x \times 0 + \bar x \times 1}}\end{array}\)

a) The dual of the given expression \(x + y\) is \(xy\)

b) The dual of the given expression \(\bar x\bar y\) is \(\bar x + \bar y\)

c) The dual of the given expression\(xyz + \bar x\bar y\bar z\) is \(\left( {x + y + z} \right)\left( {\bar x + \bar y + \bar z} \right)\)

d) The dual of the given expression\(x\bar z + x \times 0 + \bar x \times 1\) is \(\left( {x + \bar z} \right)\left( {x + 1} \right)\left( {\bar x + 0} \right)\)

The complement of an element: \(\bar 0 = 1\) and \(\bar 1 = 0\)

The Boolean sum \( + \) or \(O{\bf{ }}R\) is \(1\) if either term is \(1\) .

The Boolean product or \(A{\bf{ }}N{\bf{ }}D\) is \(1\) if both terms are \(1\) .

The dual of a Boolean expression interchanges the Boolean sum and the Boolean product, and interchanges \(0\) and \(1\) .

\(x + y\)

The dual of \(x + y\) replaces the Boolean sum \( + \) by the Boolean product.

Therefore, the dual \( = x \times y = xy\)

\(\bar x\bar y\)

Note: \(\bar x\bar y\) implies the Boolean product \(\bar x \times \bar y\)

The dual of \(\bar x \times \bar y\) replaces the Boolean product \( \times \) by the Boolean sum \( + \)

Therefore, the dual \( = \bar x + \bar y\)

\(xyz + \bar x\bar y\bar z\)

Note: \(xyz + \bar x\bar y\bar z\) implies \(\left( {x \times y \times z} \right) + \left( {\bar x \times \bar y \times \bar z} \right)\)

The dual of \(x \times y \times z + \bar x \times \bar y \times \bar z\) replaces the Boolean product \( \times \) by the Boolean sum \( + \) and replaces the Boolean sum \( + \) by the Boolean product.

Therefore, the dual \( = \left( {x + y + z} \right) \times \left( {\bar x + \bar y + \bar z} \right) = \left( {x + y + z} \right)\left( {\bar x + \bar y + \bar z} \right)\)

\(x\bar z + x \times 0 + \bar x \times 1\)

Note: \(x\bar z + x \times 0 + \bar x \times 1\) implies \(\left( {x \times \bar z} \right) + \left( {x \times 0} \right) + \left( {\bar x \times 1} \right)\)

The dual of \(\left( {x \times \bar z} \right) + \left( {x \times 0} \right) + \left( {\bar x \times 1} \right)\) replaces the Boolean product \( \times \)by the Boolean sum \( + \), replaces the Boolean sum \( + \) by the Boolean product ; replaces \(0\) by \(1\) and replaces \(1\) by \(0\) .

Therefore, the dual \( = \left( {x + \bar z} \right) \times \left( {x + 1} \right) \times \left( {\bar x + 0} \right) = \left( {x + \bar z} \right)\left( {x + 1} \right)\left( {\bar x + 0} \right)\)

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