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Q25E

Expert-verifiedFound in: Page 818

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Show that these identities hold.**

\(\begin{array}{c}a)\;x \oplus y{\bf{ = (x + }}y)\overline {(xy)} \\b)\;x \oplus y{\bf{ = (x\bar y) + }}(\bar xy)\end{array}\)

a) The given identity \(x \oplus y = \left( {x + y} \right)\overline {\left( {xy} \right)} \) is hold

b) The given identity \(x \oplus y = \left( {x\bar y} \right) + \left( {\bar xy} \right)\) is hold

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

The Boolean sum \( + \) or \(OR\) is \(1\) if either term is \(1\).

The Boolean product \( \cdot \) or \(AND\) is \(1\) if both terms are \(1\).

The \(XOR\) operator \( \oplus \) is \(1\) if one of the terms is \(1\) (but not both).

\(x\) and \(y\) can both take on the value of \(0\) or \(1\).

The XOR operator is \(1\) if one of the two elements (but not both) are \(1\).

\(\begin{array}{*{20}{r}}x&y&{x + y}&{xy}&{\overline {\left( {xy} \right)} }&{\left( {x + y} \right)\overline {\left( {xy} \right)} }&{x \oplus y}\\0&0&0&0&1&0&0\\0&1&1&0&1&1&1\\1&0&1&0&1&1&1\\1&1&1&1&0&0&0\end{array}\)

Notes that the last two columns of the table are identical.

Therefore, you get\({\bf{(x + }}y)\overline {(xy)} {\bf{ = }}x \oplus y\).

\(x\) and \(y\) can both take on the value of \(0\) or \(1\).

The \(XOR\) operator is \(1\) if one of the two elements (but not both) are \(1\).

\(\begin{array}{*{20}{r}}x&y&{\bar x}&{\bar y}&{\left( {x\bar y} \right)}&{\left( {\bar xy} \right)}&{\left( {x\bar y} \right) + \left( {\bar xy} \right)}&{x \oplus y}\\0&0&1&1&0&0&0&0\\0&1&1&0&0&1&1&1\\1&0&0&1&1&0&1&1\\1&1&0&0&0&0&0&0\end{array}\)

Notes that the last two columns of the table are identical.

Therefore, you get\((x\bar y) + (\bar xy) = x \oplus y\).

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