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Expert-verified Found in: Page 818 ### Discrete Mathematics and its Applications

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

See the step by step solution

## Step 1: Definition

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).

## Step 2:(a) Using the operator

$$x$$ 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$${\bf{(x + }}y)\overline {(xy)} {\bf{ = }}x \oplus y$$.

## Step 3: Using the operator

$$x$$ 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$$(x\bar y) + (\bar xy) = x \oplus y$$. ### Want to see more solutions like these? 