Question

Show that \({\bf{x}} \odot {\bf{y = xy + \bar x\bar y}}\).

Short answer

The given \({\bf{x}} \odot {\bf{y = xy + \bar x\bar y}}\) is proved.

Step-by-step solution

Definition

The complement of an element: \({\bf{\bar 0 = 1}}\) and \({\bf{\bar 1 = 0}}\).

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

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

The \({\bf{NOR}}\) operator \( \downarrow \) is \(1\) if both terms are \(0\).

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

The \({\bf{NAND}}\) operator \(\mid \) is \(1\) if either term is \(0\).

The Boolean operator \( \odot \) is \(1\) if both terms have the same value.

Table of values

Given is \({\bf{x}} \odot {\bf{y = xy + \bar x\bar y}}\).

Now create a table for all possible values of \({\bf{x, y}}\) and \({\bf{z}}\).

\(\begin{array}{*{20}{r}}{\bf{x}}&{\bf{y}}&{{\bf{xy}}}&{{\bf{\bar x}}}&{{\bf{\bar y}}}&{{\bf{\bar x\bar y}}}&{{\bf{x}} \odot {\bf{y}}}&{{\bf{xy + \bar x\bar y}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{1}}\end{array}\)

Then note that the last two columns of the table are identical.

Hence, it gives \({\bf{x}} \odot {\bf{y = xy + \bar x\bar y}}\).