Question

Show that \(x \odot y{\bf{ = }}\overline {(x \oplus y)} \).

Short answer

The given \(x \odot y{\bf{ = }}\overline {(x \oplus y)} \) is proved.

Step-by-step solution

Definition

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

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

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

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

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

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

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

Table of values

Given is \(x \odot y{\bf{ = }}\overline {(x \oplus y)} \).

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

\(\begin{array}{*{20}{r}}x&y&{x \oplus y}&{x \odot y}&{\overline {(x \oplus y)} }\\0&0&0&1&1\\0&1&1&0&0\\1&0&1&0&0\\1&1&0&1&1\end{array}\)

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

Therefore, it gives \(x \odot y{\bf{ = }}\overline {(x \oplus y)} \).