Question

Is it always true that \((x \odot y) \odot z{\bf{ = }}x \odot (y \odot z)\)\(?\)

Short answer

The given equation is \((x \odot y) \odot z{\bf{ = }}x \odot (y \odot z)\) always true.

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.

Using the Boolean operator

It is given that \((x \odot y) \odot z{\bf{ = }}x \odot (y \odot z)\).

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

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

\(\begin{array}{*{20}{c}}x&y&z&{x \odot y}&{y \odot z}&{(x \odot y) \odot z}&{x \odot (y \odot z)}\\0&0&0&1&1&0&0\\0&0&1&1&0&1&1\\0&1&0&0&0&1&1\\0&1&1&0&1&0&0\\1&0&0&0&1&1&0\\1&0&1&0&0&0&1\\1&1&0&1&0&0&1\\1&1&1&1&1&1&0\end{array}\)

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

Hence, it gives \((x \odot y) \odot z{\bf{ = }}x \odot (y \odot z)\).