Question

Determine whether the set \( \odot \) is functionally complete.

Short answer

The set \( \odot \) is not functionally complete.

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 \({\bf{1}}\) if either term is \({\bf{1}}\).

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

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

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

The \({\bf{NAND}}\) 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.

Contradiction method

Let’s assume that it has only one variable \({\bf{x}}\).

If \( \odot \) is functionally complete, then it should be able to write the complement of \({\bf{x}}\) using only the operator \( \odot \).

Therefore, this is to note that \({\bf{\bar x}}\) contains a \({\bf{0}}\) when \({\bf{x = }}1\), while \({\bf{x}} \odot {\bf{x}}\) contains only l's and thus it cannot write \({\bf{\bar x}}\) using only \({\bf{x}}\) and \( \odot \).