GATE CS 1999 Q1.7 — Boolean Algebra

To find which expression is not equivalent to $\bar{x}$, let us evaluate each option using Boolean algebra identities:

• Option A: $x \text{ NAND } x = \overline{x \cdot x} = \bar{x}$ (Equivalent to $\bar{x}$)
• Option B: $x \text{ NOR } x = \overline{x + x} = \bar{x}$ (Equivalent to $\bar{x}$)
• Option C: $x \text{ NAND } 1 = \overline{x \cdot 1} = \bar{x}$ (Equivalent to $\bar{x}$)
• Option D: $x \text{ NOR } 1 = \overline{x + 1} = \bar{1} = 0$ (Not equivalent to $\bar{x}$)

Therefore, the expression in Option D is not equivalent to $\bar{x}$.