GATE CS 1997 Q5.1 — Minimization (K-Maps etc.)

First, let us simplify the switching function $f(x, y, z)$ using Boolean algebra:

Now, let us evaluate the given options:

• Option A: For $\bar{y}x$ to be a prime implicant, it must be an implicant that cannot be combined with any other term to eliminate a literal. Since $\bar{y}x \le \bar{y}$ and $\bar{y}$ is also an implicant of $f$, $\bar{y}x$ is not a prime implicant.
• Option B: A minterm of a 3-variable function must contain all three variables. Since $xz$ has only two variables, it is not a minterm.
• Option C: A product term $P$ is an implicant of $f$ if $P = 1$ implies $f = 1$. If $xz = 1$, then $x = 1$ and $z = 1$. Substituting these into $f$ gives $f = \bar{1} + \bar{y} + 1 = 1$. Thus, $xz$ is an implicant of $f$.
• Option D: If $y = 1$, $f$ is not guaranteed to be $1$ (e.g., if $x=1, z=0$, then $f = 0$). Thus, $y$ is not even an implicant.