Consider a Boolean function $F$ with the following minterm expression: $$F(P, Q, R, S) = \sum m (1, 2, 3, 4, 5, 7, 10, 12, 13, 14)$$ Which of the following options is/are the minimal sum-of-products expression(s) of $F$ ?

To find the minimal Sum-of-Products (SOP) expressions, we plot the given minterms on a 4-variable Karnaugh Map ($P, Q$ on rows, $R, S$ on columns):

$$\begin{array}{c|cccc} PQ \backslash RS & 00 & 01 & 11 & 10 \\ \hline 00 & 0 & 1 & 1 & 1 \\ 01 & 1 & 1 & 1 & 0 \\ 11 & 1 & 1 & 0 & 1 \\ 10 & 0 & 0 & 0 & 1 \\ \end{array}$$

### 1. Essential Prime Implicants

• Group 1 (size 4): $m(1, 3, 5, 7) = \bar{P}S$ (covers minterms $1, 3, 5, 7$). Minterms $1, 3, 7$ make this essential.

• Group 2 (size 4): $m(4, 5, 12, 13) = Q\bar{R}$ (covers minterms $4, 5, 12, 13$). Minterm $4$ makes this essential.

### 2. Remaining Uncovered Minterms The minterms remaining to be covered are $m(2, 10, 14)$.

The possible size-2 prime implicants covering these minterms are:

• For $m(2)$: $m(2, 3) = \bar{P}\bar{Q}R$ and $m(2, 10) = \bar{Q}R\bar{S}$

• For $m(10)$: $m(2, 10) = \bar{Q}R\bar{S}$ and $m(10, 14) = PR\bar{S}$

• For $m(14)$: $m(10, 14) = PR\bar{S}$ and $m(12, 14) = PQ\bar{S}$

### 3. Covering the Remaining Minterms with Minimum Terms We can cover $\{2, 10, 14\}$ using exactly two prime implicants in the following ways:

1. Choose $m(2, 3) = \bar{P}\bar{Q}R$ (covers $2$) and $m(10, 14) = PR\bar{S}$ (covers $10, 14$).

   $$F = \bar{P}S + Q\bar{R} + \bar{P}\bar{Q}R + PR\bar{S}$$

   This matches Option (B).

2. Choose $m(2, 10) = \bar{Q}R\bar{S}$ (covers $2, 10$) and $m(12, 14) = PQ\bar{S}$ (covers $14$).

   $$F = \bar{P}S + Q\bar{R} + PQ\bar{S} + \bar{Q}R\bar{S}$$

   This matches Option (D).

• Option (A) misses minterm $14$.

• Option (C) misses minterm $2$.

Thus, the correct minimal SOP expressions are given by (B) and (D).