Given $f(w, x, y, z) = \sum_m(0,1,2,3,7,8,10) + \sum_d(5,6,11,15)$, where $d$ represents the *don't-care* condition in Karnaugh maps. Which of the following is a minimum product-of-sums (POS) form of $f(w, x, y, z)$?

To obtain the minimum Product-of-Sums (POS) form of $f(w, x, y, z)$, we find the simplified Sum-of-Products (SOP) expression for the complement function $\bar{f}$ by grouping the zeros (maxterms) and don't-care terms, and then take the complement.

1. Identify the zeros of $f$: The complete set of 4-variable minterms is $\{0, 1, \dots, 15\}$.

   • Minterms of $f$: $\{0, 1, 2, 3, 7, 8, 10\}$

   • Don't-cares ($d$): $\{5, 6, 11, 15\}$

   • Zeros of $f$ (where $f=0$): $\{4, 9, 12, 13, 14\}$

2. Group the zeros on a 4-variable K-map for $\bar{f}$:

   • Quad 1: Cells $(4, 6, 12, 14)$ with zeros at $4, 12, 14$ and don't-care at $6$. This gives the term: $x\bar{z}$.

   • Quad 2: Cells $(9, 11, 13, 15)$ with zeros at $9, 13$ and don't-cares at $11, 15$. This gives the term: $wz$.

3. Form $\bar{f}$ and complement to find $f$:

   $$\bar{f} = wz + x\bar{z}$$

   Taking the Boolean complement using De Morgan's laws:

   $$f = \overline{wz + x\bar{z}} = (\bar{w} + \bar{z})(\bar{x} + z)$$

Therefore, the minimal POS expression is $(\bar{w} + \bar{z})(\bar{x} + z)$, which corresponds to Option (A).