Product of Sums (POS) and Maxterms

Start here SOP looks at the rows where the output is 1. But what if the output is 1 almost everywhere, and 0 in just two rows? Then it’s easier to describe the 0s. That’s the idea behind Product of Sums.

Rule: for a row where $F = 0$, input 0 → plain letter, input 1 → barred letter, and OR them.

Why the opposite rule? A maxterm must be 0 only in its own row. An OR is 0 only when every literal is 0. In row $0 1 0 , A + \overline{B} + C = 0 + 0 + 0 = 0$ ✓, and in any other row at least one literal is 1.

1. Find the rows where $F = 0$.
2. Write the maxterm for each.
3. AND them all together.

Example: $F = \Sigma m \left(3 , 4 , 5 , 6 , 7\right)$, so $F = 0$ in rows 0, 1, 2:

$F = \left(A + B + C\right) \left(A + B + \overline{C}\right) \left(A + \overline{B} + C\right) =$ $\Pi M \left(0 , 1 , 2\right)$

Why it works: in a 0-row, its own maxterm is 0, so the whole AND is 0. In a 1-row, every maxterm is 1, so the AND is 1.

Another way to see it: POS of $F =$ De Morgan applied to the SOP of $\overline{F}$.

For a 3-variable function (rows 0–7):

Memory trick: switching between Σ and Π → use the other numbers. Switching F ↔︎ $\overline{F}$ → also use the other numbers. Doing both → the same numbers.

In one line: POS builds F from its 0-rows: each 0-row gives a maxterm (0 → letter, 1 → barred), and the maxterms are ANDed. $\Sigma m$ and $\Pi M$ use complementary row numbers. Exam words: product of sums (POS), maxterm, $\Pi M$ notation, OR-AND circuit, conjunctive normal form, complement of a function, Σ↔︎Π conversion.