SOP & POS — worked example

Majority vote: F=1 when ≥2 of A, B, C are 1

Three inputs, eight rows. Filling in the truth table by inspection and grouping by F-value:

#  A B C  F
 0  0 0 0  0   (0 ones)
 1  0 0 1  0   (1 one)
 2  0 1 0  0   (1 one)
 3  0 1 1  1   (2 ones) → minterm
 4  1 0 0  0   (1 one)
 5  1 0 1  1   (2 ones) → minterm
 6  1 1 0  1   (2 ones) → minterm
 7  1 1 1  1   (3 ones) → minterm

1. Read off the canonical SOP. Take each F=1 row and write a literal per variable: bit=1 → bare, bit=0 → primed. AND within a row, OR across rows.
   F = Σm(3, 5, 6, 7) = A′BC + AB′C + ABC′ + ABC
2. Read off the canonical POS. Take each F=0 row. Bit=1 → primed, bit=0 → bare. OR within a row, AND across rows.
   F = ΠM(0, 1, 2, 4) = (A+B+C)(A+B+C′)(A+B′+C)(A′+B+C)
3. Sanity check: the index sets are complementary — $\{3,5,6,7\} \cup \{0,1,2,4\} = \{0..7\}$ and disjoint. Both forms describe the same function.
4. Simplify the canonical SOP to its minimum form. Pair minterms that differ by exactly one literal: $ABC + ABC' = AB$, $ABC + AB'C = AC$, $ABC + A'BC = BC$. ABC fans out to all three pairs.
   F = AB + AC + BC

Try other input patterns yourself. Open the Simulate stage and click the F-cells to flip rows — the Σm and ΠM expressions update in real time.