Convert the following Boolean expression into its canonical POS form:

$F(A,B,C) = (B + C') \cdot (A' + B)$. Add the missing variable to each term using $X \cdot X' = 0$: $(B + C') = (B + C' + A \cdot A') = (A + B + C') \cdot (A' + B + C')$ $(A' + B) = (A' + B + C \cdot C') = (A' + B + C) \cdot (A' + B + C')$ Combining and removing the repeated term $(A' + B + C')$: $F(A,B,C) = (A + B + C') \cdot (A' + B + C) \cdot (A' + B + C')$ $= \pi(1, 4, 5)$ (verified with the boolean tool).