Boolean algebra - Sum of Products and Product of Sums - Why is the procedure defined as it is? Here's my problem: I understand how to create the sum of products (SOP) and product of sums (POS) forms of boolean functions, but I don't understand why we do it the way we do it. And I haven't found an answer anywhere online. Literally every source I've read simply tells you how to form these expressions but never explains the intuition.
Example: Suppose we have a truth table for some boolean function $F$.

The rules for forming a sum-of-products (SOP) expression say to:

*

*look at only the rows where $F = 1$

*create product terms out of $A$, $B$, and $C$, inverting any of these three variables that are $0$ and leaving the remaining variables that equal $1$ as they are

*sum all the product terms

So I would do:
$F = A'BC + AB'C+ABC'$
For the three rows where $F = 1$.
Question: why exactly do I ignore rows where $F=0$, and why do I invert any variables that are $0$?
 A: Remember that the goal is to find an expression for which $F=1$ (or, in terms of logic, for which the function is 'True').
So immediately, that's why it ignores the rows where $F=0$; we are only interested in those rows where $F=1$, which is rows3, 5, and 6
So, we are going to 'capture' $F=1$ by saying that we are dealing with  row 3, or row 5, or row 6. Since the $+$ captures the logical 'or', we thus get as our basic form:
$$F = row_3 + row_5 + row_6$$
where we want $row_3$ to be an expression that equals to $1$ if and only if we are 'dealing' with row $3$. OK, so when are we dealing with row 3? It is when $A=0$, and $B=1$, and $C=1$. Since the logical 'and' is the 'multiplication' in boolean logic, this corresponds to $A'BC$. That is, we want the inverse of $A$ to be $1$, and $B$ to be 1, and $C$ to be 1. Likewise, we get the terms $AB'C$ and $ABC'$  for rows 5 and 6 respectively, and thus end up with:
$$F = A'BC+AB'C + ABC'$$
For another example, accompanied with some further explanation, see:  Can AND, OR and NOT be used to represent any truth table?
