Please refer to this literature:
According to Naive Bayes classification algorithm:
$P(sneezing,builder|flu) = P(sneezing|flu)P(builder|flu) $
where sneezing and builder are independent events.
How do they arrive at the above conclusion mathematically?
Is it something like:
$P(sneezing,builder|flu)$ $=P(sneezing \cap builder| flu)$ $= \frac{P((sneezing \cap builder) \cap flu)}{P(flu)}$
Let $S$, $B$, and $F$ denote "sneezing", "builder", and "flu" respectively, which I suppose are events in a discrete probability space. We are given that $S$ and $B$ are independent, which means that $$ \sum_{i\in S\cap B}p(i)=\sum_{s\in S}\sum_{b\in B}p(s)p(b), $$ where $p(x)$ is the probability assigned to the element $x$ of the sample space, a.k.a. the probability mass function.
Writing out $\mathbb P(S,B\mid F)$ explicitly as a fraction yields $$ \mathbb P(S,B\mid F)=\frac{\sum_{i\in S\cap B\cap F}p(i)}{\sum_{f\in F}p(f)}.\qquad (1) $$ Note that in order for this quantity to even make sense, we are assuming that the denominator is non-zero: in other words, $\mathbb P(F)>0$.
On the other hand, we can write out $\mathbb P(S\mid F)\mathbb P(B\mid F)$ explicitly as follows: $$ \mathbb P(S\mid F)\mathbb P(B\mid F)=\frac{\sum_{i\in S\cap F}p(i)}{\sum_{f\in F}p(f)}\cdot\frac{\sum_{j\in B\cap F}p(j)}{\sum_{f\in F}p(f)}.\qquad (2) $$ Your question is asking why $(1)$ and $(2)$ are equal. By multiplying through on the denominators, we see it is the same asking why the following two expressions are equal: $$ \sum_{i\in S\cap F}p(i)\sum_{j\in B\cap F}p(j)\overset{?}{=}\sum_{i\in S\cap B\cap F}p(i)\sum_{f\in F}p(f). $$ And in fact, when written out explicitly like this, we see that the hypothesis you have been given does not imply that $(1)$ and $(2)$ are equal!! Indeed, it is easy to cook up a counterexample on a small sample space where it is false (but I will leave this exercise to you since it is instructive to work out on your own...) In other words, there is no mathematical explanation of why $(1)$ and $(2)$ are equal given your assumption... although there is a non-mathematical explanation.
How can this be? On the webpage you linked to, they were deliberately vague when expressing their assumptions:
This condition, applied to $S$ and $B$, is stronger than saying that $S$ and $B$ are independent. What they really are trying to say is that $S$ and $B$ are conditionally independent given any "reasonable" third event. Thus $(1)=(2)$ is an additional assumption that is made, but it is justified on intuitive grounds that independence of $S$ and $B$ is assumed not only on the original sample space, but also on each of the smaller sample spaces $F$ and its complement. That is the meaning of conditional independence in this case. (The general case involves a partition of the sample space into several smaller parts, depending on the values of a random variable, and imposing the condition that independence continues to hold on each of these smaller sample spaces.)