My question is about the proof of the distributive laws of boolean operations, more specifically I have some trouble understanding this equality: $B \cup (A_1 \cap \cdots \cap A_n) = (B \cup A_1) \cap \cdots \cap (B \cup A_n)$.
Denoting $(A_1 \cap \cdots \cap A_n)$ by $C$, the left hand side reads as $x \in B$ or $C$, that is $x$ is in at least one of $B$ and $C$.
Now, I interpret the right hand side the following way: If $x$ is not in $B$, then $x$ must be in all the $A_i$, because for each term, $x$ has to be in at least one of $B$ or $A_i$, and if $x \notin B$ for one term, then $x \notin B$ for all terms. That's about as far as I get, because I feel like the right hand side implies that $x$ can be in $B$ and just some of the $A_i$, which would contradict the left hand side. Like, if we look at the first term $B \cup A_1$, one valid possibility is that $x$ is in $B$ and $A_1$. At the same time, for the second term, $B \cup A_2$, it seems like it would be valid to say that $x$ is in $B$ but not in $A_2$, but this contradicts the left hand side, which implies that $x$ must be in every $A_i$ if it is in one of them.
So if I were given just the right hand side, $(B \cup A_1) \cap \cdots \cap (B \cup A_n)$, I would say, OK, $x$ can be in $B$ and $A_i$ for all $i$, $x$ can be in not $B$ and $A_i$ for all $i$, but $x$ cannot be in not $B$ and just some of the $A_i$, since for every term $x$ must be in at least $B$ or $A_i$. And then finally, the part which seems irreconcilable with the left hand side of the equation at the top: $x$ can be in $B$ and some of the $A_i$.