# Naive Set Theory: Cut and Union

Given a set X with subsets $A, B$ and $A_m$. I knew already about the following equality:

$$B \cap (A \cup A_m) = (B \cap A) \cup (B \cap A_m)$$

But I just saw the following equality:

$$B \cap (A \cup A_m) = (B \cap A) \cup ((B \setminus A) \cap A_m)$$

I wondered if this holds in general and if so, what the proof of it is.

$(B\cap A_m) = ((B\cap A)\cap A_m) \cup ((B\backslash A)\cap A_m)$ by partitionning $B$ in $B\cap A$ and $B\backslash A$.
$$B \cap (A \cup A_m) = (B \cap A) \cup ((B\cap A)\cap A_m) \cup ((B\backslash A)\cap A_m)$$
And since $((B\cap A)\cap A_m)\subset B\cap A$, you get your result.
Yes, that holds in general. It is because $$B=(B\cap A)\cup (B\setminus A)$$ so that $$B\cap A_m=(B\cap A\cap A_m)\cup ((B\setminus A)\cap A_m),$$ and after taking the union of this with $(B\cap A)$, we can forget about the $(B\cap A\cap A_m)$, because it's redundant.