# Proof in logic for arbitrary number of sets

given $n \in \mathbb{N}$ ,prove that $A \cap (B_1 \cup B_2 \cup \cdots \cup B_n) = (A \cap B_1) \cup (A \cap B_2) \cup \cdots \cup (A \cap B_n)$ ? (As HomeWork)

Now i know how to prove using induction or using "without loss of generality",

But unfortunately we did not learn them in class so "We don't know them", is there another way to prove the above equality !

If you can't use induction, then I suppose the only thing to do is to show that for any $x$:

$x \in A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)$ if and only if $x \in (A \cap B_1) \cup (A \cap B_2) \cup \cdots \cup (A \cap B_n)$

And to get started:

$x \in A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)$ if and only if

$x \in A$ and $x \in (B_1 \cup B_2 \cup \cdots \cup B_n)$ if and only if

$x \in A$ and ($x \in B_1$ or $n\in B_2$ or $\cdots$ or $n \in B_n$) if and only if (by pure logic)

($x \in A$ and $x \in B_1$) or ($x \in A$ and $x\in B_2$) or ...

• @DerekElkins Ah, I was just copying and pasting from the OP's prompt and didn't even notice that! Thanks! – Bram28 Oct 29 '17 at 14:52