Elementary set theory question 2 May I ask to have a look at the following proof I did. Thanks.
Question:

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

My Attempt:
A$\cup$(B$\cap$C):
$\forall$x[(x$\in$A)]$\vee$[(x$\in$B)(x$\in$C)]
=$\forall$x[(x$\in$A)$\vee$(x$\in$B)]$\wedge$[(x$\in$A)$\vee$(x$\in$C)]
=$\forall$x[(x$\in$A)$\cup$(x$\in$B)]$\cap$[(x$\in$A)$\cup$(x$\in$C)]
=(A$\cup$B)$\cap$(A$\cup$C)
 A: The idea of your proof is correct. But it's not written down in a correct way.
Look at the first equality sign. You write $$A \cup (B\cap C) = \forall x[(x\in A)]\vee [(x\in B)(x\in C)].$$
On the left hand side of the equality, there is a set. But on the right hand side there is not a set, but a logical expression. This is certainly not ok.
What you thought of is probably $$A \cup (B\cap C) = \{x \mid x\in A\vee (x\in B \wedge x\in C)\}.$$
Now on both sides of the equation, there is a set. And by definition of $\cap$ and $\cup$, they are the same.
I hope you get the idea. Try to rewrite your proof accordingly.
A: Very well done. The "gist" of your logic is correct. There are only a few minor problems.
One thing I'd add, and it may simply have been a typo, is in the first line of your proof, you want to add a missing $\land$ between $(x\in B)(x\in C)$:
$$x \in [A \cup(B\cap C)] \iff [(x\in A) \lor ( x\in B \land x \in C)]\tag{*}$$
Notice I also used $x \in [A\cup(B\cap C)]$ to start, with no need for the universal quantifier, because we are making claims about precisely any/every element belonging the set in question. We use this notation since we are aiming to show 
$$x \in A\cup(B\cap C) \iff x\in [(A\cup B) \cap (A \cup C)]$$ and in doing so, we will have proven the desired equality: $$A \cup(B\cap C) = [(A\cup B) \cap (A \cup C)]$$
So you want to end with $x \in [(A\cup B) \cap (A \cup C)]$ to finish the proof of the equality of the Left hand side and the right hand side.
And use $\iff$ between lines (as used in (*)). That means that "if and only if".
