# 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)

-
i see no problem – Aang Mar 6 '13 at 16:45
@Avatar: Maybe you would like to inform amWhy to make his answer a comment. – azimut Mar 6 '13 at 17:04
i would have, but people don't seem to listen to me. – Aang Mar 6 '13 at 17:07
@azimut: But you see, he had something to add. – Aang Mar 6 '13 at 17:07

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".

-

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.

-
+1 Thanks azimut – Software Mar 6 '13 at 17:36