# Proofs in set theory.

I was having trouble with the following proof: $$(X ∪ Y ) / (Y ∩ Z)=(X / Y ) ∪ (Y / Z)$$

I managed to prove the L.H.S $$\in$$ R.H.S.

Though I didn't succed to do the vice versa. So here's my go at it: $$(X / Y) \cup (Y / Z) \\ (X \cap Y') \cup (Y \cap Z')$$

At this point I observed there are three possiblities:

• $$x \in (X \cap Y')$$, which implies $$(x \in X)\cap(X\notin Y)$$. However, if x doesn't belong to Y, it means it doesn't belong to $$Y \cap Z'$$ either.
• $$x \in (Y \cap Z') \to (x \in Y) \cap (x \notin Z)$$. Same reasoning here, if it's in $$Y$$, it cannot be inside $$(X\cap Y')$$.
• Though if x belongs to both I didn't know how to appraoch it; obviously an element cannot belong to $$A$$ and $$A'$$. Their union is $$\emptyset$$.

I'd like to know if there's any approach or mind-set with set-theory. I find myself struggling with the reasoning not once in this subject even though it's quite trivial.

EDIT: I would like if someone can post their proof using algebra of sets.

You hit the nail on the head in your third point, you just didn't realize it. You're right, obviously an element can't belong to both $$A$$ and $$A'$$. And if $$x$$ were in both $$X \cap Y'$$ and $$Y \cap Z'$$, your arguments previously would show that $$x \in Y'$$ and $$x \in Y$$. That is a contradiction, which is fine - it just means that no such $$x$$ exists and your third case never happens.
The easiest way to do these is to draw a Venn diagram for the left-hand side, a Venn diagram for the right-hand side, and go "voilà, they're the same diagram". This is essentially reasoning casewise on the cases "$$x \in X, x \in Y, x \in Z$$", "$$x \in X, x \in Y, x \not \in Z$$", and so on.