# Prove $(A_1 \smallsetminus B) \cup (A_2 \smallsetminus B) = (A_1 \cup A_2) \smallsetminus B$

Prove $$(A_1 \smallsetminus B) \cup (A_2 \smallsetminus B) = (A_1 \cup A_2) \smallsetminus B$$

If I used the set identity to prove it and it seems it goes on and on, did I make a mistake or ...

$$(A_1 \cap \lnot B) \cup (A_2 \cap\lnot B)$$ and then I use the distributive law and ...

$$((A_1 \cap \lnot B) \cup A_2) \cap (A_1 \cap \lnot B) \cup \lnot B)$$

• Yes, you would have made a mistake. The proof does not 'go on and on'. It is reasonably straightforward. However, there is no telling where mistake might be without knowing what exactly you tried to do. – Graham Kemp Feb 22 at 23:14
• I am sorry that question might seem dumb or my brain is not functioning well. How do I type the union and intersection sign? – Bob Feb 22 at 23:16
• @Bob use cup and cap tags on LaTex. – RFTexas Feb 22 at 23:20
• It is a good question, @Bob . Here's a Mathjax basic tutorial and quick reference – Graham Kemp Feb 22 at 23:20
• I wish I could just write it down – Bob Feb 22 at 23:23

## 2 Answers

Isn't the distributive law just $$(M\cap N) \cup (S\cap N)= (M\cup S)\cap N$$?

So $$(A_1\setminus B)\cup (A_2\setminus B) =$$

$$(A_1\cap B^c)\cup (A_2\cap B^c) =$$

$$(A_1\cup A_2)\cap B^c =$$

$$(A_1\cup A_2)\setminus B$$.

Doesn't seem to go "on and on".

• My bad, oof I used the distributive law the other way around – Bob Feb 22 at 23:29
• Sorry I was trying to figure out how to type those symbols instead – Bob Feb 22 at 23:30

Distribution is the next step, but you have to distribute out the common factor. You may be familiar with distributing like this. $$(A_1\cup A_2)\cap\lnot B_1 ~=~ (A_1\cap\lnot B)\cup(A_2\cap\lnot B)$$ However, note that distribution is an equivalence -- it works in both directions. $$(A_1\cap\lnot B)\cup(A_2\cap\lnot B) ~= ~(A_1\cup A_2)\cap\lnot B_1$$