# Show that a necessary and sufficient condition that $(A\cap B)\cup C =A\cap(B\cup C)$ is that $C\subseteq A$.

Let $A$, $B$ and $C$ be arbitrary sets.

Show that a necessary and sufficient condition that $(A\cap B)\cup C =A\cap(B\cup C)$ is that $C\subseteq A$.

$\exists x\in (A\cap B)\cup C$ such that $x\in A\cap(B\cup C)$

If you analyze each side,

$x\in (A\cap B)\cup C \implies x\in (A \cap B) \lor x\in C \implies (x\in A \land x\in B)\lor x\in C$

$x\in A\cap (B\cup C)\implies x\in A \land x\in (B\cup C) \implies (x\in A \land x\in B)\lor (x\in A \land x\in C)$

Is it safe to conclude $C\subseteq A$?

Sorry i'm very confused with set theory... i'm not even sure if my logical statements are.. well.. logical.. kindly assist

• I think everything becomes easier if you rewrite $A\cap(B \cup C)$ to $(A\cap B)\cup(A\cap C)$. Also, you need to do it in two steps: First, show that if $C\subseteq A$, then the two sides are equal (i.e. $C\subseteq A$ is a sufficient condition). After that, you need to show that if the two sides are equal, then we have $C\subseteq A$ (i.e. $C\subseteq A$ is a necessary condition). Commented Sep 8, 2015 at 9:20
• To prove necessary and sufficient conditions means that you have to prove two things, i.e. if $C \subseteq A$ then, $(A \cap B) \cup C = A \cap (B \cup C)$ holds and also the fact that, if $C \not\subseteq A$, then the identity doesn't holds.
– DOOM
Commented Sep 8, 2015 at 9:23
• @Arthur ok let me try that now Commented Sep 8, 2015 at 9:24
• How many times will I have to retag your questions? Commented Sep 8, 2015 at 9:25
• Possible duplicate of $(A\cap B)\cup C = A \cap (B\cup C)$ if and only if $C \subset A$ Commented Jan 23, 2016 at 20:52

Use the distributive property on the left hand side and you get $(A \cup C) \cap (B \cup C) = A \cap (B \cup C)$. If $C \subset A$, $A \cup C = A$, so the condition is sufficient. If $x \in C \setminus A$, then x is in the left hand side of the equation but not the right, so it is necessary also.