Proof of the set identity $(A \cap B) \cup C=A \cap(B \cup C) \iff C \subseteq A$ I am sorry if this question might be too trivial, but this is the first time I have to do a set identity proof.
 Prove that : $$(A \cap B) \cup C=A \cap(B \cup C) \iff C \subseteq A$$
My attempt:
Lets assume the left side as true. In order to prove $C \subseteq A$ we need to prove that if $x \in C     \Rightarrow x\in A $
Let $ x\in C$. Therefore $x \in (A \cap B) \cup C$. From $(A \cap B) \cup C=A \cap(B \cup C)$ follows that $x\in A \cap(B \cup C) \Rightarrow x \in A.$
I am not sure if that's correct though, I would appreciate feedback. 
For the other way we assume that $C \subseteq A.$
Now $(A \cap B) \cup C=(A \cup C) \cap (B \cup C) $ But $C \in A \Rightarrow (A \cup C)=A \Rightarrow (A \cap B) \cup C=A \cap(B \cup C)$.
 A: $\wedge$ stands for "and".
Let $C \subseteq A$. We will first show that $(A \cap B) \cup C \subseteq A \cap (B \cup C)$, then  we will show that $A \cap (B \cup C) \subseteq (A \cap B) \cup C$. This will prove the equality.
Then note that if $x \in (A \cap B) \cup C$, then either $x \in A \wedge x \in B$, or $x \in C$.
If $x \in A \wedge x \in B$, then $x \in A \wedge x \in B \cup C$, so $x \in A \cap (B \cup C)$.
If $x \in C$, then $x \in A$, so $x \in A \wedge x \in C$, hence $x \in A \wedge x \in B \cup C$, so $x \in A \cap (B \cup C)$.
Either way,$(A \cap B) \cup C \subseteq A \cap (B \cup C)$. 
The other way, note that if $x \in A \cap (B \cup C)$, then $(x \in C \text{ or } x \in B) \wedge x \in A$.
If $x \in C$, then $x \in (A \cap B) \cup C$.
If $x \in B$, then $x \in A \wedge x \in B$, hence $x \in (A \cap B) \cup C$.
Either way,$ A \cap (B \cup C) \subseteq (A \cap B) \cup C $. 
Hence the proposition follows.
An easier way is to notice that $(A \cap B) \cup C =  (A \cup C) \cap (B \cup C)$, and see that $A \cup C = A$ if $C \subseteq A$.
A: Your argument in the forward direction looks good! Now for the backward direction try using some of the distributive identities of intersections and unions.
A: Following is an alternative proof using Boolean algebra and the Indicator functions $a, b, c$ of sets $A, B, C$. The equality on the left hand translates to:
$$
\begin{align}
a \times b + c - a \times b \times c & = a \times (b + c - b \times c) \\
& = a \times b + a \times c - a \times b \times c
\end{align}
$$
or, after canceling out the identical terms:
$$
c = a \times c
$$
which is equivalent to $c \leq a$ which is in turn equivalent to $C \subseteq A$.
A: Algebraic Proof
Let $1_X$ be the indicator function of a set $X$; that is, $1_X(x)=1$ if $x\in X$, and $1_X(x)=0$ otherwise.  We have $$1_{(A\cap B)\cup C}=1_{A\cap B}+1_C-1_{A\cap B\cap C}=1_A\cdot 1_B+1_C-1_{A}\cdot 1_B\cdot 1_{ C}$$
and
$$1_{A\cap(B\cup C)}=1_A\cdot 1_{B\cup C}=1_A\cdot\left(1_B+1_C-1_{B\cap C}\right)=1_A\cdot \left(1_B+1_C-1_B\cdot 1_C\right)\,.$$
Hence, $(A\cap B)\cup C=A\cap(B\cup C)$ if and only if
$$1_A\cdot \left(1_B+1_C-1_B\cdot 1_C\right)=1_A\cdot 1_B+1_C-1_{A}\cdot 1_B\cdot 1_{ C}\,,$$
or $$1_{A\cap C}=1_A\cdot 1_C=1_C\,.$$
However, the last condition is equivalent to saying that $A\cap C=C$, which is the same as $A\supseteq C$. 
P.S.  Just found out that dxiv offered the same solution.
