Proving a Set Theory identity Please guys help me prove some Identities.
I need to prove that:
$$B=C \Longleftrightarrow (A \cup B = A \cup C) \land (A \cap B = A \cap C)$$
and also that
$$(A \cap B)\cup C = A\cap(B\cup C) \Longleftrightarrow C\subseteq A $$
i tried everything please help me....
 A: First Statement


*

*$C\subseteq A \Rightarrow (A \cap B)\cup C = A\cap(B\cup C):$


If $C\subseteq A$, then $(A \cup C)=A$ and, it follows from
$$(A \cap B)\cup C = (A \cup C) \cap (B \cup C),$$
that $(A \cap B)\cup C = A\cap(B\cup C)$.


*

*$(A \cap B)\cup C = A\cap(B\cup C) \Rightarrow C\subseteq A:$


Suppose that $C\nsubseteq A$, then there exists $x \in C$ such that $x \not\in A$. But then, it follows that $x \in (A \cap B)\cup C$ while $x \not\in A \cap(B \cup C)$ which implies that $$(A \cap B)\cup C \neq A\cap(B\cup C) $$

Second Statement


*

*$B=C \Rightarrow (A \cup B = A \cup C) \land (A \cap B = A \cap C):$


This part is obvious.


*

*$\underbrace{(A \cup B = A \cup C)}_{(1)} \land \underbrace{(A \cap B = A \cap C)}_{(2)} \Rightarrow B=C:$


Consider $x \in B$. If $x \in A$, then $(2)$ implies that $x \in C$. Otherwise, if $x \not\in A$, then $(1)$ implies that $x \in C$. Hence, $B \subseteq B$. An exactly analogous argument works to prove that if $C \subseteq B$.
A: First statement:
($\Longrightarrow$)
$$(A \cap B)\cup C = A\cap(B\cup C) \Longrightarrow C\subseteq A $$
By contraposition: suppose $C\not\subseteq A$, so there's something $x\in C\setminus A$. Then $x \notin A\cap(B\cup C)$ because $x\notin A$. However, $x\in (A \cap B)\cup C$ because $x\in C$. So $(A \cap B)\cup C \ne A\cap(B\cup C)$.
Or directly:
We have $C\subseteq (A \cap B)\cup C = A\cap(B\cup C) \subseteq A$, because for any $X,Y$, $Y\subseteq X\cup Y$ and $X\cap Y \subseteq X$.
($\Longleftarrow$)
$$(A \cap B)\cup C = A\cap(B\cup C) \Longleftarrow C\subseteq A $$
Suppose $C\subseteq A$. As $\cap$ and $\cup$ distribute over each other, we have:
$$\begin{align}
(A \cap B)\cup C &= (A\cup C)\cap(B\cup C) \\
&= A\cap(B\cup C) \\
\end{align}$$
because $A\cup C = A$. 
The second statement is true because... identity.
