Is my proof for $A\cap B=A\cap C\Longleftrightarrow B=C$ correct? Prove or disprove: for every $3$ sets $A,B,C,: A\cap B=A\cap C\Longleftrightarrow B=C$
Proof: This proof is by case analysis.
Case I: Assuming $B=C$, I will prove $A\cap B=A\cap C$
1) $A\cap B=A\cap C$ (because $B=C$)
2) $A\cap C=A\cap B$ (because $B=C$)
3) Thus, $A\cap B=A\cap C$
Case II: Assuming $A\cap B=A\cap C$, I will prove that $B=C$
1) If $x\in A$, then $x\in A\cap B$
2) In accordance with the original assumption, $x\in A\cap B$
3) Therefore, $x\in B$
4) Since $x\in A$, then $x\in A\cap C$
5) Therefore $x\in C$
6) Since $x\in A$, $x\in B$, and  $x\in C$, we can conlusde that $B=C$
I feel that I am missing pertinent information to make this proof complete and correct. It's difficult to take my thoughts and organize them in the form of a proof. I would like constructive criticism to make this proof correct and to later use what I learn from it to improve the proofs I write in the future. 
 A: It can't be because the statement isn't true.  Let A = {1,2,3} B={2,4,6,8} C = {2, 7, 94, "pink honk honk", $\pi$, my grandmother's socks}.  Then $A \cap B = \{2\} = A \cap C$ but $B \ne C$.
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Okay, less flippant answer:
Your proof:
"Case I: Assuming B=C, I will prove A∩B=A∩C"
"1) A∩B=A∩C (because B=C)"
Okay, you are done.  Stop.
"1) If x∈A, then x∈A∩B"
Um.  No.  if $x \notin B$ then $x \notin A \cap B$.  $5 \in ${integers}.  But $5 \notin ${integers} $\cap$ {even integers}$.
Same problem with 4.
6) x in A, B, C so B = C.  
Um, no.  $3$ in {primes}, {odd numbers}, {real numbers} ergo {real numbers} = {odd numbers} = {prime}?  You've only shown $A\cap B \subset B$ and $A\cap B \subset C$
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Formal disproof:
$B / A\cap B$ and $C /A \cap C$ will have no bearing on $A \cap B = A \cap C$ and $B / A\cap B$ and  $C /A \cap C$ can be utterly different.  Let A, B, C, w be such that $w \notin A$, $w \in B$ but $w \notin C$ and $A \cap B = A \cap C$.  Then $B \ne C$.  Simplest example is $A = \emptyset; B = \emptyset; C =\{1\}$.  Then $A \cap B = A \cap C = \emptyset$.
A: $\textbf{Hint}$: Consider the case $A= \varnothing$.
You proved $B=C \Rightarrow A \cap C = A \cap B$ correctly. Just note that step 1 and 2 are equivalent.
For proving that $B=C$, you have to prove $B \subseteq C$ and $C \subseteq B$, i.e. take an element in $B$ and prove it is also in $C$ (and conversely).
Step 1 in your second case is false. Just because an element is in $A$ does not imply that it is in $B$. $A$ is potentially "bigger" than $A \cap B$ (for example $B=\varnothing$ and $A \neq \varnothing$).
It might be a good idea to write out what it means for an element to be in the intersection of two sets: By definition, $x$ is in $A \cap B$ if and only if $x$ is in $A$ $\textbf{and}$ $x$ is in $B$.
