Proof of set membership by analytic proof 
Given the sets $A,B,C,D$ and the axioms:
  
  
*
  
*$A\cap B\ne\emptyset$
  
*$A\cap C\ne\emptyset$
  
*$B\cap C\ne\emptyset$
  
*$A\cap B\cap C\ = \emptyset$
  
*$D\subsetneq A$
  
*$B\cap D = \emptyset$
  
*$C\cap D\ne\emptyset$
  
  
  prove analytically that $A\nsubseteq C$

I can easily show this using a Venn diagram, but that's not "analytic" solution. Is it correct that analytic means transforming the statements into "set builder notation", for example:
$A\cap B = \{x\mid x\in A \vee x\in B\}$
and
$D\subsetneq A = \{x\mid x\in D\rightarrow x\in A \wedge \exists x(x\in A \wedge x\notin D)\}$
and then using propositional logic somehow? But also I don't know how to handle the intersected sets equaling or not equaling the empty set. I'm hoping for a hint to get me started or a good reference that talks about this kind of proof.
 A: From 1. we see that since $A \cap B \neq \emptyset$ that there is
some $x \in A \cap B \subset A$.
From 4. we see that $(A \cap B) \cap C = \emptyset$, hence $x \notin C$.
Hence $ A \not\subset C$.
A: Suppose otherwise that $A \subseteq C$. Then $D \subsetneq A \subseteq C$. Since $B \cap D \ne \varnothing$, there exists an element in $B$ that is also in $A$ and $C$, which violates condition 4, a contradiction.

Proof without contradiction: By conditions 5 and 7 there exists an element in $A \cap B$. It must not lie in $C$ because of condition 4. We have found an element of $A$ that does not lie in $C$, so $A \not\subseteq C$.
A: First of all, I asked a colleague of mine what an 'analytic' proof is, and they told me this is how some people refer to 'real' proofs, or 'rigorous' proofs. For me, this is just what a proof is, so I am going to proceed under this assumption of the interpretation of 'analytic'. If you need general advice on how to construct proofs, I have found this article very useful, especially Part II. 
So, my first approach would be a proof by contradiction, assuming that $A\subseteq C$. From axiom 4, we would have that $\emptyset=A\cap B \cap C=B\cap C$, where the second equality follows from symmetry of $\cap$ and our assumption (since $A\subseteq C$ implies $A\cap C=A$). This, however, contradicts axiom 3, that $B\cap C\neq\emptyset$. 
Of course, whether this proof counts as rigorous or not depends on what sort of assumptions you are allowed to make in your set theory course. Do you take the Zermelo-Fraenkel axioms, or are you using naive set theory? The proof above uses the latter. 
A: 1) $A \cap B \ne \emptyset$
Let $x \in A \cap B \implies x \in A$.
4) $A \cap B \cap C  = \emptyset$
$x \in C => x \in (A \cap B) \cap C = A \cap B \cap C = \emptyset$ but that's a contradiction.  So $x \in C$.  So $A \not \subseteq C$.  2,3,5,6 and 7 were superfluous.
