Elementary Set Theory: Proof of Distributive Property I'm new to proofs.  While working through a finite math book, I've been asked to prove the distributive properties of set operations.  I'll use the distributive property of union over intersection as the example for my question.  
Prove that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
The proofs I see for this property aim to show, in two steps, that each side is a subset of the other side.  For example, the first step usually looks something like this:


*

*Let x ∈ A ∪ (B ∩ C). If x ∈ A ∪ (B ∩ C) then x is either in A or in (B and C). 

*x ∈ A or x ∈ (B and C)

*x ∈ A or {x ∈ B and x ∈ C}

*{x ∈ A or x ∈ B} and {x ∈ A or x ∈ C}

*x ∈ (A or B) and x ∈ (A or C)

*x ∈ (A ∪ B) ∩ x ∈ (A ∩ C)

*x ∈ (A ∪ B) ∩ (A ∪ C)

*x ∈ A ∪ (B ∩ C) => x ∈ (A ∪ B) ∩ (A ∪ C)

*Therefore, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C)


Then the second step would do something similar to show that (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C).
My question in this--In jumping from lines 3 to 4, aren't we relying on the very property we're trying to prove?  If so, how is that considered as proof?
I don't have trouble following the steps, and I don't doubt that the property is valid.  But as a newcomer to proofs I'm confused because it looks (to me) like we're using the property to prove itself.  To my non-mathematician mind, drawing a couple of Venn diagrams provides a more intuitive "proof" than manipulating these equations.
 A: Jumping from $3$ to $4$ needs to be justified.
At $3$ you have $x \in A$ or $(x \in B \;\text{and} \; x \in C).$
Then we form two cases. One in which we see what happens when $x \in A$ and second in which $x \in B \cap C$.
Case 1): $x \in A$ -
Since $A \subseteq A \cup B$ and $A \subseteq A \cup C$,
$x \in A \implies x \in A \cup B$ and $x \in A \implies x \in A \cup C$. Thus $x \in A \cup B$ and $x \in A \cup C$. That is $x \in (A \cup B) \cap (A \cup C).$
Case 2):$x \in B \cap C$ -
Since $x \in B$ and $x \in C$, we have $x \in A \cup B$ and $x \in A \cup C$. $(\because B \subseteq A \cup B, C \subseteq A \cup C).$
Thus $x \in (A \cup B) \cap (A \cup C)$ in this case too.
Oops we jumped to $7$ directly. :P  
A: From lines 3 to 4 you are using the property $\text{A or (B and C)} \Leftrightarrow  \text{(A or B) and (A or C)}$ which is not related to the properties of the $\in$ operator. This property can be proved using truth tables, for example. 
A: Let's prove that $(A\cup B)\cap C = (A\cup C)\cap(B\cup C)$.
Firstly, let $x \in (A\cup C)\cap(B\cup C)$, then $x \in A\cup C$ and $x \in B\cup C$.
Secondly, there may be two options:


*

*$x \in C$, and therefore $x \in (A\cap B)\cup C$.

*$x \notin C$. Then $x \in A$ and $x \in B$. So $x \in A\cap B$, and therefore $x \in (A\cap B)\cup C$.


Finally, $x \in (A\cap B)\cup C$ in both options, and $x \in (A\cup C)\cap (B\cup C)$ is given. Therefore, $(A\cap B)\cup C = (A\cup C)\cap(B\cup C)$.
