A different approach to prove Distributive Properties Is there any wrong with proving it like this (instead of proving it with logic arrows).
Q: Prove that $(A \cup (B \cap C)=(A \cup B) \cap (A \cup C)$.
Ans:
If $\ x \in A \cup (B \cap C)$ is true. Then there are 3 cases:
\begin{eqnarray*}
&x \in A \\ 
&x \in B\ \text{and}\ x \in C \\
&x \in A\ \text{and}\ x \in B\ \text{and}\ x \in C.
\end{eqnarray*}
Does not contradict $x \in A \cup B) \cap (A \cup C)$.
$$\therefore A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C).$$
Then do the same thing with $(A \cup B) \cap (A \cup C)$.
 A: I wouldn't accept this answer, because it avoids answering the question! For example, why is it that there are those 3 cases as you claim under the first assumption? And then after that why do none of them contradict $x \in ( A \cup B ) \cap ( A \cup C )$? The gap in the explanation is precisely the logic that you want to avoid, which is impossible because set union and intersection are defined logically.
Your attempt at the other question is reasonable, although not quite precise. (It is actually incorrect to use "$\Rightarrow$" because it does not mean what you think it means.) Here is a precise version:
Given $x \in A \cup ( B \cap C )$:
  $x \in A$ or $x \in B \cap C$.
  $x \in A$ or ( $x \in B$ and $x \in C$ ).
  $( x \in A$ or $x \in B$ ) and ( $x \in A$ or $x \in C$ ). [This is by boolean algebra.]
  $x \in A \cup B$ and $x \in A \cup C$.
  $x \in ( A \cup B ) \cap ( A \cup C )$.
Given $x \in ( A \cup B ) \cap ( A \cup C )$:
  [Exactly the same as above but in reverse order. Check that each step actually works!]
Whenever you have two such expressions involving set operations that you want to prove equal, you can expect that this phenomenon would happen. Another way that is often shorter is to literally expand the definition:
$A \cup ( B \cap C )$
$ = \{ x : x \in A \cup ( B \cap C ) \}$
$ = \{ x : x \in A \lor x \in ( B \cap C ) \}$
$ = \{ x : x \in A \lor ( x \in B \land x \in C ) \}$
$ = \{ x : ( x \in A \lor x \in B ) \land ( x \in A \lor x \in C ) \}$
$ = \{ x : x \in A \cup B \land x \in A \cup C \}$
$ = \{ x : x \in ( A \cup B ) \cap ( A \cup C ) \}$
$ = ( A \cup B ) \cap ( A \cup C )$.
