I know this is a simple proof, but how do I go about address the if then statements in a set proof using element arguments? I'm having a hard time wrapping my head around the right way to approach these proofs.
Here are two examples
Example #1
\begin{align}
&\text{Prove: for all sets}\,A\,\text{and}\,B\,\text{if}\,A \subseteq B\,\text{then}\,A \cup B \subseteq B \\
&\text{by definition of Union}\,x \in A\,\text{or}\, x \in B\\
\end{align}
Example #2
\begin{align}
&\text{Prove: if}\,B \cap C \subseteq A,\,\text{then}\,(C-A)\cap(B-A)=\emptyset\\
\end{align}
I get stuck after this. What is the right way to approach the next step in addressing the if then statement of the proof?
Best Answer
If i understand your question correctly, take example #1, the if..then statement is basically an implication, that is, you need to prove $$ A \subseteq B \implies A \cup B \subseteq B $$ therefore, you can just assume that the premise holds (i.e. $A \subseteq B$) and try to show the validity of the consequent (i.e. $A \cup B \subseteq B$).
In this case, the proof boils down to applying the definition of set inclusion and union, sketching the approach:
Assume that for all $x$, $x \in A \implies x \in B$ (that is, $A \subseteq B$, our if..then hypothesis)
then under this assumption we show $A \cup B \subseteq B$ by taking a generic element $y \in A \cup B$ and showing that $y \in B$ (again, by applying the definition of set inclusion)
This is trivial since by definition of set union $y \in A \vee y \in B$, and thus we distinguish the two cases
The approach for example #2 is essentially the same, you just need to apply other set theory definitions, such as intersection, set difference, complement. I leave the details and the other example to you, hope this helps!