Set Theory Proofs - If then statements 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?
 A: 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

*

*$y \in B$ trivially implies the consequent we are looking for (that is, exactly $y \in B$)

*$y \in A$ implies $y \in B$ since our if..then hypothesis lets us infer so.

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!
A: For your question, I believe its important to know the structure of your proof.
Proof: If (A\subset B), then (A\cup B) \subset B.
Since (A\subset B), every (x\in A) is in B).
To show (A\cup B \subset B), we must verify if (y\in A\cup B) then (y\in B).
Since (y\in A\cup B), we know (y\in A) or (y\in B).
If (y\in A) then we know (by A\subset B), (y\in B).
Thus, (y\in B).
Ergo, If (A\subset B), then (A\cup B) \subset B.
