How do you prove things with null

[pwhitey86]

For instance, the following seems obvious but I don't know how to state the proof formally (and directly):

Show $$A \cap (B-A) = \{\}$$

Here is a try:
For any $$x \in U \ if \ x \in A$$ then $$x \notin (B-A)$$
therefore $$A \cap (B-A) = \{\}$$

there is something missing...

 

[Manchot]

Well, by definition (of set exclusion),
$$x \in B-A \leftrightarrow x \in B \wedge x \notin A$$
Also by definition (of set intersection),
$$x \in A \cap (B-A) \leftrightarrow x \in A \wedge x\in B-A$$
Substituting the first statement into the second, you get
$$x \in A \cap (B-A) \leftrightarrow x \in A \wedge x \in B \wedge x \notin A$$
But the RHS is clearly false (x is not simultaneously in A and not in A), meaning that the LHS is also false. Therefore, for all x,
$$x \notin A \cap (B-A)$$

Therefore, $A \cap (B-A)$ satisfies the defining property of the empty set.

 

[HallsofIvy]

For instance, the following seems obvious but I don't know how to state the proof formally (and directly):

Show $$A \cap (B-A) = \{\}$$

Here is a try:
For any $$x \in U \ if \ x \in A$$ then $$x \notin (B-A)$$
therefore $$A \cap (B-A) = \{\}$$

there is something missing...

You have shown that if x is in A then it is not in $$A \cap (B-A)$$ What if x is not in A? That's what's missing. (Yes, it's trivial but you should say it.)