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.)