Negating statements / Finding $(A \cap B)',A \oplus B$ if $A=\{x \in\Bbb R \mid -3\le x\le0\}$ and $B=\{x \in \Bbb R\mid -1 < x < 2\}$ I am a bit new on this field and I am trying to solve some questions. I don't really think they are hard but there are some key points that I don't get it or I am stuck.
Lets see.

1) Write the negation of the sentences:
  1a: "If the teacher is missing, some students will not do their homework."

My approach: Let $p=$ The teacher is missing; $q=$ some students; $r=$ do their h/w. That would be  $(p \to (q \land \neg r) )$. I searched for "De Morgan's law", but didn't find anything about what to do with the negation of "$\to$".

1b. Find the negation of: $(\forall x \in D)(x+4 \le 8)$. 

I really have no idea, but the obvious would be $(\not\exists x \notin D)(x+4 \le 8)$.
Also the negation of $\exists$ would be $\not\exists$, right?   

2) $A=\{x \in\Bbb R \mid -3\le x\le0\}$ and $B=\{x \in \Bbb R\mid -1 < x < 2\}$
2a) Find $(A \cap B)'$.

I think this is simple, so the answer would be all the numbers $\in \Bbb R$, except the numbers $-1 < x \le 0$ (That would be their common numbers). That makes sense to me, I mean a lot of sense.

2b) $A \oplus B$. 

No I cant even visualize this. I know the truth table of xor at least, but this cant really help me to continue.
I would really appreciate some input, at least for my answers if possible.
Thanks in advance!   
 A: Re Ia) "Some students" is not a proposition is it? So you can't use a propositional variable to represent that phrase!

If the teacher is missing, some students will not do their homework

can be rendered

$(P \to Q)$

where $Q$ represents the proposition "some students will not do their homework", and $\to$ represents the conditional construction. The issue of the negation of conditionals is a vexed one. If the conditional can be read as the material truth-functional conditional then

$\neg(P \to Q)$

is equivalent to

$(P \land \neg Q)$.

But if not, then not!
1b) 

$(\forall x \in D) (x+4 \leq 8) $

is short for

$\forall x(x \in D \to x+4 \leq 8) $

where this time the material conditional is definitely intended. Which is equivalent to

$\neg\exists x\neg(x \in D \to x+4 \leq 8) $

So its negation is

$\exists x\neg(x \in D \to x+4 \leq 8) $

which is equivalent to 

$\exists x (x \in D \land \neg(x+4 \leq 8)) $

or

$\exists x (x \in D \land x+4 > 8) $

which abbreviates to

$(\exists x \in D)(x+4 > 8) $

which is -- if you think about it -- intuitively the right answer.
A: Since Peter already addressed question 1, let me comment on 2.
Your idea for 2a. is correct. So let us try to be a little more specific. We currently have:
$$(A\cap B)' = \{x \in \Bbb R\mid \text{not}(-1 < x \le 0)\}$$
The statement $\text{not}(-1<x\le 0)$ can be simplified: it must be that either $-1 \not <x$, i.e. $-1 \ge x$, or $x \not\le 0$, i.e. $x > 0$. In conclusion:
$$(A\cap B)' = \{x \in \Bbb R \mid x \le -1 \text{ or } x > 0\}$$

Now for 2b. By definition of $\oplus$, we have that: $$A \oplus B = \{x \mid x \in A \text{ or } x \in B, \text{ but not both}\}$$
which can be written as $$A \oplus B = (A\cup B)\setminus (A \cap B)$$ 
with $\setminus$ denoting set difference, $C \setminus D = \{x \in C: x \notin D\}$.
We had already determined $A \cap B$. Can you also do $A \cup B$ and after that, $A \oplus B$? I advise you to stay close to the definitions, write out everything meticulously, and simplify your answer as much as possible. That's the best way to get to grips with it.
