Do probability relations use the same rules as formal logic? I tried to show the probability of event $E$: Either $A$ is true, or $B$ is true, but $A$ and $B$ can't be true at the same time, given that all we know are quantities $P(A \cap B)$ and $P(A \cup B)$:
$$\Pr(E) = \Pr((A \cup B) \cap \lnot(A \cap B))$$
$$\Pr(E) = \Pr((A \cup B) \cap (1 - (A \cap B)))$$
$$\Pr(E) = \Pr((A \cup B) - (A \cup B)\cap(A \cap B))$$
$$\Pr(E) = \Pr((A \cup B) - (A \cap B))$$
Is this a correct proof? Does it make sense to use this notation or to treat it like a logic equation with $\lor, \land$, De Morgan's laws, etc?
 A: Taking the probability of $P(A \cup B)$ is really the equivalent of asking about $P(x \in A \vee x \in B)$. Probabilities are nothing more than just operations on sets. Therefore, anything you're doing would just be a logical operator on sets. Therefore, you can apply the rules of mathematical logic on it. 
A: Close.   Your proof is more correctly written using set operators rather than logical connectives:
Events are considered sets of outcomes.   Probability is a measure of these sets with certain properties; and one of these is that probability measure is: additive for unions of disjoint sets.
$$\begin{align}\Pr(E) ~=~& \Pr\big((A \cup B) \cap (A \cap B)^\complement\big) &~=~& \Pr(X\cap Y^\complement)
\\[1ex] ~=~& \Pr\big((A \cup B)  \setminus((A\cup B)\cap  (A \cap B))\big) &~=~& \Pr(X\setminus (X\cap Y))\\[1ex] ~=~& \Pr(A \cup B) - \Pr\big((A \cup B)\cap(A \cap B)) &~=~& \Pr(X)-\Pr(X\cap Y)\\[1ex] ~=~& \Pr(A \cup B) - \Pr(A \cap B) &~=~& \Pr(X)-\Pr(Y)
 \end{align}$$

Or perhaps more clearly:
When you have some events (sets) $X,Y$, where $Y\subseteq X$, then $Y$ and $X\setminus Y$ are disjoint; and also their union is $X$.    Hence by the afore mentioned rule that the measure is additive for disjoint sets we have:
$$\begin{align}\Pr(X) ~=~ &\Pr(Y\cup (X\setminus Y))\\[1ex] ~=~& \Pr(Y)+\Pr(X\setminus Y)\\[2ex]\therefore ~\Pr(X\cap Y^\complement)~=~& \Pr(X)-\Pr(Y) 
\end{align}$$
Now when $X=A\cup B, Y=A\cap B$, then $Y\subseteq X$, so we have established that:
$$\begin{align}\Pr\big((A\cup B)\cap(A\cap B)^\complement\big)~=~&\Pr(A\cup B)-\Pr(A\cap B)
\\[2ex]\therefore ~ \Pr(E)~=~&\Pr(A\cup B)-\Pr(A\cap B)
\end{align}$$
$\blacksquare$
