A simple problem involving natural deduction - so what am I doing wrong? The problem is straight-forward. I'm supposed to derive the following in propositional logic, using natural deduction:
$$
\lnot (A \wedge B) \vdash \lnot A \vee \lnot B
$$
Here is my initial attempt at a derivation, using a tree structure:
$$
\dfrac{\dfrac{\dfrac{\lnot (A \wedge B) = (A \wedge B) \rightarrow \bot \qquad (A \wedge B)}{\bot}\small{\rightarrow E}}{\lnot A}\small{RAA}}{\lnot A \vee \lnot B}\small{\vee I}
$$
When I arrive at the top-line, I realize that the only way (?) for me to construct $(A \wedge B)$ is to use and-introduction. If I assume $A$ above this line, the noted $RAA$ discharges this particular assumption. However, I'm not quite sure how to handle $B$ in this instance. Have I worked myself into a corner here? Is there a better way to do this? Any feedback would be greatly appreciated!
 A: So far, so good.  But first, you want to assume $A$ and $B$ seperately, so that they are the assumptions you discharge, not $A\land B$.
Add an assumption of $\lnot(\lnot A\lor\lnot B)$ and you can derive another contradiction to discharge the assumed $B$ in exactly the way you did with the assumed $A$.
$$\dfrac{\dfrac{\dfrac{\dfrac{\dfrac{\dfrac{\dfrac{\neg(A\land B)}{(A\land B)\to\bot}{\small\textsf{def }\lnot}\quad \dfrac{[A]^1\quad[B]^2}{A\land B}{\small\land\mathsf I}}{\bot}{\small{\to}\mathsf E}}{\lnot A}{\small\lnot\mathsf I^1}}{\lnot A\lor \lnot B}{\small\lor\mathsf I}\quad\lower{1ex}{\lnot(\lnot A\lor\lnot B)}}{\bot}}{\lnot B}{\small\lnot\mathsf I^2}}{\vdots}$$
Note: These are not RAA, but Indirect Proof $IP$ also know as Negation Introduction ($\lnot\mathsf I$).  You assume a proposition, derive a contradiction, therefore discharge to deduce the negation of the proposition.  Your proof system might also be using conditional introduction and negation's definition to implement this (as they do with negation elimination).
$$\dfrac{\dfrac{\dfrac{\lower{5ex}\vdots\quad{[A]^1\\~~~\vdots}}{\bot}}{A\to\bot}{\small{\to}\mathsf I^1}}{\lnot A}{\small\text{def }\lnot}$$
Reduction to Absurdity is where you assume a negation of a proposition, derive a condition, and therefore discharge to deduce the proposition.
