Is this a correct natural deduction proof for $\{(\phi\vee\psi),\neg\phi\}\vdash\psi$? I'm not sure I used RAA correctly by putting $\neg\psi$ next to $\bot$ and discharging it.

 A: I see the idea (the notation is different to what I'm used to).
You want to use $\lor$-elimination to conclude $\psi$ from $\phi \lor \psi$. 
My write-up would be (the left hand clauses I call axioms)


*

*$\phi \lor \psi$ (axiom)

*$\psi$ (assumption 1)

*$\psi \Rightarrow \psi$ (assumption 1 is dropped), by $\Rightarrow$-introduction.

*$\phi$ (assumption 2)

*$\lnot \phi$ (axiom) 

*$\bot$ (from 4 and 5), introduction of $\bot$.

*$\psi$ (from 6 and Ex Falso Quodlibet, EFQ)

*$\phi \Rightarrow \psi$ by $\Rightarrow$ introduction. Asumption 2 dropped. 6,7 go away.

*$\psi$, by 1., 3. and 8. and $\lor$-elimination. 


(normally I'd use indentation for subproofs of implication, RAA etc. as well)
If I understand your proof correctly your strategy is a bit different: you first assume $\phi$, from which you get $\bot$ (using the axiom $\lnot \phi$), and then (maybe you don't have EFQ as a rule) you use $\lnot \psi$ as a temporary second assumption to get $\psi$ with RAA (reductio ad absurdum). The assumption $\psi$ goes away and then $\phi$ goes away to get the implication. This is also possible, of course. The RAA is used correctly, AFAIK. But I'd use Ex Falso Quodlibet, which is a bit more direct.
