fitch style proof problem regarding rule of inferences 
Okay I have the following fitch proof almost figured out, from here I know I have to move forth in conlcuding but I don't know what I am missing
(2 premises)
S → (R ∨ P)
P → (¬R → Q) 
my work below
new scope assume S   Rule: > elim
R ∨ P
New scope assume R
Q∨R     Rule: V intro
End scope – (new scope assume P)
¬R → Q         Rule: > Elim
my confusion comes where I go from here
Q∨R            rule Taut Con ??
Goal:├ S → (Q ∨ R)
 A: Do not assume $p$, rather assume $\neg r$ and use the Law of Excluded Middle.
$\begin{array}{r|l:l} 1& s \to r\vee p \\[-1ex]
2 & p\to(\neg r\to q)\\[-1ex]
3 & \underline{\vert\quad s\qquad} \\[-1ex]
4 & \vert\quad r\vee p\\[-1ex]
5 & \vert\quad\underline{\vert\quad r\qquad}\\[-1ex]
6 & \vert\quad\vert\quad q\vee r \\[-1ex]
7 & \vert\quad r\to q\vee r\\[-1ex]
8 & \vert\quad\underline{\vert\quad \neg r\qquad}\\[-1ex]
9 & \vert\quad\vert\quad p\\[-1ex]
10 & \vert\quad\vert\quad \neg r\to q\\[-1ex]
11 & \vert\quad\vert\quad q\\[-1ex]
12 & \vert\quad\vert\quad q\vee r\\[-1ex]
13 & \vert\quad \neg r\to q\vee r\\[-1ex]
14 & \vert\quad r\vee\neg r & \star\\[-1ex]
15 & \vert\quad q\vee r\\[-1ex]
16 & s\to q\vee r
\end{array}$

Revision: No, of course you may assume $p$ and move the LEM into the subproof.   It makes the proof a little longer, but Disjunctive Syllogism may not be available as a basic rule for inference.
$\begin{array}{r|l:l} 1& s \to r\vee p \\[-1ex]
2 & p\to(\neg r\to q)\\[-1ex]
3 & \underline{\vert\quad s \qquad}\\[-1ex]
4 & \vert\quad r\vee p\\[-1ex]
5 & \vert\quad\underline{\vert\quad r\qquad}\\[-1ex]
6 & \vert\quad\vert\quad q\vee r \\[-1ex]
7 & \vert\quad r\to q\vee r\\[-1ex]
8 & \vert\quad\underline{\vert\quad p\qquad}\\[-1ex]
9 & \vert\quad\vert\quad \neg r\to q\\[-1ex]
11 & \vert\quad\vert\quad\underline{\vert\quad \neg r\qquad}\\[-1ex]
12 & \vert\quad\vert\quad\vert\quad q\\[-1ex]
13 & \vert\quad\vert\quad\vert\quad q\vee r\\[-1ex]
14 & \vert\quad\vert\quad \neg r\to q\vee r\\[-1ex]
15 & \vert\quad\vert\quad r\vee\neg r & \star\\[-1ex]
16 & \vert\quad\vert\quad q\vee r\\[-1ex]
17 & \vert\quad p\to q\vee r\\[-1ex]
18 & \vert\quad  q\vee r\\[-1ex]
19 & s\to q\vee r
\end{array}$
