I would like to deduce $R$ from the following set of well formed formulas $\Gamma =\{S\rightarrow R, \neg R \rightarrow P, S \lor \neg P\}$. The only rules that I am allowed to use for deduction are as follows:
(1) I may define any tautology e.g. write $(S \lor \neg P) \rightarrow (P \rightarrow S)$.
(2) I may write any formula either previously deduced or in $\Gamma$
(3) I may use modus ponens e.g., if I know $\beta_1 \rightarrow \beta_2$ and $\beta_1$, then I also know $\beta_2$.
My thinking and attempts prior to asking this question are as follows: I am able to reach $\neg P \rightarrow R$ and $P \rightarrow S$, $S \rightarrow R$. I know that in a traditional first-order setting I might approach this by assuming $P$ and reaching $R$ and subsequently assuming $\neg P$ and also reaching $R$ and by exhaustion of cases reaching $R$. However, given the rules outlined above for this kind of deduction, I am not allowed to do this.
Additionally, my instinct is to come up with some kind of tautology of my own invention like $R \lor \neg R$ and use that somehow, but my attempts haven't amounted to anything. It has been difficult for me that $\Gamma$ doesn't include an assumption of the truth of any propositional atom, e.g. $S$, etc.
Any thoughts, hints, etc.?
If you can really define any tautology you want, then how about using the following tautology:
$$(S \rightarrow R) \rightarrow ((\neg R \rightarrow P) \rightarrow ((S \lor \neg P) \rightarrow R))$$
3 times Modus Ponens, and Presto!