I have a question about the following proposition as argument, I need to get to an argument that has no number, from 1-2-3 premises. derived: $$ \begin{align} &(P \vee Q)\vee M \\ & R\supset S \tag{1} \\ & (P\equiv M) \,\&\, ( P\vee (R\vee M)) \tag{2} \\ & S\supset \,\sim\! R \tag{3} \\ \end{align} $$
I appreciate any cooperation
Outline of deduction:
From (1) and (3), $R \to \neg R$, which is equivalent to $\neg R \vee \neg R$, which is to say, $\neg R$.
From (2), $P \vee (R \vee M)$, so $P \vee M$ (because not $R$).
From this, it follows that $(P \vee Q) \vee M$, as $P \to (P \vee Q)$.
Note that the equivalence $P \equiv M$ wasn't used.