MHB Can the premise P∨Q be ignored in a propositional logic proof?

[Guest2]

I've to derive the following proposition in PL using the system in http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/propositional-logic-8386.html (in which Evgeny.Makarov has explained everything ever so kindly to me).

I'm trying to prove $\displaystyle P \vee Q, ~(R ~ \& ~ P) \to \neg Q, ~R, ~ R \to P: \neg Q$. I tried using disjunctions by assuming P to get the conclusion - but then when assumed Q to get the conclusion (its negation) I got stuck. Someone told me that I can ignore $ P \vee Q$ and get the conclusion without. Is this really allowed? In that case I could do the following, I think:

$ \begin{aligned} & \left\{1\right\} ~~~~~~~~~ 1. ~ P \vee Q\ldots \ldots \ldots \ldots \text{premise}
\\& \left\{2\right\} ~~~~~~~~~ 2. ~ \left(R ~ \& ~ P\right) \to \neg Q \ldots . \text{premise}
\\&\left\{3\right\} ~~~~~~~~~ 3. ~R \ldots \ldots \ldots \ldots \ldots \ldots \text{premise}
\\&\left\{4\right\} ~~~~~~~~~ 4. ~R \to P \ldots \ldots \ldots \ldots\text{premise}
\\&\left\{3, ~ 4\right\} ~~~~~ 5. ~P \ldots \ldots \ldots\ldots\ldots .. 3, ~4 \text{MP}
\\&\left\{3, ~4\right\} ~~~~~ 6. ~ R ~ \& ~ P\ldots \ldots \ldots \ldots. 3,5 \text{& I}
\\&\left\{2,~3,~4\right\} ~7. ~ \neg Q \ldots\ldots\ldots\ldots\ldots \text{2, 6 MP} \end{aligned} $

But I'm not sure whether I can really do that. My book has something it calls 'augmentation' and I suspect it may have something to do with that.

 

[Ackbach]

It looks good to me. I don't see any reason why you can't ignore the premiss $P\lor Q$.