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Propositional logic problem


Active member
Jan 4, 2014
I've to derive the following proposition in PL using the system in this thread (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.
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Indicium Physicus
Staff member
Jan 26, 2012
It looks good to me. I don't see any reason why you can't ignore the premiss $P\lor Q$.