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Propositional logic problem

Guest

Active member
Jan 4, 2014
199
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.
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
It looks good to me. I don't see any reason why you can't ignore the premiss $P\lor Q$.