MHB Understanding Ke Logic Rules & Finding Contradictions

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The discussion revolves around understanding Ke logic rules and identifying contradictions within a set of premises. The user attempts to derive a conclusion, specifically $\neg S$, from the given premises but struggles to find a contradiction. It is clarified that one cannot derive $P \land Q$ from the premises provided, and to prove $\neg S$, one must apply the law of excluded middle or the branching rule. The conversation emphasizes the importance of counterexamples to demonstrate the validity of arguments, particularly in relation to the truth values of the variables involved. Ultimately, the user seeks clarity on how to ascertain the validity of an argument when dealing with negated conclusions.
lyd123
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Hi, the question and Ke logic rules are attached.

This is my attempt at the question.

$1. P \land (R\implies Q) $ Premise
$2. ( P \land Q ) \implies \lnot S) $ Premise
$3. ( P \land S) \implies R) $ Premise
$4. \lnot S $ Conclusion
$5. P \land Q$ $ \beta 2,4$
$6. P $ $ \alpha 5$
$7. Q$ $ \alpha 5$
$8. R\implies Q $ $ \alpha 1$

I don't think the lines I wrote after this make a lot of sense. Usually a contradiction would be found, but in this case I don't seem to find a contradiction. I think maybe I have to negate the conclusion, I thought it was already negated because of the \lnot. But how do I know when the argument form is valid (invalid being if there is a contradiction).Thank you for any help. :)

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I assume that $\neg S$ is the original conclusion, not its negation.

You cannot derive $P\land Q$ from $P\land Q\implies \neg S$ and $\neg S$.

To prove $\neg S$, one must use the law of excluded middle, or the branching rule. From premise 1 we have $P$ and $R\implies Q$. If $S$, then we get $R$ from premise 3, $Q$ from premise 1 and finally $\neg S$ from premise 2. If $\neg S$, then nothing is left to do.
 
Thank you, I understand now.If an argument was valid, how would we know? For example, in this case if the the original was S and the negated conclusion is ¬S ?

1.P∧(R⟹Q) Premise
2.(P∧Q)⟹¬S Premise
3.(P∧S)⟹R Premise
4.¬S Negated Conclusion
 
These premises do not imply $S$. The easiest way to see this is to find a counterexample, i.e., an assignment of truth values to variables that makes all premises true and the conclusion false. In this case it is $R=Q=S=F$ and $P=T$.
 
Evgeny.Makarov said:
To prove $\neg S$, one must use the law of excluded middle, or the branching rule

I suppose you mean : $ S\vee\neg S$

Evgeny.Makarov said:
If $S$, then we get $R$ from premise 3

I suppose you mean from P and S we get : $P\wedge S$ and then using premise 3 and α rules (Modus Ponens) we get $R$

If yes, there is no rule in α or β rules to account for: $P$,$S$ $\Rightarrow P\wedge Q$
 
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