# KE Deduction

#### lyd123

##### New member
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 dont 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.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
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.

#### lyd123

##### New member
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

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
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 $P=T$.