- #1

lyd123

- 13

- 0

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. :)

View attachment 8735

View attachment 8736

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. :)

View attachment 8735

View attachment 8736