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  1. $(p \land q ) \iff (r \implies s)$

  2. $(s \vee \neg t)$

So, $\neg s \implies ((r \implies \neg p) \vee \neg t)$

My lecturer has written in our notes that this statement is invalid, but I'm not so sure. I've attached my workings and have found there to be a contradiction whilst using the 'no counterexample' method (i.e. assume the premise to be T whilst the conclusion F, if there is a contradiction then the statement is valid) to find the validity of this statement. Thus, I think this inference is valid.

My question is: is the above inference valid or not?

Thank you so much in advanced for your help! :)

My workings out

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We will follow the instructor's notes and assume that the inference is invalid.

This means : assume the premises T and the conclusion F.

We may rewrite the conclusion as :

$s \lor \lnot r \lor \lnot p \lor \lnot t \ $;

in order to have it F we need :

$s=$ F and $r=p=t=$ T.

With this truth-assignment, the second premise : $s \lor \lnot t$ would be F.

Thus, the argument is valid.


If instead the sought conclusion is : $¬s \to ((r \to ¬p) \land ¬t)$, with the same approach we have two possibilities in order to "falsify" it; either :

(i) $s=$ F and $t=$ T, or :

(ii) $s=$ F and $r=p=$ T.

In the first case, again, we cannot satisfy the second premise.

In the second case, we have two variables "undefined" : $t$ and $q$.

The two premises are equivalent to :

$($ T $\land q) \leftrightarrow$ F

F $\lor \lnot t$.

Thus, if we set : $q=t=$ F, we can satisfy both premises and we have shown that the argument is invalid.



Conclusion :

Is this a valid or invalid inference?

It depends on the formula (in this case : the conclusion) ...

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If s is not true, then (not t) has to be true, thus the inference is automatically true.

On a side note, I think that you may have made a typo...

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  • $\begingroup$ Yes there is a typo in the notes. You can see I've circled the t:F as I keep getting t:T $\endgroup$ – Andy Oct 13 '16 at 8:37
  • $\begingroup$ p.s the typed notes are by my lecturer and the hand written notes are mine. :) $\endgroup$ – Andy Oct 13 '16 at 8:40
  • $\begingroup$ If you do not change the "and" to "or", the interference would be invalid and the counterexample is given in step 5... I think. $\endgroup$ – Ginger88895 Oct 13 '16 at 8:43
  • $\begingroup$ Hmmm, my lecturer emailed and told us to do exactly that: change the 'and' to the 'or'... I've emailed him and asked what's going on. Will update you when I have a reply from him! Thanks @user370007 for your reassurance! Been trying for ages to figure this one out :P $\endgroup$ – Andy Oct 13 '16 at 8:49
  • $\begingroup$ I'm quite interested in this too :D $\endgroup$ – Ginger88895 Oct 13 '16 at 9:05

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