1
$\begingroup$

I have read some of the answers on similar questions but I can't really get my head around this.

So, here are 2 questions I need to answer.

Show using a truth table:

That the inference from:

$ p\rightarrow(q \land r),\neg q$ to $ \neg p$ is valid

And the inference from:

$p\rightarrow(q\lor r),\neg q$ to $\neg p$ is not valid.

Ok so if I'm correct a truth table contains all possible situations (or valuations) of an argument. An inference would be getting a conclusion out of two pieces of information $ p\rightarrow(q \vee r),\neg q$ and $\neg p\rightarrow(q\wedge r),\neg q$ respectively for both questions, and drawing a conclusion based on that.

An argument or inference is valid if the conclusion is true as well as the "two pieces of information" or "premises", correct? This would mean that, a situation or "valuation" (only one out of a possible 8 in the case of my examples) where $ p\rightarrow(q \vee r),\neg q, \neg p$ should be true?

So all I would have to do is make a truth table, where all the true-false values are present for the given formulas, and look for a "line" (aka valuation) in the truth-table that has that property?

Because I feel I'm making things way more complex than they have to be. I was looking into modus tollens, but that only applies to a situation where you have 2 formulas and not 3 (as is the case in both of my questions). However if it is possible to answer these questions with modus tollens, how do I do this? Or could you show me how?

Thanks in advance, Rope.

PS. If you could also tell me if my assumption about the definitions of some of the keywords in this question I stated are corect, that would be awesome :P.

$\endgroup$
  • $\begingroup$ I do actually, thanks for pointing that out. But I still don't grasp the concept, I think.. $\endgroup$ – Byebye Aug 15 '14 at 12:15
1
$\begingroup$

An inference (or argument) from given premises to a conclusion is valid if and only if there is no possible situation in which the premises are true and the conclusion is false. So, talking about propositional logic and truth-tables, the inference is valid (or the premises entail the conclusion) if and only if there is no valuation of the atoms (p, q, r,...) that makes the premises true and the conclusion false.

Check that your definition is not correct, since you say that both premises and conclusion must be true for some valuation. But the correct definition above means that the only case in which an inference is invalid is when there is a line in the truth-table (so there is a truth-valuation) which makes the promises true and the conclusion false. Otherwise, the inference is valid.

I would recommend checking the book An introduction to formal logic, by Peter Smith. Or A modern formal logic primer, by Paul Teller, freely available from the author (check Chapter 4)

$\endgroup$
  • $\begingroup$ Thanks @Zamu, I reread the defintion in the chapter I'm reading and it says the same thing :). $\endgroup$ – Byebye Aug 15 '14 at 14:18
3
$\begingroup$

Modus tollens does indeed apply. In each scenario, you are given two premises, and asked if you can infer, from the two given premises, the conclusion, $\lnot p$. It is also the case that we need to determine whether from $\lnot q$ we can then infer $\lnot(p\land q)$ in the first case, or $\lnot(p \lor q)$ in the second.

In the first case, put $t = (q\land r)$. Now, we have $p \rightarrow t$. If we can assert $\lnot t$ from the second premise $\lnot q$, then by modus tollens, we can validly infer $\lnot p$.

So when can we assert $\lnot t = \lnot(q \land r) \equiv \lnot q \lor \lnot r$?

Since we have $\lnot q$ as a premise, we can certainly infer $\lnot q \lor \lnot r \equiv \lnot (q \land r)$. That is, we can validly conclude, given $\lnot q$, that it is not the case that $q\land r$, i.e. that $\lnot (q\land r)$ holds. So $p \rightarrow (q \land r), \quad \lnot (q \land r)\,$ together imply $\;\lnot p$.


However, in the second case, put $t = q\lor r$. Having $\lnot q$ tells us nothing about whether we can negate $t$. It might be the case that $r$ is true, in which case $t$ is true. So we cannot negate $t =q\lor r$ knowing only $\lnot q$. Since we cannot negate $(q\lor r)$ based only on $\lnot q$, we cannot validly infer $\lnot p$.

$\endgroup$
  • $\begingroup$ Thanks for the answer @amWhy, I get it now. I have just one question about the first answer. How is it that when you have no premise about $r$ (so you know nothing about r) you can infer that since $\neg q$ is a premise $\neg r$ must follow from $q \wedge r$? Is it because the binary operator must still hold for $\neg q$? And the only solution to that is to negate $r$? $\endgroup$ – Byebye Aug 15 '14 at 12:44
  • $\begingroup$ It doesn't follow that $\lnot r$. Look at the truth table for $q \land r$. What happens when we know that $\lnot q:$ i.e. when q is false? A conjunction is true if and only if every conjunct is true. When one proposition of a conjunction is false, the entire conjunction is false, irregardless of whether the other propositions are true or false. If we have $q \land r$, and $\lnot q$, we can infer $\lnot(q \land r)$. The truth-value of $r$ is irrelevant. $\endgroup$ – Namaste Aug 15 '14 at 12:51
  • $\begingroup$ Aaah ok I get. So the assurtion is that because $\neg q$ is a premise $\neg (q \wedge r)$ follows as both q and r must be true for the proposition to be true. Thanks! $\endgroup$ – Byebye Aug 15 '14 at 12:56
  • $\begingroup$ You're welcome, Rope! $\endgroup$ – Namaste Aug 15 '14 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.