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Why is modus ponens in prepositional logic considered a valid form?

I can think of an example where a true premise leads to a false conclusion:

If the kid is wet in the winter, then it was raining on him

The kid is wet in the winter

Conclusion: it was raining on him

Where, in fact,

The kid is wet in the winter

A passing car splashed water on him

Therefore, it is not the case that it was raining on him.
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    $\begingroup$ In the example you give, the conditional statement "If the kid is wet in the winter, then it was raining on him" is clearly false as demonstrated by the following statements you make. Modus ponens requires the implication in question to be true. $\endgroup$ Jun 23 at 20:59
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    $\begingroup$ The result of an argument is true if its form is valid and its propositions are true. So here we have an argument with a valid form but false propositions, so the result being wrong doesn't mean that the form is invalid. $\endgroup$ Jun 23 at 21:00
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    $\begingroup$ @Dasem But one of your premises is not true in the example you give. Both premises, the conditional statement and the hypothesis, have to hold in order to apply modus ponens: MP says that $Q$ is true whenever $P$ and $P\rightarrow Q$ are true. $\endgroup$ Jun 23 at 21:07
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    $\begingroup$ Not sure why this is getting downvoted. It's an honest question with an attempt. I see questions with absolutely no attempt get upvotes. $\endgroup$ Jun 23 at 21:09
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    $\begingroup$ I think the issue is that you've missed that the conditional itself, as a required bit of knowledge, is itself a premise in your argument. If there's another way that the child could have gotten wet other than getting rained on, then the conditional in your argument is simply not true. $\endgroup$ Jun 23 at 21:11
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Maybe writing it formally helps. To deduce $q$ through modus ponens, we need:

$(p \implies q) \wedge p$

i.e. the premise p and the implication itself must be true. Remember that $(p \implies q)$ is itself a proposition, either true or false. In your case, $p \implies q$ is NOT true, since a kid being wet does not imply that it has rained.

Modus ponens is the inference rule itself, but it doesn't tell you about whether the premises are true or not.

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  • $\begingroup$ Thanks i will study your answer, but as far as i understand in general, p imply q or if p then q is a given definition (sorry i don't know hot to type in formal syntax and I'm typing from my phone) or a Premise, and then we are given that p is true, according to my understanding we can't conclude q because there isn't additional information that attach p to q in order for us to conclude q, and that why i chose this example in the post, but i assume i missing something cause right now it isn't make sense to me, but as i said i will study your answer carefully, thanks. $\endgroup$
    – Dasem
    Jun 23 at 22:31
  • $\begingroup$ Even though i excepted the answer my question still remains unanswered since they closed my other post/question on this subject, which mean they have created this wired situation by closing my other question and tagging it as duplicates even though they didn't even understand the question, but never mind. $\endgroup$
    – Dasem
    Jun 25 at 9:36
  • $\begingroup$ But i will summarize my findings after dealing with it for a few days: modus ponens is a weak inference rule the only place it's solid is in math where it's represents there a neccessary connection any other places like in natural language as my example shows it can give a false conclusion and therefore it's invalid. $\endgroup$
    – Dasem
    Jun 25 at 14:50
  • $\begingroup$ Your example is not a correct application of the inference rule called modus ponens. So no, it does not provide a counterexample and modus ponens is completely valid and is used extensively in both formal and natural language. Please study my answer and the comments below your question more carefully. $\endgroup$
    – WhiteLake
    Jun 25 at 14:59
  • $\begingroup$ @ Lili i understand your answer, and i accepted it because at that time the truth table didn't show me a contradiction when i assumed q is false, and that happened because my logic book had a style of presenting the argument in 3 separated lines or columns and with your answer i manged to understand that the (and) operator was missing, after that the truth table worked. $\endgroup$
    – Dasem
    Jun 25 at 15:06

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