Translating NO and NOT ALL into Predicate Logic

[@iliyasone]

9:14 I think it's incorrect. We know, that no professor fails every student, it's mean that all professors not fail AT LEAST one student, so it would be something like that: Ɐx[Px-> Ǝy[Sy ∧ ¬Fxy]]

[@micaeltchapmi]

I agree as well. I wrote the alternative translation as Ɐx[Px -> ¬Ɐy[Sy -> Fxy]] which I believe is equivalent to ∀x[Px→∃y[Sy∧¬Fxy]]

[@naiko1744]

I agree as well

[@gooddeedsleadto7499]

Thanks for the explanation with sketches. Could u also draw sketches related to the four questions in the end?

[@madelinelong6136]

Thanks for the videos! I would love to see a video on FOL to CNF conversions. :)

[@ali_rauf660]

Thanks for the videos, appreciate a lot !!

[@rhino_for_free]

than you sir it was really helpful

[@Mentalcheez]

I'm confused about the placement of the brackets for the last question. Wouldn't it be NotAllx [ [...] -> Rxy] ? Why is the implication for "becoming richer" not directly consequential to "every kid who writes"?

[@raphaelgomes2947]

For #4, could you write ~∀x[Kid(x) & Writes(x) → Richer(x, ~x)] That sort of makes sense to me semantically but seems to have a contradiction there.

[@abdelkaderbensaid432]

Thank you so much, sir!

[@nickevans8551]

For the alternative method of "No professor fails every student" it's written in the video as ∀x[Px→∀y[Sy→¬Fxy]]. Doesn't this translate as "for all x, if x is a professor, then for all y, if y is a student, then the professor (x) does not fail the student (y)". Wouldn't this be equivalent to saying "Every professor doesn't fail any students" or "No professor fails any students"? It seems like it should be written ∀x[Px→∃y[Sy∧¬Fxy]] so to say "for all x, if x is a professor, then there exists a student (y) that is not failed by professor (x)". Wouldn't this be the be equivalent to "No professor fails every student" or am I missing something?

[@albertobriceno416]

I came to the same conclussion.

[@micaeltchapmi]

I agree as well. I wrote the alternative translation as Ɐx[Px -> ¬Ɐy[Sy -> Fxy]] which I believe is equivalent to ∀x[Px→∃y[Sy∧¬Fxy]]

[@naiko1744]

I agree as well

[@adept2814]

Hey man! What happened to your website? It was nice having an overview of the different discrete maths videos :

[@Trevtutor]

The site is being transformed to contain full courses with lessons, topics, quizzes, and supplemental materials. Unfortunately the playlists will be the best way to find material at the moment. Sorry!

[@DaiMoscv]

Not every professor is evil, maybe the sentence isn't but that's fine, lol

[@tulikamishra9708]

for no dog is happy what if we write.. for all x NOT Dx OR NOT Hx

[@Trevtutor]

That works since it’s equivalent to the paraphrase under the conditional law.

[@xx_foxpvp_xx374]

i love you can i have you and can i please have some more