[Logic] Free and Bound Variables

[@beaucarrel2778]

You've explained this very well! Better than my teacher! I understand now, thank you TheTrevTutor!

[@mariyamulla8843]

Thank you very much for this video -gratitude from last minute studying student🙏

[@ABDULRAHMANYAKOOB-b7u]

you've explained this in a very simple manner, all my confusion regarding the free and bound variables is clear. Thank You so much!!!!!

[@Trevtutor]

Ah this is an older version from an older playlist but I’m glad it still helped!

[@coffee-syrup]

I'm starting studying 4 days prior to my exam, definitely anxiety and tdah acting up. I hope I can pass this exam! By far thanks to you I'm having better understanding, today I'm revising all the theory and next days I will be doing exercises. Man I fear natural deduction so much!

[@SonGoku-fx4pg]

Unlike my professor's explanation, yours is easy for a human to comprehend. Thank you!

[@ThisIsNotMyHandle]

Great job. Would love to see a video on natural deduction and predicate logic too.

[@RondellKB]

OMG me too :'(

[@eljuntero1]

That's exactly what I wanted to ask him! It's the only thing missing in his lectures. It would really be of great help. Anybody knows a good lesson explaining natural deduction for first order logic? I reaally need help!

[@gothfrog69]

You sir are an excellent teacher!

[@physics_philosophy_faith]

Yes please do modal logic! This is great and this whole series has been. It's the only time I can say that I learned something faster and more efficiently than with a book

[@teinili]

The Explanation of the negation blew my mind :D It's so obvious this way that I dont understand why my prof didn't tell us that but let us just remember some weird rule

[@jpnesseth]

Love more expansion on the "problem" / ontology of free variables beyond mere syntax & wff formation.

[@kylechurch6296]

I thought this was an excellent series, too, and I hope you take up teaching logic again, Trevtutor. Thank you very much.

[@shaneri]

thank you ! helped me a lot best wishes to your channel!

[@Peter-bg1ku]

You cleared all my confusion within the first 40 odd seconds

[@jasdeep4000]

Excellent video

[@lucasCulhaci]

Thank you for this vid! you explained this better than my teacher ngl

[@nachocapic]

That proof was amazing, thank you.

[@SeekingUltimateSynthesis]

These last two videos of the series were great....still didn't get the soundness and completeness sections for now. Will have to go get an actual logic book... Since you did say that this should be a companion to a logic course/book... not a standalone course I should have probably gotten the book to begin with....

[@waterbottlexd1298]

Thank you!

[@nosborwilliansen8750]

Gosh, you're amazing! Thank you!

[@eljuntero1]

Great lecture TheTrevTutor, thanks a lot! Do you know a good lesson for natural deduction in predicate logic? You did a lecture for propositional logic that was really great, but I can't find clear lessons about predicate logic on the internet....

[@TechVenturer]

really helpful

[@rioferianto1278]

Thanks !!!

[@kevinbenavides92]

Thank you!

[@EM-fb8zg]

For your own sanity has me dead

[@kakomakoliko9742]

never got those proofs

[@farah-bq3rg]

wow so helpful

[@GeorgeTFilm]

I am confused about the last slide. Surely ∃𝑥(P𝑥) ¬∀(Px) ?? Why must we always say that ∃𝑥(P𝑥) ¬∀¬Px?? Both are true; Not all x are Px, and not all x are NOT Px... So why must me always say the latter?

[@ausoniacalabrese7916]

Let's put these formulae into natural language: ∃𝑥(P𝑥) There exists some x with the property P. ¬∀𝑥(P𝑥) It is not the case that for all 𝑥, 𝑥 has property P. So, the negation of the first statement: ¬∃𝑥(P𝑥) It is not the case that there exists some 𝑥 with P. Is equivalent to: ∀𝑥(¬P𝑥) For all x, it is not P. The first statement, ¬∃𝑥(P𝑥) says that there is no such thing as an 𝑥 with the property P. Thus, for all x, none of them have the property P. Thus, they are equivalent statements. Here's an example in practice: ∃𝑥(P𝑥) There exists some person with blonde hair. ¬∀𝑥(P𝑥) It is not the case that all people have blonde hair. ¬∃𝑥(P𝑥) There does not exist any person with blonde hair. ∀𝑥(¬P𝑥) For all people 𝑥, they do not have blonde hair. Last thing: I think I answered your question, but it is kind of vague since the formulae you reference in your comment (like "¬∀¬P𝑥") don't make sense (i.e., is not a well-formed formula.) The quantifier ∀ does not have a variable.

[@markomaricofficial]

Cover modal logic

[@njmusama]

ecellant

[@linazeghlache5251]

so we're basically racist towards free variables

[@escargol173]

THANK YOU!