From 100 level - 300 level, wednesday 13th of March 2024 will be the day I will be writing my last symbolic logic exam in Philosophy department, university of Lagos, Nigeria. All thanks to you Prof. Mark Thorbsy for making symbolic logic easy for me all through the years✨🙇🏾‍♂️😮‍💨

Mr. Thorsby, you have completely saved me from my terrible logic professor, thank you so much

It's so helpful to just hear/see another way of explaining the rules and how to approach the proofs. And it's especially helpful for me since we are using this same book. Thank you!

Typo @ 19:08 the conclusion is (x)(Px -> Cx)

@mrguysnailz4907

8 жыл бұрын

+Cameron Jones bless you, I came right to the comments looking for something to prove I'm not mad

@stevenf5902

3 жыл бұрын

I noticed that too! Phew!

You the man Thorsby

dude, you made this sound not as complicated as my professor made it

This was SO helpful for my final, thank you!

this is a gem....thank you...thank you so much

Still helpful in 2021

very helpful lecture. glad i found this video. you make things alot more understandable then my professor. keep up the good work :)

Great lecture, thank you!

Thank you so much for this it helped a lot

Great lesson!

thaaannnnnnnnkkk yyooouuuu sooooooooooo mcchhh for this video. it was of great help. this will help me score full.❤

You are amazing!!!

You’re the man !!!

Your psychiatrist/doctor/college grad example confused me a bit since you wrote (x)(Px->Dx) in the conclusion instead of (x)(Px->Cx)

can someone explain how he used constructive dilemma in example #9?

What's the difference between a constant and a variable? How do I distinguish them?

Thanks!!

How come you don't use universal quantifier simple, you just got x

Great video. But how can we recognize the scope of qualifiers. Like Q12, ∃x Ax → ∀x(Bx → Cx), why doesn't ∃ cover the first arrow? In short, why is ∃x( Ax )→ ∀x(Bx → Cx) correct instead of ∃x( Ax → ∀x(Bx → Cx) )?

Can you do a video on invalidity ( predicate logic )

lavang elaichi..tuya maichi

28:10 instead of calling the constant implied by Ex Fx a 'mechanical device' it might be better to call it an auxiliary constant, since the existential instantiation gives the (previously not used) constant an auxiliary or perfunctory (stand-in) role. The referent of this auxiliary constant is a real thing, though we may not be able to pinpoint what or who exactly it is. Whatever it is, it exists and we call it 'a'. But i wonder how the axioms of predicate logic actually allow for this, since in a given model the constants refer to specific fixed objects. The constant 'a' refers to some specific individual or object, it is not available as a 'constant placeholder' so to speak. I guess you could make a 'without loss of generality' meta-logical argument - ignore what 'a' previously referred to and use it now to label the referent of Ex Fx.

how do you prove whether something is wrong, or the argument is invalid? I've been having a hard time grasping this with PL.

Dziękuję

prof. Mark can you please do # 6 in the homework because i tried to do it and i got stuck after line 4 . thanks

Oh, Hi Mark!

@0cards0

5 жыл бұрын

what does he say at 35:05 ? "you can never instantiate an..?"

Prove that ~ B → ~ (A and (A → B)) by Rules of inference

You really didn't have to instantiate twice for #9. Rather, it can look like this: 1. (x)(Ax>Bx) 2. ~Bm / (3x)~Ax 3. Am>Bm (1 UI) 4. ~Am (2,3 MT) 5. (3x)~Ax (4 EG)

@mrguysnailz4907

8 жыл бұрын

+Papa Schulz your profile pic makes serious comments coming from you strangely hilarious.

@Austin_Schulz

8 жыл бұрын

MrGuySnailz Right back atcha.

@mrguysnailz4907

8 жыл бұрын

I guess :^)

36:17 clearly invalid...