8.2 Predicate Logic: Using the Rules of Inference

Поділитися
Вставка
  • Опубліковано 6 лис 2024

КОМЕНТАРІ • 39

  • @dt-gamer4021
    @dt-gamer4021 8 місяців тому +1

    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✨🙇🏾‍♂️😮‍💨

  • @ShaneFolden
    @ShaneFolden 5 років тому

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

  • @lisarobertson630
    @lisarobertson630 11 років тому +1

    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!

  • @cameronjones8648
    @cameronjones8648 9 років тому +20

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

    • @mrguysnailz4907
      @mrguysnailz4907 9 років тому +3

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

    • @stevenf5902
      @stevenf5902 4 роки тому

      I noticed that too! Phew!

  • @TomideAdeleke
    @TomideAdeleke 9 років тому +15

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

  • @Scurvied
    @Scurvied 10 років тому +5

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

  • @thephoenix_lab
    @thephoenix_lab 3 роки тому +1

    Still helpful in 2021

  • @thomassomerville198
    @thomassomerville198 10 років тому +1

    You the man Thorsby

  • @EightTwoFourWeddingStudios
    @EightTwoFourWeddingStudios 11 років тому

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

  • @CandidGestures
    @CandidGestures 8 років тому

    This was SO helpful for my final, thank you!

  • @maxpercer7119
    @maxpercer7119 2 роки тому

    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.

  • @naharulhayat
    @naharulhayat 7 років тому

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

  • @filipalaureano1868
    @filipalaureano1868 9 років тому

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

  • @sanomi4492
    @sanomi4492 6 років тому

    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) )?

  • @natashaoliveira1293
    @natashaoliveira1293 9 років тому +1

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

  • @bobbyjones8938
    @bobbyjones8938 9 років тому +4

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

  • @anikethinge5904
    @anikethinge5904 6 років тому

    lavang elaichi..tuya maichi

  • @francescopiazza4882
    @francescopiazza4882 4 роки тому

    Great lesson!

  • @israali5414
    @israali5414 7 років тому

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

  • @draspotnuk
    @draspotnuk 4 роки тому

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

  • @PhilosophySama
    @PhilosophySama 3 роки тому

    You’re the man !!!

  • @Garb0e
    @Garb0e 12 років тому

    Great lecture, thank you!

  • @madeachanneljusttocomment2632
    @madeachanneljusttocomment2632 8 років тому

    Thank you so much for this it helped a lot

  • @sallytamer90
    @sallytamer90 11 років тому

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

  • @JohnnieCYP
    @JohnnieCYP 5 років тому +1

    Oh, Hi Mark!

    • @0cards0
      @0cards0 5 років тому

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

  • @darianjh722
    @darianjh722 10 років тому

    You are amazing!!!

  • @learnwithsid2044
    @learnwithsid2044 7 років тому

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

  • @Austin_Schulz
    @Austin_Schulz 9 років тому +1

    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
      @mrguysnailz4907 9 років тому

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

    • @Austin_Schulz
      @Austin_Schulz 9 років тому +2

      MrGuySnailz Right back atcha.

    • @mrguysnailz4907
      @mrguysnailz4907 9 років тому

      I guess :^)

  • @kevinsweeney2809
    @kevinsweeney2809 6 років тому

    Thanks!!

  • @agatau8190
    @agatau8190 6 років тому

    Dziękuję

  • @francescopiazza4882
    @francescopiazza4882 4 роки тому

    36:17 clearly invalid...