Symbolic Logic Lecture #6: Derivations in SL, part II

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  • Опубликовано: 10 сен 2016

Комментарии • 9

  • @jacksanders2611
    @jacksanders2611  7 лет назад +1

    In order to work along with the lectures, you'll find it useful to check out John Halpin's Logic Café: thelogiccafe.net/PLI/. This week we're working with Chapter Five again.

  • @ellieisright692
    @ellieisright692 5 лет назад

    Thank you! I missed some lectures in my philosophy of logic class and this helped SO MUCH!
    Also I 100000% agree the mind isn't confined in the CNS or body!

    • @user-wl2rb3rh5c
      @user-wl2rb3rh5c Год назад

      Hey Ellie, what were you studying? BA in philosophy?🙂

    • @user-gs7uu1rm8t
      @user-gs7uu1rm8t 5 месяцев назад

      Mind 100% does reside in nervous system

  • @MrDabs16
    @MrDabs16 7 лет назад +1

    Excellent. This is useful all the way from UCL (London).

    • @jacksanders2611
      @jacksanders2611  7 лет назад +1

      Glad to know! Thanks!

    • @user-wl2rb3rh5c
      @user-wl2rb3rh5c Год назад

      @@jacksanders2611
      Jack, in the example of 31:58 the student uses the "new rule of association", yet not a single rule of replacement was shown in the previous tutorials ("binged" them for every second 😅) was there any part of the lecture thats missing by any chance? Only thing remotely close taught in previous lectures was equilivancy of propositions in lecture 2..

  • @wilcoxandrew88
    @wilcoxandrew88 Год назад

    In the example at 23:05 , If we assume A+B (or just B) and then assume ~J, we can derive ~J ---> B. You mentioned if you know something is true then it can be implied by anything.
    Does it also work the other way around? Could we derive B --> ~J? (along with ~J --> B)

    • @jacksanders2611
      @jacksanders2611  Год назад +1

      All such derivations are conditional on the assumptions made. Thus if you assume all of A, B, and ~J, then each of those can subsequently be derived no matter what else you assume. That's because you've assumed them. But you aren't finished with a conditional proof until you back out of the assumptions, and given the way you propose running the proof, all you can back out to is "if A&B, and if ~J, then anything implies any one of the three".