Systems of Modal Logic | Logic Tutorial | Attic Philosophy

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  • เผยแพร่เมื่อ 5 ก.ค. 2024
  • There isn't just one modal logic - there are many different systems, each with their own interesting behaviour. In this tutorial, we'll take a look at some of the main systems of modal logic from a semantic point of view.
    00:00 - Intro
    00:49 - Systems of Modal Logic
    01:15 - Modal Frames
    01:50 - Reflexivity, Symmetry, Transitivity
    03:09 - Serial, Euclidian
    05:27 - Entailment
    07:27 - Partition Models
    10:38 - Exercise
    11:56 - Wrap-up
    If there’s a topic you’d like to see covered, leave me a comment below.
    Links:
    My academic philosophy page: markjago.net
    My book What Truth Is: bit.ly/JagoTruth
    Most of my publications are available freely here: philpapers.org/s/Mark%20Jago
    Get in touch on Social media!
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    Twitter: / philosophyattic
    #logic #philosophy #modallogic

ความคิดเห็น • 37

  • @gbswann
    @gbswann 3 หลายเดือนก่อน

    Best explanation I have ever heard.

  • @drapala97
    @drapala97 3 ปีที่แล้ว +7

    Best intro on the subject i could find. Gonna subscribe and keep digging into your logic content!

  • @vitusschafftlein77
    @vitusschafftlein77 3 ปีที่แล้ว

    Dear Marc (if I may),
    your videos on logic are just spot on - you got yourself a new subscriber! Your explanation for Euclidean models was especially helpful to me; it was the one I had always been baffled about. Do you know why Euclidean frames are called "Euclidean"? The triangle situation reminds me of Euclidean geometry but that's everything I know. I'll make sure to recommend you and your channel :)

    • @AtticPhilosophy
      @AtticPhilosophy  3 ปีที่แล้ว +2

      Thanks! I think the name “Euclidean” comes from Euclid’s axiom, that things both equal to the same thing are equal to one another. In our terminology, this says identity is Euclidean.

  • @nasreddinemerabtene7597
    @nasreddinemerabtene7597 ปีที่แล้ว

    Put them in a playlist, very interesting

  • @Nicoder6884
    @Nicoder6884 ปีที่แล้ว +1

    10:36 Just KT5 will suffice in fact, as T and 5 together show that B is true.

    • @AtticPhilosophy
      @AtticPhilosophy  ปีที่แล้ว

      Good point, this always catches people out!

  • @cristhianoduarte8320
    @cristhianoduarte8320 3 ปีที่แล้ว

    Hey Mark,
    Let me ask you something. Suppose your model is Euclidean and transitive, and you have a frame looking like that triangle you showed in your video. Can I say, or even better, when can I say that the bottommost arrow equals the composition (that's not well defined right?) of the two other arrows? There is a key notion in quantum dynamics, that goes by the name of Markovianity (or divisibility), and that expresses something very similar to this property. Simply put, a particular dynamics is represented by a set of arrows whose origin are at the same point - just like the white arrows in your diagram. The dynamics is said to be divisible, or Markovian, whenever there are intermediate arrows (like the blue ones) such that the diagram commutes, in other words, such that the composition holds true (A -f-> B and A -g- > C is markovian iff there exists B -h-> C such that g=h \circ f).

    • @AtticPhilosophy
      @AtticPhilosophy  3 ปีที่แล้ว +1

      One difference here is that, in your context, the arrows denote functions, whereas in modal logic graphs, the arrows are relations. These arrows aren't functional: Rxy & Rxz doesn't imply y = z. So there isn't a straightforward notion of composition for relations. Second, in modal logic diagrams, each arrow denotes the *same* relation, whereas in category theory diagrams, each arrow can be a different function. (That's why it's interesting when the diagrams commute.) There are some similarities, I guess. Transitivity allows you to go directly along a path in one step - this feels similar to treating the composition of many functions as a single function. And Euclidianness allows you to go across the triangle - a bit like when you see the commuting triangle diagram. I'm not sure how deep those similarities are though.

    • @cristhianoduarte8320
      @cristhianoduarte8320 3 ปีที่แล้ว

      ​Again, thank you very much! =)

  • @platosbeard3476
    @platosbeard3476 2 ปีที่แล้ว

    Is the serial relation for modelling induction?

    • @AtticPhilosophy
      @AtticPhilosophy  2 ปีที่แล้ว +1

      No, but I can see why you say that! There are lots of serial relations: seriality is a property of relations, saying that everything is related to something.

  • @dylanrodriguezmallorquin3464
    @dylanrodriguezmallorquin3464 2 ปีที่แล้ว

    In the symmetrical relation, why it's a if-then implication instead of iff implication?
    i.e:
    ¿¿¿for all x,y(Rx→Ry and Ry→Rx)???

    • @AtticPhilosophy
      @AtticPhilosophy  2 ปีที่แล้ว +1

      Good question! They’re actually equivalent. The iff implies the ‘if then’ and (less obviously) the ‘if then’ implies the iff. To see why, take any pair of things a,b. Symmetry gets you from Rab to Reba, and also from Rba to Rab (since it holds for ANY things x,y). So Rab iff Rba and, generalising, for any x,y, Rxy iff Ryx.

    • @dylanrodriguezmallorquin3464
      @dylanrodriguezmallorquin3464 2 ปีที่แล้ว

      @@AtticPhilosophy Alright, now I understand it perfectly!!! Thanks so much for your response and for your videos!!!

  • @rodgerwilliams7238
    @rodgerwilliams7238 8 หลายเดือนก่อน

    @AtticPhilosophy The Euclidian Relation expressed as ∀x∀y∀z(Rxy ∧ Rxz → Ryz) the conclusion of the implication inside the bracket only has Ryz which kind of makes me think one way arrow only between y and z. But the diagrammatic explanation is a two way arrow does this not then mean Ryz ∧ Rzy ? I am battling a bit relating the diagram to the statement / sentence. If I look at the diagram in the video I am thinking
    ∀x∀y∀z(Rxy ∧ Rxz → Ryz Ryz ∧ Rzy ∧ Ryz ∧ Rzz ∧ Ryy) where am I going wrong in my interpretation of the diagram o the mathematical / logical expression ?
    PS Thanks for the great videos. The explanations and simplicity makes it very easy to understand and apply the material.

    • @AtticPhilosophy
      @AtticPhilosophy  8 หลายเดือนก่อน +1

      Euclidian relations are tricky! The conditional has just Ryz in the consequent. But from Rxy ^ Rxz we can also infer Rxz ^ Rxy and hence (using the Euclidian conditional) Rzy as well. So we get both Ryz and Rzy, in that circumstance, from the same conditional. Ion other words, whenever you 'complete the triangle' for a Euclidian relation, you can complete it in either direction.

    • @rodgerwilliams7238
      @rodgerwilliams7238 8 หลายเดือนก่อน

      @@AtticPhilosophy Thank you that makes understanding it better yes if I think of it that you can go clockwise or anticlockwise. Another question please, do you or anyone else on this thread maybe know of a good detailed Tutorial Series on TH-cam with a similar method of teaching as this series? In other words detailed but taking the viewer right from the beginning to advanced detailed topics for final year undergrads ?

  • @bennyharvey703
    @bennyharvey703 2 ปีที่แล้ว

    i like

  • @ravenecho2410
    @ravenecho2410 ปีที่แล้ว

    euclidean requires a bidirected relation, which is a stricter criterion for all connected nodes. this is stronger than a directional node, so symetry + directional + reflex = symmetry + (directional relation + symetry) -- okay that was the "easy part"
    for the connectivity, if for all connections every relation is symetric, then for any node k, in model m, node j would have to be connected for [0,k]
    this entails fully connected graphs
    okay, so now presume we have a model space (universe?), we know that for all nodes which are connected are fully connected -> but we haven't shown all worlds are fully connected. so we'll create a model X and X-not, world X has X as true world X not has X not as true.
    world X cannot draw valid inferences contining X as an .... what's a relation -> negation is a relation -> scratch that
    we have shown that for nodes which are connected connect a fully formed graph, we have not shown implication, as counter example presume to nodes which are merely reflexive. these are fully connected graphs which satisfy our critereon.
    Draw some diagrams and formalize, that's why partitions exist as fully connected graphs within a Universe or Model (?) and how they exist separately.
    This is fun! tho i'm confused on a couple of items (i have seen model logic twice now XD -> both from your videos!)

    • @ravenecho2410
      @ravenecho2410 ปีที่แล้ว

      i'll formalize it maybe hahaha, but i'm still on trying to understand what the objects are

    • @AtticPhilosophy
      @AtticPhilosophy  ปีที่แล้ว

      Reasoning with graphs & diagrams is great, isn’t it?! It’s a good way to make sense of modal logic.

  • @nasreddinemerabtene7597
    @nasreddinemerabtene7597 ปีที่แล้ว

    q : I watch one of your modal logic videos
    M, w1 |= ◻️q