Best intro on the subject i could find. Gonna subscribe and keep digging into your logic content!
@AtticPhilosophy3 жыл бұрын
Thanks very much! That made my day.
@gbswann4 ай бұрын
Best explanation I have ever heard.
@nasreddinemerabtene7597 Жыл бұрын
Put them in a playlist, very interesting
@Nicoder6884 Жыл бұрын
10:36 Just KT5 will suffice in fact, as T and 5 together show that B is true.
@AtticPhilosophy Жыл бұрын
Good point, this always catches people out!
@vitusschafftlein773 жыл бұрын
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 :)
@AtticPhilosophy3 жыл бұрын
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.
@cristhianoduarte83203 жыл бұрын
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).
@AtticPhilosophy3 жыл бұрын
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.
@cristhianoduarte83203 жыл бұрын
Again, thank you very much! =)
@ravenecho24102 жыл бұрын
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!)
@ravenecho24102 жыл бұрын
i'll formalize it maybe hahaha, but i'm still on trying to understand what the objects are
@AtticPhilosophy2 жыл бұрын
Reasoning with graphs & diagrams is great, isn’t it?! It’s a good way to make sense of modal logic.
@rodgerwilliams723810 ай бұрын
@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.
@AtticPhilosophy10 ай бұрын
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.
@rodgerwilliams723810 ай бұрын
@@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 KZfaq 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 ?
@platosbeard34762 жыл бұрын
Is the serial relation for modelling induction?
@AtticPhilosophy2 жыл бұрын
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.
@dylanrodriguezmallorquin34642 жыл бұрын
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)???
@AtticPhilosophy2 жыл бұрын
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.
@dylanrodriguezmallorquin34642 жыл бұрын
@@AtticPhilosophy Alright, now I understand it perfectly!!! Thanks so much for your response and for your videos!!!
@bennyharvey7032 жыл бұрын
i like
@nasreddinemerabtene7597 Жыл бұрын
q : I watch one of your modal logic videos M, w1 |= ◻️q