George Levy wrote:

## Advertising

>Reflexive, Transitive and Symmetric applies only to the relation R that >define accessibility. Exactly. >So: > >Reflexive: > >W ---->| > <------| I will assume this is a courageous attempt to draw a curl loop. Nice. >Transitive: > >W1 ------> W2 --------> W3 Yes. More precisely W1 ------> W2 --------> W3 entails W1 ------> W2 --------> W3 - - - - ------->------- That is w1 R w2 and w2 R w3 entails w1 R w3. I also suggest using little letter (w, w1, w2, ...) for the worlds and capital letters (W) for the set of worlds. As you do yourself below. >Symmetric > >W1 <-----> W2 OK. w1 R w2 entails w2 R w1. >And the Goedel-like formula > ><>p --> -[]<>p > >means: if p is true in at least one world accessed from w, then it is >false that >in any world x accessed from w, that p is true in at least one world >accessed from >x: This is correct for "<>p --> -[]<>p is true in w". But remember that when we try to characterize a formula by a frame, we want the formula being true in all the worlds of the frame, and this for all the valuation i.e. independently of the truth value of p. > |---->w121 p false > |---->w12: p true ---->|---->w122 p false > |---->w13: p false |---->w123 p false >w1--|---->w14: p false > |---->w15: p false |----->w161 p false > |---->w16 p false---->|----->w162 p false > |----->w163 p false This is a nice little model where <>p --> -[]<>p is true at w1. But it is not a frame, because if you change the truth value of p in some world of this model, the formula will not remain true. Try to convince yourself that the formula <>p --> -[]<>p is true in *all* realist models. >I have a little bit of trouble with reading -[]<>p from left to >right....Is it - ( >[] ( <>p ) )? Yes. -[]<>p is - ( [] ( <> p)) >Also does the frame of reference changes when you talk about [] i.e., any >world x? >In other words you start with world w but then, as you express [] all (any) >world(s) x accessible from w do you have to change the frame to x when you >pursue >the reasoning to <>? It depends what you want to do. In the semantical characterisation the frame are never changed. >If what I assumed is correct, then the Goedel-like formula makes sense. > >Now reinterpreting [] and <> to mean provable and consistent we get for ><>p --> -[]<>p > >if p is consistent in at least one world accessed from w, then it is false >that in >any world x accessed from w, that p is provable in at least one world >accessed >from x. Here again you are just saying that <>p --> -[]<>p is true in a world. Remember that <>p --> -[]<>p is respected by a frame if and only if the frame is realist. Recall that saying that "<>p --> -[]<>p is respected by a frame" means <>p --> -[]<>p is true in *all the worlds* of *all models*. You must keep in mind that we have three semantical levels. A can be true in a world w of a model (W,R,V) based on a frame (W,R). A can be true in all worlds w of a model (W,R,V) based on a frame (W,R). A can be true in all models (W,R,V) based on a frame. >Now going to your LASE > >p-> []<>p > >if p is true in w, then in any world x accessed from w, it is possible to >access >one world where p is true. > >This statement seems to be correct only when R is symmetric. Indeed p-> []<>p will be true in all worlds on any symmetric frame, independently of the truth value of p in any world. The converse is true also. >or if there is a transitive loop back to w. No. That is true only for some special models, and is not relevant for the general case. >I assume the following: >A terminal world is defined as a world from which no other world is >accessible. OK. >---->w --| >A transitory world is one from which at least one other world is >accessible w---->w1 OK. Note that w could be equal with w1. >An ideal world is one with no access to a terminal >world w---->w1 Be careful. transitory and terminal applies to worlds. Ideal and realist applies to frames and models. Note that, for exemple, the reflexive frame are all ideal. >A realist world is one with access to at least one terminal >world. w---->w1 --| You mean a realist frame, I guess. >Your theorem > >5. (W,R) respects []p -> <>p iff (W,R) is ideal >means >if (if p is true in all worlds accessible from w, then p is true in at >least one >world accessible from w.) is true then w does not lead to a terminal world. There are a little to much nesting for me there! I prefer to decompose the statement. "(W,R) respects []p -> <>p" means that "[]p -> <>p" is true in all world of all models based on (W,R). Now the theorem says that this entails the frame (W,R) is ideal, and conversely if the frame is ideal then "[]p -> <>p" is true in all world of all models based on (W,R). >In a terminal world nothing is possible, everything is false so p cannot >be true. Not at all. You are using unconsciously p-><>p, which is a residual leibnizian intuition! In a terminal world nothing is possible, but all classical tautologies will be true for exemple. In particular the constant sentence TRUE is true in a terminal world. Only <>TRUE is false, because in a terminal world there is no world to go. >The premise doesn't even hold. Since an ideal world does not lead to a >terminal >world, then from an ideal world (W,R) respects []p -> <>p is always true. Ideal frame (not world). Note that this shows only one half of the theorem. You must still show that if (W,R) respect []p -> <>p, then (W,R) is ideal. Let us do it. Let us suppose (W,R) respect []p -> <>p, and let us suppose that (W,R) is *not* ideal. We must shown that this is contradictory. Now if (W,R) is not ideal, it means that there is a terminal world w in W. Now, []p is true in w (for we have seen that all statement with the form []# is true in terminal world). "(W,R) respect []p -> <>p" entails []p -> <>p is true in w. By modus ponens in w, we have that <>p is true in w. Contradiction (because all statement with the form <># are false in terminal world). >6. (W,R) respects <>p -> -[]<>p iff (W,R) is realist >means >if (if there is at least one world accessible from w where p is true, then >it is >false that from any world x accessible from w, it is possible to access at >least >one world in which p is true) is true, then w leads to at least one terminal >world. You forget the frame level. "(W,R) respects <>p -> -[]<>p" first means that "<>p -> -[]<>p" is true in all worlds of all models build on (W,R). To show that if (W,R) is realist it respects <>p -> -[]<>p, is "easy". To show the converse is a little longer. Let us show that if (W,R) is realist then (W,R) respect <>p -> -[]<>p. Let us suppose that (W,R) is realist and that it does not respect <>p -> -[]<>p. And let us show that this entails a contradiction. (As you can see, classical logician are fond of proof by "reduction ad absurdo"!). If (W,R) does not respect <>p -> -[]<>p, it means there is a world w, in W, such that <>p -> -[]<>p is false at w. This means (just remember the truth table of the implication) that <>p is true at w and -[]<>p is false at w. This means that both <>p and []<>p are true at w. But (W,R) has been supposed to be realist, which means that there is a terminal world w1 accessible from w. So (by Kripke semantics), because we have just show that []<>p is true at w, <>p must be true in w1. But w1 is a terminal world, so we got a contradiction (in a terminal world we never have the truth of a proposition with the form <>#). >hmmm.. All this reasoning could be quickly shown with one >drawing......When I read >this statement I suffer from a mental stack overflow. :-) Take it easy :-) Bruno