George Levy wrote:

>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

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