On 08 Mar 2014, at 06:12, LizR wrote:
On 6 March 2014 21:44, Bruno Marchal <[email protected]> wrote:
On 05 Mar 2014, at 23:06, LizR wrote:
On 5 March 2014 20:59, Bruno Marchal <[email protected]> wrote:
You have to show two things:
1) R is transitive -> (W,R) respects []A -> [][]A
and
2) (W,R) respects []A -> [][]A -> R is transitive
Let us look at "1)". To show that "R is transitive -> (W,R)
respects []A -> [][]A", you might try to derive a contradiction from
R is transitive, and (W,R) does not respect []A -> [][]A.
What does it mean that (W,R) does not respect a formula? It can
only mean that in some (W,R,V) there is world alpha where that
formula is false.
To say that "[]A->[][]A" is false in alpha means only that []A is
true in that world and that [][]A is false in that world.
OK. I'm not sure where V came from, but anyway...
W = the set of worlds
R = the binary relation (of accessibility)
(W, R) = the multiverse, or the "frame"
(W, R, V) is the same as the multiverse, except that now, in each
worlds of W, the sentence letters p, q, r, ... got a value 1, or 0.
And so, all formula can be said to be true or false in each world,
by the use of classical logic and of the semantic of Kripke (the
fact that []A is determined in alpha by the value of A in its
accessible worlds).
So V is an illumination?
Yes. Also called "valuation".
That is the function which gives, for each world, the true values of
the atomic letters (p, q, r, ...).
A formula A is a law in a multiverse (W,R), or (W,R) respects A, if
and only if for all V, (W,R,V) satisfies A. And this means that for
all V, A is true in all worlds in that multiverse.
In modal logic the semantics is three-leveled: a proposition can be
true in a world (in some (W,R,V)), it can be true in all worlds, and
it can be true in all worlds whatever V is, that is true in all
(W,R,V), with some W and and R.
So as you say a contradiction is t -> f (because f -> x is always
true, as it t -> t)
Like a tautology is true in all worlds, a contradiction is a
proposition false in all worlds, like f, or (A & ~A), or "0 = 1" (in
arithmetic). f -> x is a tautology, yes, and x -> t also.
So []A is true in a world alpha.
I guess you assume []A -> [][]A is false in alpha, which belong to a
transitive multiverse, and you want to show that we will arrive at a
contradiction.
Hence if alpha is transitive,
I understand what you mean. But of course it is R which is transitive.
and if []A is true in all worlds reachable from alpha, let's call
one beta, then []A is also true in all worlds reachable from beta.
It looks like now you suppose []A -> [][]A is true in alpha. So I am
no more sure of what you try to prove.
We don't know if alpha is reachable from beta, but we do know that
if []A is true in beta then it's true in all worlds reachable from
beta.
I let you or Brent continue, or anyone else. I don't want to spoil
the pleasure of finding the contradiction. Then we can discuss the
"2)".
Surely the pleasure of NOT finding a contradiction?
No, the pleasure of finding a contradiction from
[]A -> [][]A is false in alpha
and
R is transitive
I was suggesting you to prove P -> Q, by showing that P & ~Q implies
a contradiction. it is the easiest way, although there are
(infinitely many) other ways to proceed.
Oh dear I don't think my brain can take this!
Maybe a diagram would help. Anyway I have to go now :)
Diagram would help a lot. I teach this basically every years since a
long time, and I only draw diagrams on the black board, not one
symbols, except for the sentence letters true or false in the worlds.
Take it easy, we have all the time. My feeling is that you are
impatient with yourself. Just calm down.
It seems like I have no time! (But you sound like Charles :-)
And you sound like the White Rabbit: "No time, no time ...". :)
You will eventually NOT understand why you ever did find this
difficult. But this takes times and work, that's normal.
I hope you're right.
No problem, unless you try hard to convince yourself that you will
never learn, which is a good technic to not learn.
But as long as you find the courage to be mistaken, you will learn,
and eventually get the familiarity. But you need to be indulgent and
patient with yourself.
I am more anxious about Brent, who solved well the reflexive
multiverse case, but seems mute on the transitive multiverse case.
Well, I know what it is, all those mails, ....
Bruno
PS I have to go now. Will answer other mails later. Next week is also
busy, and I expect some black hole in the basement. Thanks for your
patience.
http://iridia.ulb.ac.be/~marchal/
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