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 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.
You will eventually NOT understand why you ever did find this
difficult. But this takes times and work, that's normal.
Bruno
It is almost more easy to find this by yourself than reading the
solution, and then searching the solution is part of the needed
training to be sure you put the right sense on the matter.
Keep in mind the semantic definitions. We assume some illuminated
(W,R,V)
Atomic proposition (like the initial p, q, r, ...) is true in a
world alpha , iff V(p) = 1 for that word alpha.
Classical propositional tautologies are true in all worlds.
[]A is true at world alpha iff A is true in all worlds accessible
from alpha.
(W,R,V) satisfies a formula if that formula is true in all worlds in
W (with its R and V, of course).
(W,R) respects a formula if that formula is satisfied for all V. So
the formula is true in all worlds of W, whatever the valuation V is.
Courage!
http://iridia.ulb.ac.be/~marchal/
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