On 05 Mar 2014, at 01:36, LizR wrote:
On 5 March 2014 04:18, Bruno Marchal <[email protected]> wrote:
Good.
To prove that P -> Q, you can prove that P & ~Q leads to a
contradiction, or you can prove that ~Q leads to ~P.
But it helps a lot if you start from what you want to prove, up to
the conclusion, so that not only you prove it, but you know exactly
what you discovered. In this case a necessary link, in Kripke
semantics, between a binary relation (reflexivity) and a modal
formula []A->A.
I had to get my head around ... well, everything ... again. So I may
have sneaked up on the result.
You learned that the fact that (W, R) respects []A -> A is
equivalent with the fact that R is reflexive.
OK?
OK.
OK.
So, the next question was
A Kripke multiverse (W, R) is said transitive if R is transitive.
That is
alpha R beta, and beta R gamma entails alpha R gamma, for all alpha
beta and gamma in W.
Show that
(W, R) respects []A -> [][]A if and only R is transitive,
Damn. This looks too complicated for me to fake it!
Hmm.....
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.
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)".
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!
Bruno
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