On 25 Mar 2014, at 03:02, LizR wrote:
Thank you for the above, for my diary!
On 24 March 2014 20:14, Bruno Marchal <[email protected]> wrote:
New exercise:
show
(W,R) respects A -> []<>A
iff
R is symmetrical.
OK, symmetrical means for all a and b, a R b implies b R a.
A -> []<>A can (I hope) be read as "the truth of A in one particular
world (which I will call this world) implies that for all worlds
accessible from this world, there exists at least one world in which
A is true".
I was about to write that you did that error again, but looking twice,
I saw you are entirely correct, OK.
Well, there is indeed one world accessible from those other worlds,
in which A is true - this one! Because all worlds accessible from
this one can access this world (due to symmetry) and in this world A
is true.
Excellent.
You proved that
If R is symmetrical then (W, R) respects A -> []<>A.
What about finishing the work and prove the reciprocal? Hmm... Please
look at the iff in the quoted quote above.
You should still show that
if (W, R) respects A -> []<>A then R is symmetrical.
The mere fact that A -> []<>A is true in all worlds, whatever the
valuation is, imposes the symmetry for the binary accessibility.
To show that, you can reason by absurdum. You imagine that (W, R)
respects A -> []<>A, and you consider that R is not symmetrical. Then
you have to find a valuation leading to a counterexample, a world in
which A is true and []<>A is false.
Let me do it, so that you can rest after the good work :)
*
If (W, R) is not symmetrical, there is two worlds a and b so that a R
b, and ~(b R a). OK?
Let us choose the valuation V which assign 0 to p in all the worlds
accessible from beta.
Well, but then if p is true in alpha, []<>p is true in alpha (as we
assume that (W, R) respects A -> []<>A). But then <>p must be true in
beta, OK?
But beta accesses only to worlds with p false. Contradiction.
We say that the illuminated multiverse (W, R)
with W = {a, b, c},
together with the non symmetrical relation explicitly defined by aRb,
bRc. (so we have 'not bRa'), together with the valuation: p true in
a, and false in c, constitutes a counterexample, to the idea here that
a (W,R) with R non symmetrical can respect A -> []<>A. Indeed, in that
illuminated multiverse the assymmetry break makes it possible to break
the law, and in the world a p is verified and []<>p is not
contradicting the "law" A -> []<>A.
All right?
Bruno
PS I send this to FOAR as this is part of an answer to his question,
and a key (albeit tiny) part of the derivation of the "physical laws",
notably giving clues on the reversibility on the bottom of the domain
of indeterminacy ( the true sigma_1 sentences).
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