On 05 Mar 2014, at 23:31, LizR wrote:
Let's take 3 worlds A B C making a minimal transitive multiverse.
ARB and BRC implies ARC. So if we assume ARB and BRC we also get ARC
Right.
(if we don't assume this we don't have a multiverse or at least not
one we can say anything about.
This, or something like this ...
[]p in this case means the value of p in A is the same as its value
in B and C (t or f).
What if p is false in A, and true in all worlds accessible from A?
This also means that in A B and C, []p is true, hence we can also
say that in all worlds [][]p.
Correct.
(And indeed [][][]p and so on?)
Sure. at least in a multiverse where []A -> [][]A is a law. In that
case it is true for any A, and so it is true if A is substituted with
[]A, and so [][]A -> [][][]A, and so []A -> [][][]A, and so on.
So it's true for the minimal case that []p -> [][]p
But then adding more worlds will just give the same result in each
set of 3... so does that prove it?
Not sure.
No, hang on. Take { A B C } with p having values { t t f }. []p is
true in C, because C is not connected to anywhere else, which makes
it trivially true if I remember correctly. But []p is false in A and
B. So [][]p is false, even though []p is true in C. So []p being
true in C doesn't imply [][]p.
I might need to see your drawing. If C is not connected to anywhere
else, C is a cul-de-sac world, and so we have certainly that [][]p is
true in C (as []#anything# is true in all cul-de-sac worlds).
So that seems to disprove it, because C is in its own little
multiverse. There's nothing in the definition that says ARB and BRC
entails CRA or CRB, is there?
No, indeed.
Unless I have the "trivially true" thing wrong...
Yes, in the cul-de-sac world, [][]p is automatically "vacuously" true.
Good work Liz. I will provide "clean" solutions, but I wait a bit for
Brent. Brent?
Liz, meanwhile you might try this one, which is a bit more easy than
the transitivity case:
Show that (W,R) respects []A -> <>A if and only if R is ideal.
(I remind you that R is ideal means that there is no cul-de-sac world
at all in (W,R)).
Do you see that (W, R) is reflexive entails that (W,R) is ideal? If
all worlds access to themselves, no world can be a cul-de-sac world,
as a cul-de-sac world don't access to any world, including themselves.
Bruno
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