On 03 Mar 2014, at 05:40, meekerdb wrote:
On 3/2/2014 9:57 AM, Bruno Marchal wrote:
Brent, Liz, others,
I sum up the main things, and give a lot of exercises, or
meditation subject.
Liz we can do them one at a time, even one halve. Ask questions if
the question asked seems unclear.
***
A Kripke frame, or multiverse, is a couple (W, R) with W a non
empty set of worlds, and R a binary relation of accessibility.
An illuminated, or valued, multiverse (W,R, V), is a Kripke
multiverse together with an assignment V of a truth value (0, or 1)
to each propositional letter for each world. We say that p is true
in that world, when V(p) = 1, for that world. If you want V is a
collection of functions V_alpha in {0, 1}, one for each world alpha.
***
Some class of multiverses will play some role.
A Kripke multiverse (W, R) is said reflexive if R is reflexive.
alpha R alpha, for all alpha in W.
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.
A Kripke multiverse (W, R) is said symmetric if R is symmetric.
alpha R beta entails beta R alpha, for all alpha in W.
A Kripke multiverse (W, R) is said ideal if there are no cul-de-sac
worlds. For all alpha, there is beta such that alpha R beta.
A Kripke multiverse (W, R) is said realist if all non cul-de-sac
worlds can access to a cul-de-sac world.
***
Finally: (The key thing)
I say that a Kripke multiverse (W,R) respects a modal formula if
that formula is true in all worlds in W, and this for any valuation
V.
***
Show that
(W, R) respects []A -> A if and only if R is reflexive,
R is reflexive implies (alpha R alpha) for all alpha. []A in alpha
implies A is true in all beta where (alpha R beta), which includes
the case beta=alpha. So R is reflexive implies (W,R) respects []A->A.
OK.
Assume (W,R) respects []A->A,
OK.
so that []A->A is true in all W.
Precisely: []A->A is true in all world in W, and this whatever the
valuation or illumination of the atomic sentences (p, q, r, ...), and
thus whatever is the truth value of A.
That means that every world has another R-accessible world and
whatever valuation any formula A has in the world, it has the same
valuation in the R-accessible world.
Hmm? I don't see how that implies R is reflexive, unless I can say
that any two worlds whose valuation is the same for every formula
are just the same world.
I let you think a bit more. May be another will have an insight. You
might try to reason by absurdum.
You must show that: [(W,R) respects []A->A] -> R is reflexive.
(and I recall that "respects" means that []A -> A is true, whatever A
is, and whatever his truth value is in any (W, R, V)).
Suppose that this is false.That means (definition of "->" actually)
that we have both (W, R) respects []A -> A, *and* R is not reflexive.
But if R is not reflexive, there is some world alpha such that ~(alpha
R alpha). OK?
Consider all worlds beta so that alpha R beta, and consider the
valuation V where V(p) = true in all those worlds beta, and V(p) =
false in alpha.
In that case, []A is true in alpha, and A is false in alpha. OK?
That gives the counter-example, and we are done. The fact that R is
not reflexive makes it possible to find a valuation and a world such
that []A -> A is false in that world, and so []A -> A cannot be
respected by a non reflexive frame (multiverse).
Note that if the set of beta is empty, then []p is automatically true,
and V(p) = true will do. It is in fact taken into account above.
Oops, I have solved it. Are you OK? Liz?
If this is OK, try the next one, again in both direction, perhaps.
(W, R) respects []A -> [][]A if and only R is transitive,
Bruno
Brent
(W, R) respects []A -> [][]A if and only R is transitive,
(W, R) respects A -> []<>A if and only R is symmetrical,
(W,R) respects []A -> <>A if and only if R is ideal,
(W, R) respects <>A -> ~[]<>A if and only if R is realist.
You can try to find small counter-examples, and guess the pattern
of what happens when you fail.
Of course proving that (W, R) respects []A -> A if and only if R is
reflexive, consists in proving both
(W, R) respects []A -> A if R is reflexive,
and
(W, R) respects []A -> A only if R is reflexive, that is
R is reflexive if (W, R) respects []A -> A
That's a lot of exercises. 10 exercises.
We can do them one at a time. Who propose a proof for
(W, R) respects []A -> A if R is reflexive, That is:
R reflexive -> (W, R) respects []A -> A
?
Bruno
Oh! I forget this one:
Show that all the Kripke multiverses (W, R), whatever R is, respect
[](A -> B) -> ([]A -> []B).
http://iridia.ulb.ac.be/~marchal/
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