On 02 Apr 2014, at 04:45, meekerdb wrote:
On 4/1/2014 7:40 AM, Bruno Marchal wrote:
BTW, are you OK in the math thread? Are you OK, like Liz
apparently, that the Kripke frame (W,R) respects A -> []<>A iff R
is symmetrical?
Should I give the proof of the fact that the Kripke frame (W,R)
respects []A -> [][]A iff R is a transitive?
Bruno
Here's the ones I've done so far. One more to go. Hold off on that
proof (or put a warning in the subject line so I can avoid reading
it).
?
It looks like you did it below.
Liz, try to see if you are convinced by Brent, before reading this post.
Brent
> *******************
> 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 R is not reflexive. Then there exists at least one world
beta such that (alpha R beta) and ~(beta R alpha). Consider a
valuation such that p=f in alpha and p=t in all beta.
in all beta different from alpha (of course). OK.
Then []p is true in alpha but p is false so []A->A is false in alpha
for some A. R not reflexive implies []A->A is not respected for all
alpha and all valuations.
OK.
> (W, R) respects []A -> [][]A if and only R is transitive,
R is transitive means that for all beta such that (alpha R beta) and
all gamma such that (beta R gamma), (alpha R gamma). So every []A
implies A=t in all beta and also A=t in all gamma. But A=t in all
gamma means []A is true in beta, which in turn means [][]A is true
in alpha. So R is transitive implies (W,R) respects []A->[][]A.
Nice direct proof.
People can search an alternate proof using the reduction ad absurdum.
Suppose R is not transitive, so for all beta (alpha R beta) and
there are some gamma such that [(beta R gamma) and ~(alpha R gamma)].
I cannot parse that sentence, I guess some word are missing. R is not
transitive means that there exist alpha, beta and gamma, such that
alpha R beta, and beta R gamma, and ~(alpha R gamma). I will guess
that this is what you meant.
Let A=t in beta,
OK. Or A=t in all the beta such that alpha R beta, but you can also
assume alpha accesses only beta, to build the counterexample.
A=f in gamma.
Good choice, to build the counterexample.
Then []A is true in alpha but []A isn't true in beta, so [][]A isn't
true in alpha. So (W, R) respects []A -> [][]A implies R is
transitive.
Very good, so the transitive case is closed!
You should no more worry reading my posts :)
OK Liz? Others? Feel free to ask definitions or explanations.
The next one is important, as it plays a role in the 'derivation of
physics'.
> (W, R) respects A -> []<>A if and only R is symmetrical,
R symmetrical means that if (alpha R beta) then (beta R alpha).
Yes, for all alpha and beta in W.
Suppose A is true in alpha; then <>A is true in beta (by symmetry of
R) and this holds for all alpha and beta so []<>A in alpha.
And so A -> []<>A is true in alpha. (Here we are using the deduction
rule in the CPL context, which is valid. Later we will see it is not
valid in the modal context).
Suppose R is not symmetrical, so there is a pair of worlds (alpha R
beta) and ~(beta R alpha). So consider V such that A=t in alpha and
A=f in all worlds gamma such that (beta R gamma) then ~<>A in beta.
So it would be false that []<>A in alpha.
Liz told me this already! OK.
> (W,R) respects []A -> <>A if and only if R is ideal,
R is ideal, means that for every alpha there is a beta such that
(alpha R beta). Suppose []A is true in alpha, then A must be true
in every world beta (alpha R beta) and there is a least on such
beta, so <>A is true in alpha.
OK.
Suppose R is not ideal, then there is a cul-de-sac alpha. For alpha
[]A is vacously true for all A, but <>A is false so []A-><>A is false.
Yes, all cul-de-sac world are counterexample of []A -> <>A. In the
Kripke semantics, they are counterexamples of <>#, with # put for any
proposition.
> (W, R) respects <>A -> ~[]<>A if and only if R is realist.
R is realist means that for every world alpha there is a world beta
such that (alpha R beta) and beta is cul-de-sac.
For every *transitory* world alpha. OK. The cul-de-sac world are still
world!
Suppose A is true in beta, then <>A is true in alpha but <>A=f in
beta so []<>A cannot be true in alpha. Hence <>A->~[]<>A in alpha
where alpha is any non cul-de-sac world. Then consider a cul-de-sac
world like beta; <>A is always false in beta so <>A->X is true in
beta for any X, including ~[]<>A.
OK. Nice.
So you proved that R is realist implies that (W, R) respects <>A ->
~[]<>A.
But you have still not prove that if R is *not* realist, (W,R) does
not respect <>A -> ~[]<>A (unlike all other cases). OK?
You proved: "(W, R) realist" implies "respects <>A -> ~[]<>A", but not
yet the converse, that "respects <>A -> ~[]<>A" implies " (W, R)
realist".
I let you search, and might justify this (with pre-warning to avoid
spoiling!).
And what about the euclidian multiverse? May be you did them?
R is euclidian, or euclidean, if (aRb and aRc) implies bRc, for all
a, b and c in W. (I use "a" for the greek alpha!)
Proposition: (W,R) respects <>A -> []<>A iff R is euclidian.
OK?
Bruno
***********************
--
You received this message because you are subscribed to the Google
Groups "Everything List" group.
To unsubscribe from this group and stop receiving emails from it,
send an email to everything-list+unsubscr...@googlegroups.com.
To post to this group, send email to everything-list@googlegroups.com.
Visit this group at http://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.
http://iridia.ulb.ac.be/~marchal/
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to everything-list+unsubscr...@googlegroups.com.
To post to this group, send email to everything-list@googlegroups.com.
Visit this group at http://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.
