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
***********************
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