On 04 Mar 2014, at 01:18, LizR wrote:
OK, so ignoring Brent who I'm sure is way ahead of me...
The problem is to show that
(W, R) respects []A -> A if and only if R is reflexive,
Where reflexive means for all alpha, { alpha R alpha } (and nothing
more is implied!)
OK.
(better not to use the accolade though, as you are just saying that
for all alpha, alpha R alpha (and then if you represent R by a set, R
will contain all couples like (alpha, alpha), so R = {(alpha, alpha),
(alpha beta), (beta beta), ... }.
And []p means that p is true in all worlds reachable from the world
being considered.
OK.
You could have said more precisely that
[]p means, in a world alpha, that p is true in all worlds reachable
from the world alpha.
(...I think. I just checked my diary and was told that "[]p means
that p is a law". Maybe that was the wrong page...)
Yes, that's was in the Leibniz semantics. Something similar will
happen with kripke, but if I explain now, it can be confusing.
OK, so anyway, before I get too confused
You should never allow this to happen. It happens because you allow
slight confusion, and then they add up. I know it is not easy.
let's consider world alpha which is part of W.
A part of W means usually a subset of W.
A world is an element of W.
If W = {a, b}, a and b are elements of W. The parts of W are { }, {a},
{b}, {a, b}. If W has n elements, we have seen that W has 2^n parts.
I hope you don't mind I help you to use the standard terminology, as
it will help us a lot later.
We know { alpha R alpha }.
?
At this stage I am not sure if you try to prove:
(W, R) respects []A -> A -> R is reflexive,
or
R is reflexive -> (W, R) respects []A -> A
I will have to guess. And here I guess you assume R is reflexive, and
so you intent to deduce from this that (W, R) respects []A -> A.
[]p means p is true in all worlds reachable from alpha (I think)
which includes alpha itself, hence it means that p has to be true in
alpha, hence it means "[]p -> p".
That's correct, but still a bit fuzzy. To say that (W, R) respects a
formula, like []A -> A, means that (W,R, V) satisfies the formula, for
all valuations V.
May be I am just nitpicking, but what if p is not true in alpha. Do we
still have []p -> p?
Your fuzziness, or perhaps my own imperfect brain, makes consistent
that you did treat that case, or that you did not.
(Conversely, if alpha wasn't reachable from itself, then p being
true in all worlds reachable from alpha wouldn't entail that p is
true in alpha.)
Very good. Just a bit lazy. When you say that "p being true in all
worlds reachable from alpha wouldn't entail that p is true in alpha",
you might give us the counterexample, like chosing a valuation
(illumination) V with p false in alpha and true in all worlds
accessible from alpha.
We love counter-examples, you know.
QED, perhaps? Did I just prove something?
Yes. You did.
You have proved that
1) if R is reflexive, []A -> A is automatically true in all worlds in
any reflexive illuminated multiverse.
And then, three lines above, beginning by "Conversely, ..." you have
proved that, conversely indeed:
2) ~ (R is reflexive) -> ~(For all V (W,R,V) respects []A ->
A)) (by showing one V with a world in which []A -> A is false,
when R is not reflexive).
If so, I'still m not sure that proves "if any only if"...
Oh! You might say that once you proved 2) you did prove that
2') (For all V (W,R,V) respects []A -> A)) -> (R is reflexive)
But you can derived "P -> Q" from a derivation of "~Q -> ~P". All
right.
You did it, and it would have been simpler for you, and for me, if you
just started from what you were asked to prove.
Although... maybe it does.
Sure it does.
For []p to imply p in a world alpha, where []p means "p is true in
all worlds reachable from alpha", it can only imply p is true if
alpha is "reachable from alpha".
This applies to all worlds in (W, R) hence it must be reflexive.
I think.
Good.
To prove that P -> Q, you can prove that P & ~Q leads to a
contradiction, or you can prove that ~Q leads to ~P.
But it helps a lot if you start from what you want to prove, up to the
conclusion, so that not only you prove it, but you know exactly what
you discovered. In this case a necessary link, in Kripke semantics,
between a binary relation (reflexivity) and a modal formula []A->A.
You learned that the fact that (W, R) respects []A -> A is equivalent
with the fact that R is reflexive.
OK?
Bruno
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