On 23 Jan 2014, at 07:42, LizR wrote:
On 23 January 2014 00:58, Bruno Marchal <[email protected]> wrote:
On 22 Jan 2014, at 04:23, LizR wrote:
I'm going to take a punt and assume the order in which things are
ANDed together doesn't matter, in which case the above comes out as
equal (equivalent). Did I blow it?
Not sure that I understand what you mean by blowing it. But you are
correct in all answers.
"Blow it" means to fail. I was worried that assuming that the order
of ANDing didn't matter would turn out to be wrong...
You read too much quantum mechanics.
Oh, it looks we are later:
So we are.
Even more so now.
Actually, you will have to remind me what [] and <> mean before I
go any further.
(...snip...)
OK?
Yes, so far so good.
I hope so.
Now, your question, what does mean []A, for A some formula. And what
does mean <>A
(...snip...)
<>p is true if ~[]~p is true, if []~p is false, which means that
there is a world in which p is true.
So <>p means p is true in at least one world.
Unfortunately all this might not seem helpful for a formula which
mix modal compounds with non modal compounds, like
[]p -> p
Here, there is no more truth table available, and so you have to
think. The Leibniz semantic (the only semantic we have defined)
provides all the information to solve the puzzle.
I read this as "p is true in worlds implies that p is true in a
particular world" - is that right?
"p is true in worlds" is a bit sounding like weird to me.
[]p means that p is true in all worlds, and that implies indeed that p
is true in each particular world. keep in mind that we have fix the
entire multiverse, by the set of the propositional variables (like {p,
q, r}, for example).
You might reread my explanation for []p -> p. Which happens indeed
to be a law in this Leibnizian setting.
And the question is now: which among the following are also
Leibnizian laws:
p -> []p
No. True in this world doesn't imply true in all worlds
Correct. For example (assuming the order "p, q r"):
p is true in 111 does not entail that p is true in 000.
[]p -> [][]p
p is true in all worlds implies that it's true in all worlds that p
is true in all worlds (and so on). A law.
Exact.
[]p -> <>p
Yes that follows
Indeed. (and I suspect this is what made Leibniz saying that we are in
the best possible world, although he should have said that we are in
the best possible multiverse. The multiverse of Leibniz satisfies the
deontic axiom. If something is necessary, then it is possible. We will
see that this is not the case in computerland, or in the arithmetical
platonia.
p -> []<>p
Yes,
Nice.
it's true in all worlds that p is true in at least one world.
Er, I first wrote that this justification was slightly wrong, but you
make me realize that this is true with the notion of world that I have
defined (which is not the most common one, both for Leibniz and
Kripke). But with the definition given that is 100% correct).
(Later, we will stop asking that all worlds (in the sense given)
belongs in the multiverse. We can decide to suppress all worlds in the
multiverse in which p is true. And keep Leibniz semantics in that new
sort of multiverse (meaning that []x is true = x is true in all worlds
of that multiverse).
Question: how to add one word in your justification above, so that "p -
> []<>p" is still justified as a law. Which word?)
<>p -> []<>p
Yes, it's true in all worlds that p is true in at least one world
OK. (in a sense, you are too quick, but that is entirely my fault.
You make me progress in the pedagogy. I learned something, but it is
too late, for you). No problem, as in all case, we will slightly
change the meaning of the word "worlds".). No problem, you are 100%
correct.
<>p -> ~[]<>p
This is getting hard to follow!
It looks as though the right hand side is "it's not true in every
world that there is a world where p is true" which - if so - is
false, or not implied by <>p
Very good. Of course you can deduce it from the preceding line. If
both "<>p -> []<>p" and "<>p -> ~[]<>p" where true, then by the truth
of <>p, we would have both []<>p and ~[]<>p, and that would be a
contradiction.
[]p & ([](p -> q) ->. []q (sometimes p ->. q is more readable
than (p -> q). The comma makes precise which is the main connector.
Eek! (Isn't there a bracket missing?)
Correct! It should be []p & ([](p -> q)) ->. []q
better (in readability): ([]p & [](p -> q)) ->. []q, or even just
[]p & [](p -> q) ->. []q
I think that's probably a law...if I read it right. p is true in all
worlds, and p->q in all words implies that q is true in all worlds.
Exact.
Or maybe I am misreading that completely...
False. You first mistake. In a meta-statement of doubt. That's a good
mistake, and it means you lack a bit of trust apparently. It is normal
in the beginning.
[](p -> q) ->. ([]p -> []q)
it's true in all words that p->q, this implies that p being true in
all worlds implies that q is true in all worlds. Which sounds like
it should be a law...?
Exact. But you don't use the hint I gave (and even don't quote it). or
perhaps I made the hint below. Well don't mind to much.
In fact you have already shown that ((p & q) -> r) is equivalent with
(p -> (q -> r)). That makes
[]p & [](p -> q) ->. []q equivalent with [](p -> q) ->. ([]p ->
[]q), by pure CPL.
You can verify or guess the result by looking at each world in the
little 8 worlds multiverse. Keep in mind that "p -> q" is false (in
some world) only when p is true in that world, and q false in that
world. Look at the truth table of "p -> q".
OK, I will come back later and check the "multiverse". I have to
stop for now.
OK.
Bruno
http://iridia.ulb.ac.be/~marchal/
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