On 24 Jan 2014, at 00:01, LizR wrote:
On 24 January 2014 00:33, Bruno Marchal <marc...@ulb.ac.be> wrote:
[]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.
Oops. I meant to say p is true in all worlds.
[]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).
OK. That was actually what I meant!
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.
A bit like Smullyan's recipe for immortality - "When I wake up, I
say truthfully - tomorrow when I wake up, I will repeat these
words" (or something similar).
[]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.
OK...
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?)
I don't know. I'm confused.
I give then answer.
Well I repeat the question first. With the definition I gave of
"world" the Multiverse has to contain
1) all possible valuation (assignment of 0 or 1) of the letters (=
propositional variables).
2) cannot contain two worlds with the same valuation.
But we can generalize a bit the semantics of Leibniz, by abandoning 1)
and 2).
Then your argument that p-> []<>p, which was
"it's true in all worlds that p is true in at least one world." becomes
if p is true (in this world, say) then it's true in all worlds that p
is true in at least one world.
You need just use a conditional (if). The word asked was "if".
OK?
(same for the formula below)
<>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.
Good point, it's just the negation of the previous statement! So if
the previous statement is true, this one has to be false.
Just to be sure: ~[]<>p is the negation of []<>p. But of course "<>p -
> ~[]<>p" is not the negation of
"<>p -> []<>p". OK?
By the way what is the negation of (p -> q)? is it (~p -> ~q)? Or is
it (~q -> ~p)?
[]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.
Except I typed "word" instead of world.
My brain automatically supplies that kind of thing. That's why I am
very bad to find the typo. Bah! In different worlds people uses
different words!
I kept doing that (my fingers think they know best) but I missed
correcting that one.
I am astigmatic, so when by a miracle I see the typo, usually I
correct it on the wrong line, and just add a new typo. Yes my hands
are also far quicker than my poor brain. Luckily, I don't use a gun!
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.
I am very good at meta-statements of self-doubt!
That's a good sign of sanity, but please, like anything else, you have
to use this with moderation :)
[](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.
I did a lot better than my meta-doubt led me to expect :)
It seems indeed.
To sum up:
Are you convinced that in a Leibnizian multiverse (even with our
generalized definition) we have the following laws (with their usual
name):
[]p -> p (T)
[]p -> [][]p (4)
[](p -> q) ->. ([]p -> []q) (k)
[]p -> <>p (D)
p -> []<>p (B)
<>p -> []<>p (5)
and that we don't have (as law)
p -> []p (Triv)
<>p -> ~[]<>p (g)
Of course, if some formula are not laws in a multiverse, it remains
possible that such formula are true in *some* world of the multiverse.
Can you build a little multiverse in which those last two formula are
true in some world? or is the negation of such formula laws?
Bruno
http://iridia.ulb.ac.be/~marchal/
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