On 24 January 2014 00:33, Bruno Marchal <[email protected]> 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.
> <>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.
>
>
>> []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. I kept doing that (my fingers think
they know best) but I missed correcting that one.
> 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!
>
>
>> [](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 :)
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