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... > Oh, it looks we are later: > > So we are. > Actually, you will have to remind me what [] and <> mean before I go any > further. > > (...snip...) > OK? > Yes, so far so good. > > 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? > > 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 > []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. > []p -> <>p > Yes that follows > p -> []<>p > Yes, it's true in all worlds that p is true in at least one world. > <>p -> []<>p > Yes, it's true in all worlds that p is true in at least one world > <>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 > []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?) 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. Or maybe I am misreading that completely... > [](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...? > > 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. > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
