On 23 January 2014 00:58, Bruno Marchal <marc...@ulb.ac.be> 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 everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
