On 24 January 2014 23:05, Bruno Marchal <[email protected]> wrote: > > On 24 Jan 2014, at 00:01, LizR wrote: > > On 24 January 2014 00:33, Bruno Marchal <[email protected]> wrote: > >> (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? >
OK. I think I see. p becomes "if p is true" rather than "p is true" > > 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 -> q) ? Or if ~ Or even if 1... I guess p -> q is 'the truth of p implies the truth of q" or "if p is true then q is true". If p is false, q could be true or false. It's equivalent to (~p V q) which means if p is false, the result is true regardless of q, if p is true, the result is q. The negation of that would be if p is false, q is true, wouldn't it? So (~p -> q) ? Also it would be ~(~p V q) which maybe comes out to - at a guess (p V ~q). Or does it? ok I need a truth table! p,q=00 = 0, 01=0, 10=1, 11=0. Call that 0010 (That looks familiar...) So the negation would have to be 1101. p V ~q gives 00=1 01=0 10=1 11=1 so not quite the same! So you can't "multiply through" by ~ (somehow that doesn't surprise me) Let's try ~p -> q, that's (~~p V q) or p V q which is true except for 00. SO not that either! OK let's stop trying to work out the answer and use the hint you gave. Is it (~p -> ~q)? Or is it (~q -> ~p)? The first one is (p V ~q) which gives 00=1 01=0 10=1 11=1 or 1011, which isn't the negation I got above 1101 The second gives (q V ~p) which gives 00=1 01=1 10=0 11=1 which is 1101, so that's the right answer. (~q -> ~p) "if q is false that implies p is false" > 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) > > OK, I believe I proved those (but now I need to read them as English to understand them again, I am no good with the symbols). > 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? > OK, p -> []p The negation is ~[]p -> ~p "if it's not true that p is true in all worlds, that implies p is false in this world" (?) That isn't a law. Nor is the truth of p in this world proof of it being true in all worlds. I can make a multiverse in which p is always true, say one world where p is true. In that "multiverse" p -> []p <>p -> ~[]<>p "if p is true in some world that implies that in not all worlds, p is true in some world" - which isn't true. The negation is []<>p -> <>p which is just []p -> p, which is "T" hence a law. I think! :-) -- 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.
