On 27 January 2014 06:11, Bruno Marchal <[email protected]> wrote: > > On 26 Jan 2014, at 01:56, LizR wrote: > > On 25 January 2014 23:56, Bruno Marchal <[email protected]> wrote: > >> >>> 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" >> >> Yes. >> >> Rereading a previews post I ask myself if this is well understood. >> > > I have tended to work on the basis that 'p' means 'p is true' > > That is correct. > > - to make it easier to get my head around what an expression like "[]p -> > p" means. > > ? > > p -> q means: if p is true then q is true. (or means, equivalently 'p is > false or q is true') > > In fact "p -> q" is a sort of negation of p. It means "p if false (unless > q is true)". >
OK, I think I misunderstood something you said which made me think I'd previously misunderstood ... but actually I hadn't. I got it right the first time. > I realise it could also mean "if p is false in all worlds, that implies it > is false in this one" > > Here you talk like if "p -> q" implies "~p -> ~q". > > But "p -> q" is equivalent with "~q -> ~p", not with "~p -> ~q" > > "Socrates is human -> Socrates is mortal" does not imply "Socrates is not > human -> Socrates is not mortal". Socrates could be my dog, for example. > > But "Socrates is human -> Socrates is mortal" does imply "Socrates is not > mortal -> Socrates is not human" > > Keep in mind that p -> q is ~p V q. Then (if you see that ~~p = p, and > that p V q = q V p). > > ~p -> ~q = ~~p v ~q = p V ~q = ~q V p = q -> p. (not "p -> q"). OK? > Yes. > You said that we cannot infer anything from Alicia song as we don't know >> if his theory/song is true. >> But the whole point of logic is in the art of deriving and reasoning >> without ever knowing if a premise is true or not. Indeed, we even want to >> reason independetly of any interpretation (of the atoical propositions). >> > > Yes, I do appreciate that is the point. I was a bit thrown by the word > usage with Alicia, "if A is singing...everybody loves my baby...can we > deduce..." I mean, I often sing all sorts of things that I don't intend to > be self-referential (e.g. "I am the Walrus") so I felt the need to add a > little caveat. > > OK. > > Let me try to be clear. > > From the truth of "Everybody loves my baby & my baby loves nobody but me" > you have deduced correctly the proposition "everybody loves me". (with me > = Alicia, and, strangely enough, = the baby). > > From the truth of "Alicia song "Everybody loves my baby & my baby loves > nobody" ", we can only deduce that everybody loves Alicia or Alicia is > not correct. In that last case either someone does not love the baby, or > the baby does not love only her, maybe the baby loves someone else, > secretly. > OK. > That error is done by those who believe that I defend the truth of comp, >> which I never do. >> In fact we never know if a theory is true (cf Popper). That is why we do >> theories. We can prove A -> B, without having any clues if A is false (in >> which case A -> B is trivial), or A is true. >> I will come back on this. It is crucially important. >> > > I agree. I think psychologically it's hard to derive the results from a > theory mechanically, without at least having some idea that it could be > true. But obviously one can, as with Alicia. > > You are right. Most of the time, mathematicians are aware of what they > want to prove. They work topdown, using their intuition and familiarity > with the subject. To be sure, very often too, they will prove a different > theorem than the one they were thinking about. In some case they can even > prove the contrary, more or less like Gödel for his 1931 result. He thought > he could prove the consistency of the Hilbert program, but the math reality > kicked back. > Ooh, really?! Well that really IS maths "kicking back big time". I must remember that as an example of how maths really can kick back unexpectedly. > > Nevertheless, the level of rigor in math today is such that in the paper, > you will have to present the proof in a way showing that anyone could > extract a formal proof of it, whose validity can be checked mechanically in > either directly in predicate first order calculus, or in a theory which > admits a known description in first order predicate calculus, like ZF, > category theory. > > All physical theories admits such description (like classical physics, > quantum mechanics, cosmology, etc.). > Yes you need what I would call a formal theory, or whatever I should call it. > Actually those theories does not even climb very high on the ordinal > vertical ladder (of set theory). > ??? > > So, the concrete rational talk between scientists consists in "proofs" > amenable to the formal notion of proofs, which is indeed only a sequence of > formula obtained by the iteration of the modus ponens rule. > technically, some proofs in analysis can be obtained or analysed in term > of iterating that rule in the constructive transfinite, but this will be > for another day. > > But for now, we are not really concerned with deduction, as we look only > at the semantics of CPL and propositional modal formulas. > > A good example is Riemann Hypothesis (RH). We don't know if it is true, >> but thousand of papers study its consequence. >> If later we prove the RH, we will get a bunch of beautiful new theorem. >> If we discover that RH leads to a contradiction, then we refute RH, and >> lost all those theorems, but not necessarily the insight present in some of >> the proofs. >> > > Yes, I understand. (But I bet some of those people really, really wish > that the RH will turn out to be true!) > > You can bet on that. > > >> The negation of (p -> q) = ~(p -> q) = ~(~p V q) = ~~p & ~q = p & ~q. >> That's all. It describes the only line where (p -> q) is false. p must be >> false and q true. >> > > Ah, so ~(~p V q) is ~~p & ~q. I would have naively assumed it was ~~p V ~q > (though obviously using a truth table would show the error) > >> I will have to come back on this later! > > Many logical laws have names. Here are the laws of de Morgan: > > ~(A & B) = (~A V ~B) > Aha. > > ~(A V B) = (~A & ~B) > Aha again! This is important to know when using a rule like (p -> q) = (~p V q), so the negation is ~(~p V q) which is (~~p & ~q) or (p & ~q) > > It is similar with ~ExP(x) = Ax ~P(x), and ~AxP(x) = Ex~P(x), or with ~[]A > = <> ~A, and ~<>A = [] ~A. > > Drawing exercise (which I will not solve, thus) in "modern math": > > Compare with, A and B being arbitrary subset of some big set. > > The complement of (A intersection B) = the union of the complement of A > with the complement of B. > The complement of (A union B) = the intersection of the complement of A > with the complement of B. > > Can you verify this by drawing potatoes? > Yes, I just did it (on paper). -- 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. 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