Re: Modal Logic (Part 1: Leibniz)
On 21 Jan 2014, at 11:14, Alberto G. Corona wrote: Thanks for the info. It is very interesting and It helps in many ways. You are welcome. The problem with mathematical notation is that it is good to store and systematize knowledge, not to make it understandable. The transmission of knowledge can only be done by replaying the historical process that produces the discovery of that knowledge, as Feyerabend said. And this historical process of discovery-learning-transmission can never have the form of some formalism, but the form of concrete problems and partial steps to a solution in a narrative in which the formalism is nothing but the conclussion of the history, not the starting point. Doing it in the reverse order is one of the greatest mistake of education at all levels that the positivist rationalsim has perpetrated and it is a product of a complete misunderstanding that the modern rationalism has about the human mind since it rejected the greek philosophy. Another problem of mathematical notation, like any other language, is that it tries to be formal, but part of the definitions necessary for his understanding are necessarily outside of itself. Mathematics may be a context-free language, but philosophy is not, as well as mathematics when it is applied to something outside of itself. but that is only an intuition that I have not entirely formalized. You are right, and I hope you will understand that machine understand this, in some technical sense. For example, you have done in the above mail an excellent work of narrative pedagogy, unlike in other cases. But sorry I don´t want to deviate you from the subject of modal logic, that is very interesting. Please forget my responses for now. Oops, I forget to forget :) Bruno 2014/1/21, Bruno Marchal marc...@ulb.ac.be: On 20 Jan 2014, at 23:47, LizR wrote: On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: If you remember Cantor, you see that if we take all variables into account, the multiverse is already a continuum. OK? A world is defined by a infinite sequence like true, false, false, true, true, true, ... corresponding to p, q, r, p1, q1, r1, p2, q2, ... I assume it's a continuum, rather than a countable infinity because if it was countable we could list all the worlds, but of course we can diagonalise the list by changing each truth value. Very good. (Those who does not get this can ask for more explanations). On 21 Jan 2014, at 01:32, LizR wrote: On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: Are the following laws? I don't put the last outer parenthesis for reason of readability. p - p This is a law because p - q is equivalent to (~p V q) and (p V ~p) must be (true OR false), or (false OR true) which are both true (p q) - p using (~p V q) gives (~(p q) V q) ... using 0 and 1 for false and true ... (0,0), (0,1) and (1,0) give 1, (1,1) gives 1 ... so this is true. So it is a law. I think. (p q) - q Hmm. (~(p q) V q) is ... the same as above. p - (p V q) (~p V (p V q)) must be true because of the p V ~p that's in there (as per the first one) q - (p V q) Is the same...hm, these are all laws (apparently). I feel as though I'm probably missing something and getting this all wrong. Have I misunderstood something ? No, it is all good, Liz! What about: (p V q) - p and p - (p q) What about (still in CPL) the question: is (p q) - r equivalent with p - (q - r) Oh! You did not answer: ((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE)) So what? Afraid of the logician's trick? Or of the logician's madness? Try this one if you are afraid to be influenced by your intuition aboutCOLD, WET and ICE: ((p q) - r) - ((p - r) V (q - r)) Is that a law? And what about the modal []p - p ? What about the []p - [][]p, and p - []p ? Is that true in all worlds? Let me an answer the first one: []p - p. The difficulty is that we can't use the truth table, (can you see why) but we have the meaning of []p. Indeed it means that p is true in all world. Now, p itself is either true in all worlds, or it is not true in all worlds. Note that p - p is true in all world (as you have shown above, it is (~p V p), so in each world each p is either true or false. If p is true in all worlds, then p is a law. But if p is true in all world, any A - p will be true too, given that for making A - p false, you need p false (truth is implied by anything, in CPL). So if p itself is a law, []p - is a law. For example (p-p) is a law, so [] (p-p) - (p-p) for example. But what if p is not a law? then ~[]p is true, and has to be true in all worlds. With this simple semantic of Leibniz, []p really simply means that p is true in all world, that is automatically true in all world. If p is not a law, ~[]p is true, and, as I said, this has to be true in all world (in all world we have that p is not a law). So []p is false in all worlds. But false - anything in CPL. OK? So []p
Re: Modal Logic (Part 1: Leibniz)
On 27 Jan 2014, at 23:57, LizR wrote: On 27 January 2014 06:11, Bruno Marchal marc...@ulb.ac.be wrote: On 26 Jan 2014, at 01:56, LizR wrote: On 25 January 2014 23:56, Bruno Marchal marc...@ulb.ac.be 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. Yes. I talk too much :) Consider all typo and incorrect statements of me as exercises! (OK, it is a bit easy for me to say this, but expect errors. It is frequent and normal). 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. OK. 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. 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. Another famous example is the proof of the irrationality of sqrt(2). Although the fact that Pythagorus killed 100 cows, and one disciple (!) belongs plausibly to a legend, it is still a kicking back result, showing that not all length on the plane are commensurable. of course there are many others. 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,
Re: Modal Logic (Part 1: Leibniz)
On 26 Jan 2014, at 01:56, LizR wrote: On 25 January 2014 23:56, Bruno Marchal marc...@ulb.ac.be 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). 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? 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. 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. 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.). 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,
Re: Modal Logic (Part 1: Leibniz)
On 27 January 2014 06:11, Bruno Marchal marc...@ulb.ac.be wrote: On 26 Jan 2014, at 01:56, LizR wrote: On 25 January 2014 23:56, Bruno Marchal marc...@ulb.ac.be 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
Re: Modal Logic (Part 1: Leibniz)
On 24 Jan 2014, at 21:52, LizR wrote: On 24 January 2014 23:05, Bruno Marchal marc...@ulb.ac.be wrote: On 24 Jan 2014, at 00:01, LizR wrote: On 24 January 2014 00:33, Bruno Marchal marc...@ulb.ac.be 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 Yes. Rereading a previews post I ask myself if this is well understood. 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). 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. 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. 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! There is no shame to come back to a truth table. May be some day I can explain you the tableau method, which is handy to verify CPL statements, and can be extended to modal logic, but this is not really part of the course ... 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 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. What you show above is that (p - q) is equivalent with (~q - ~p). It is the correct answer to a wrong question. My fault. I am not sure why I add that negation. Sorry. There was a trap, but I did not intent it. 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). Yes. OK. You can
Re: Modal Logic (Part 1: Leibniz)
On 25 January 2014 23:56, Bruno Marchal marc...@ulb.ac.be 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' - to make it easier to get my head around what an expression like []p - p means. I realise it could also mean if p is false in all worlds, that implies it is false in this one 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. 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. 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!) 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! -- 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.
Re: Modal Logic (Part 1: Leibniz)
On 24 Jan 2014, at 00:01, LizR wrote:
On 24 January 2014 00:33, Bruno Marchal marc...@ulb.ac.be 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.
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?
(same for the formula below)
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.
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 ([](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.
My brain automatically
Re: Modal Logic (Part 1: Leibniz)
On 24 Jan 2014, at 00:20, LizR wrote: On 24 January 2014 01:06, Bruno Marchal marc...@ulb.ac.be wrote: On 23 Jan 2014, at 08:57, LizR wrote: Everybody loves my baby. Therefore my baby loves my baby. But my baby loves nobody but me. Therefore - the only way this can be true - is if Alicia is her baby. So the answer is yes! Excellent. And that was predicate logic! So you are in advance! I don't know what predicate logic is (but I watch a lot of TV detective shows. Perhaps that helps!) Lol. In propositional logic, we have atomic proposition, like p, q, r, ... and we can combine them with the logical connectors , V, -, ~, -, etc. (By the way, how many connectors can we have?) The syllogism Humans are all mortal Socrates is human, Thus Socrates is mortal cannot be studied with propositional logic. We need first order logic. In first order logic we have no propositional variables, but we do have predicate like love(x), greater(x,y), etc... , we can have also terms, like f(x), or g(x, y) (like s(x) in Peano, or +(x,y), usually written x + y, etc. We can have constant symbol, also, like 0. And we have the quantifier A (read for all). (E is defined by the duality below). The syllogism above becomes: Ax (human(x) - mortal(x)) human(Socrates) Thus mortal(humans) Just a little question, useful for later considerations, do you find this obvious? Ax((x 5) - x ≠ 10) ? More on this soon or later. To give a taste of first order logic, it is: Alicia theory: (with Ax = for all x). Ax (x loves MyBaby) (everybody loves my baby) Ax ((MyBaby loves x) - (x = Me)) (my baby loves nobody but me) You deduce correctl, in that theory, that MyBaby = Me, and that everybody loves Me. Nice! It seemed to make more sense as a puzzle in English than with symbols! That happens! And now, given that we talk first order logic (the logic with quantifier like A and E (it exists)), I suggest a little meditation on duality. Ah, Stephen will be happy :) is the [] versus duality related to the Stone duality? I ask myself. Where is my Stone Spaces, by Johnstones? I guess (and hope) in some remaining closed boxes. Do you agree that the Ex in ExP(x) (it exists some x such that it is the case that P(x)) is a dual of Ax, in a similar sense that is a dual of [] in propositional modal logic? ...and you lost me completely. OK, I will take a deep breath and try and break down the problem... Ex means some x exists, which is like saying x perhaps (in some world, x is true) Ax means for all x I think which is like saying []x (in all worlds, x is true) These seem kind of parallel, except I guess Ex and Ax operate within a single world, not a logical multiverse ? Or do they? (Or does it matter?) You are right. It is a single world. The worlds here, will be a different matter. To get the multiverse, we will have to study first order modal logic, but we will not do that a lot. We have defined A by ~[]~A. Can we define ExP(x) by ~Ax ~P(x) ? ExP(x) means that for at least one x, P(x) is true - P(x) is some proposition regarding x, so if P is loves and x is my baby ExP(x) would be Someone loves my baby OK. So ~P(x) is doesn't love my baby So Ax ~P(x) is Nobody loves my baby So ~Ax ~P(x) is Somebody loves my baby - which is the same!! :) Exact. So the answer is yes. Do you agree with the following: ~[]p -~p p isn't true in all words - there is a world in which p is false - yes ~p - []~p there is not a world in which p is true - in all worlds p is false - yes []p - ~~p p is true in all worlds - it isn't true that there is any world in which p is false - yes (the inverse of the one I did above) Can you write those equivalence for A and E in predicate logic? Are they intuitively valid? ~AxP(x) - Ex~P(x) It isn't the case that for all x, P(x) is true ... hence there exists an x for which P(x) is false ~Ex P(x) - Ax ~P(x) There doesn't exist an x for which P(x) is true ... hence for all x, P(x) is false Ax P(x) - ~Ex ~P(x) For all x, P(x) is true ... hence there doesn't exist an x for which P(x) is false. Everything is correct. Let us come back on modal logic. The idea of the modal box [] is an idea of necessity. The dual () is read possible. Can you find the most common english term for the following possible modalities: [] = necessary, = possible [] = obligatory, = ? preferable? Aaah no. Come on. If the military service is not obligatory, then not doing a military service is ... Well, you will say preferable, but that might be only your opinion. It is *permitted . OK? Not (obligatory p) = Permitted (not p) That's why []p - p is a deontic law. Only tyrant can forbid something obligatory, so as to put in jail everyone they want. [] = everywhere, = ? somewhere OK. [] = always, = ? sometime OK. And what about the most important modality
Re: Modal Logic (Part 1: Leibniz)
On 24 January 2014 23:05, Bruno Marchal marc...@ulb.ac.be wrote: On 24 Jan 2014, at 00:01, LizR wrote: On 24 January 2014 00:33, Bruno Marchal marc...@ulb.ac.be 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 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.
Re: Modal Logic (Part 1: Leibniz)
On 23 Jan 2014, at 07:42, LizR wrote:
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...
You read too much quantum mechanics.
Oh, it looks we are later:
So we are.
Even more so now.
Actually, you will have to remind me what [] and mean before I
go any further.
(...snip...)
OK?
Yes, so far so good.
I hope so.
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?
p is true in worlds is a bit sounding like weird to me.
[]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).
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.
[]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.
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?)
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.
[]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.
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.
[](p - q) -. ([]p - []q)
it's true in all words that p-q,
Re: Modal Logic (Part 1: Leibniz)
On 23 Jan 2014, at 07:44, LizR wrote: I think after looking at your next post that I have messed up []p - p and therefore, no doubt, everything else. I need to do the truth table business ... later! No, you were 100% right. You confirms my feeling (when going in my bed yesterday evening, that I might have been wrong by adding too much explanation in the second post). Too much explanation can be anti-pedagogical. My fault. Bruno -- 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. http://iridia.ulb.ac.be/~marchal/ -- 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.
Re: Modal Logic (Part 1: Leibniz)
On 23 Jan 2014, at 08:57, LizR wrote: On 23 January 2014 08:18, Bruno Marchal marc...@ulb.ac.be wrote: OK. A last little exercise in the same vein, for the night. (coming from a book by Jeffrey): Alicia was singing this: Everybody loves my baby. My baby loves nobody but me. Can we deduce from this that everybody loves Alicia? Surely we can't deduce anything about A and her baby, unless we know that the song is true! :-) Oh, but I am not asking if everybody loves Alicia. Only if we can deduce that everybody loves Alicia from Alicia's theory. For this, we can be agnostic on that theory. We need just to assume it. But if it is... Yes, that is what we can assume, if only for the sake of the argument. Everybody loves my baby. Therefore my baby loves my baby. But my baby loves nobody but me. Therefore - the only way this can be true - is if Alicia is her baby. So the answer is yes! Excellent. And that was predicate logic! So you are in advance! To give a taste of first order logic, it is: Alicia theory: (with Ax = for all x). Ax (x loves MyBaby) (everybody loves my baby) Ax ((MyBaby loves x) - (x = Me)) (my baby loves nobody but me) You deduce correctl, in that theory, that MyBaby = Me, and that everybody loves Me. Nice! And now, given that we talk first order logic (the logic with quantifier like A and E (it exists)), I suggest a little meditation on duality. Do you agree that the Ex in ExP(x) (it exists some x such that it is the case that P(x)) is a dual of Ax, in a similar sense that is a dual of [] in propositional modal logic? We have defined A by ~[]~A. Can we define ExP(x) by ~Ax ~P(x) ? Do you agree with the following: ~[]p -~p ~p - []~p []p - ~~p Can you write those equivalence for A and E in predicate logic? Are they intuitively valid? Let us come back on modal logic. The idea of the modal box [] is an idea of necessity. The dual () is read possible. Can you find the most common english term for the following possible modalities: [] = necessary, = possible [] = obligatory, = ? [] = everywhere, = ? [] = always, = ? And what about the most important modality which plays the key role in our comp context (and which is the reason why we do all this): [] = provable, = ? Bruno http://iridia.ulb.ac.be/~marchal/ -- 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.
Re: Modal Logic (Part 1: Leibniz)
On 24 January 2014 00:33, Bruno Marchal marc...@ulb.ac.be 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
Re: Modal Logic (Part 1: Leibniz)
On 24 January 2014 01:06, Bruno Marchal marc...@ulb.ac.be wrote: On 23 Jan 2014, at 08:57, LizR wrote: Everybody loves my baby. Therefore my baby loves my baby. But my baby loves nobody but me. Therefore - the only way this can be true - is if Alicia *is* her baby. So the answer is yes! Excellent. And that was predicate logic! So you are in advance! I don't know what predicate logic is (but I watch a lot of TV detective shows. Perhaps that helps!) To give a taste of first order logic, it is: Alicia theory: (with Ax = for all x). Ax (x loves MyBaby) (everybody loves my baby) Ax ((MyBaby loves x) - (x = Me)) (my baby loves nobody but me) You deduce correctl, in that theory, that MyBaby = Me, and that everybody loves Me. Nice! It seemed to make more sense as a puzzle in English than with symbols! And now, given that we talk first order logic (the logic with quantifier like A and E (it exists)), I suggest a little meditation on *duality*. Ah, Stephen will be happy :) Do you agree that the Ex in ExP(x) (it exists some x such that it is the case that P(x)) is a dual of Ax, in a similar sense that is a dual of [] in propositional modal logic? ...and you lost me completely. OK, I will take a deep breath and try and break down the problem... Ex means some x exists, which is like saying x perhaps (in some world, x is true) Ax means for all x I think which is like saying []x (in all worlds, x is true) These seem kind of parallel, except I guess Ex and Ax operate within a single world, not a logical multiverse ? Or do they? (Or does it matter?) We have defined A by ~[]~A. Can we define ExP(x) by ~Ax ~P(x) ? ExP(x) means that for at least one x, P(x) is true - P(x) is some proposition regarding x, so if P is loves and x is my baby ExP(x) would be Someone loves my baby So ~P(x) is doesn't love my baby So Ax ~P(x) is Nobody loves my baby So ~Ax ~P(x) is Somebody loves my baby - which is the same!! :) So the answer is yes. Do you agree with the following: ~[]p -~p p isn't true in all words - there is a world in which p is false - yes ~p - []~p there is not a world in which p is true - in all worlds p is false - yes []p - ~~p p is true in all worlds - it isn't true that there is any world in which p is false - yes (the inverse of the one I did above) Can you write those equivalence for A and E in predicate logic? Are they intuitively valid? ~AxP(x) - Ex~P(x) It isn't the case that for all x, P(x) is true ... hence there exists an x for which P(x) is false ~Ex P(x) - Ax ~P(x) There doesn't exist an x for which P(x) is true ... hence for all x, P(x) is false Ax P(x) - ~Ex ~P(x) For all x, P(x) is true ... hence there doesn't exist an x for which P(x) is false. Let us come back on modal logic. The idea of the modal box [] is an idea of necessity. The dual () is read possible. Can you find the most common english term for the following possible modalities: [] = necessary, = possible [] = obligatory, = ? preferable? [] = everywhere, = ? somewhere [] = always, = ? sometime And what about the most important modality which plays the key role in our comp context (and which is the reason why we do all this): [] = provable, = ? possible? Bruno http://iridia.ulb.ac.be/~marchal/ -- 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. -- 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.
Re: Modal Logic (Part 1: Leibniz)
On 22 Jan 2014, at 00:16, LizR wrote: On 21 January 2014 22:29, Bruno Marchal marc...@ulb.ac.be wrote: Oh! You did not answer: ((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE)) So what? Afraid of the logician's trick? Or of the logician's madness? Try this one if you are afraid to be influenced by your intuition aboutCOLD, WET and ICE: No, I will get back to you on the rest when I have time. I ran out of time (plus I thought maybe I was going up the wrong path...you have reassured me about that now!) No problem. I ran out of time myself. Bruno -- 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. http://iridia.ulb.ac.be/~marchal/ -- 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.
Re: Modal Logic (Part 1: Leibniz)
On 22 Jan 2014, at 04:23, LizR wrote:
On 21 January 2014 22:29, Bruno Marchal marc...@ulb.ac.be wrote:
No, it is all good, Liz!
What about:
(p V q) - p
Using the same formula this is equivalent to(~(p V q) V p), which
for (0,1) is 0, hence not a law.
and
p - (p q)
And this is (~p V (p q)) which is 0 for (1,0), hence also not a
law :-)
What about (still in CPL) the question:
is (p q) - r equivalent with p - (q - r)
is (~(p q) V r) equal to (~p V (~q V r)) ?
or is
~((p q) ~r) equal to ~(p ~~(q ~r))
i.e. is
~((p q) ~r) equal to ~(p (q ~r))
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.
Oh! You did not answer:
((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE))
Eek! That is even more difficult. Luckily you provided something
that didn't involve so much typing...
((p q) - r) - ((p - r) V (q - r))
Expanding furiously and trying not to make any mistakes...
~((p q) ~r) - ( ~(p ~r) V ~(q ~r))
~(~((p q) ~r) ~(~(p ~r) V ~(q ~r)))
Um! Assuming for a moment that's correct, we have 8 possible
combinations of values for p,q,r
r = 0 gives 1 (so that's half the values sorted)
r = 1 also gives 1 (so that's the other half)
So assuming I expanded it correctly, it's a law.
Very good. And so
((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE))
Is true in all worlds!
Of course it is non sense if you interpret the arrow in p - q as a
causal implication. I use that formula to explain that - is not a
causal relation.
I will try more later...
OK.
Oh, it looks we are later:
Actually, you will have to remind me what [] and mean before I go
any further.
OK.
Let us take only 3 propositional variable, or letters, in our
language; p, q, and r, say.
A world (in that context) is given when we say which (atomic)
proposition is true, and which is false.
So, with three propositional variable we get 8 worlds, in the
multiverse associated to {p, q, r} (our language).
They are the one in which p, q, and r are all true, the one in which
p, and q are true, but r is false, ..., the one in which p, q and r
are all false.
OK?
If we fix the order p, q, r on {p, q, r}, we can represent a world by
a sequence of 0 and 1 (which by the way are often used to represent
true and false)? The 8 worlds of the multiverse are given by
000
100
010
110
101
001
111
011
OK?
Let A, B, C range over arbitrary formula. I recall that there are two
kind of formula. the atomic formula and the compound formula. (A, B, C
are metavariable. A-B is NOT a formula, unless A and B designate some
formula (which can contains only the formal p, q, r (and the logical
symbols, parentheses, etc.)
It is the same in algebra. x is not number, unless x designate some
number, like when x = 42. OK?
An atomic formula is just a letter from our set of propositional
letter. We call it an atomic proposition when we think about it in the
company of some truth or false assignment (a proposition can be said
true, or false, not a letter!). OK?
CPL is truth-functional. It means that the truth value (in some world,
thus) of a compound formula is determined by the the truth value of
its subformula, that is eventually by its atomic components.
So the semantic here is very easy, and can be described by the truth
tables, where the truth value of a compound formula is put under the
main connector of the formula :
~ p
0 1
1 0
p q
1 1 1
0 0 1
1 0 0
0 0 0
p V q
1 1 1
0 1 1
1 1 0
0 0 0
p - q (~p V q)
1 1 1
0 1 1
1 0 0
0 1 0
OK?
Now, your question, what does mean []A, for A some formula. And what
does mean A
Well, for Leibniz and Aristotle, it means that A has the value true
in all worlds, or A is true in all worlds, or that all worlds (in the
multiverse) satisfy A.
In particular, given that you have shown that (p - p) is a law, true
in all worlds, we have that
[] (p - p)
OK? For each CPL laws A, sometime called tautology, we will have
that []A.
examples are easly derived from you work. We will have
[] (p - p)
[] (p - (q - p))
[] ((p q) - r) - ((p - r) V (q - r))
etc.
OK? That is easy. tautologies, that is laws, are universal, so they
are verified or satisfied in all worlds, and so the fact that they are
laws is true in all worlds.
This will remain valid for the more general Kripke semantics so you
might try to remind this:
If A is true in all worlds, then []A is true in all worlds.
OK?
from this, and the semantic of the not ('~'), can you find the
semantics for the diamond ?
p is true if ~[]~p is true, if []~p is false, which means that there
is a world in which p is true.
Unfortunately all this might not seem helpful for a formula which mix
modal compounds with non modal
Re: Modal Logic (Part 1: Leibniz)
Hi Liz,
May be I am to quick.
On 22 Jan 2014, at 12:58, Bruno Marchal wrote:
On 22 Jan 2014, at 04:23, LizR wrote:
On 21 January 2014 22:29, Bruno Marchal marc...@ulb.ac.be wrote:
No, it is all good, Liz!
What about:
(p V q) - p
Using the same formula this is equivalent to(~(p V q) V p), which
for (0,1) is 0, hence not a law.
and
p - (p q)
And this is (~p V (p q)) which is 0 for (1,0), hence also not a
law :-)
What about (still in CPL) the question:
is (p q) - r equivalent with p - (q - r)
is (~(p q) V r) equal to (~p V (~q V r)) ?
or is
~((p q) ~r) equal to ~(p ~~(q ~r))
i.e. is
~((p q) ~r) equal to ~(p (q ~r))
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.
Oh! You did not answer:
((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE))
Eek! That is even more difficult. Luckily you provided something
that didn't involve so much typing...
((p q) - r) - ((p - r) V (q - r))
Expanding furiously and trying not to make any mistakes...
~((p q) ~r) - ( ~(p ~r) V ~(q ~r))
~(~((p q) ~r) ~(~(p ~r) V ~(q ~r)))
Um! Assuming for a moment that's correct, we have 8 possible
combinations of values for p,q,r
r = 0 gives 1 (so that's half the values sorted)
r = 1 also gives 1 (so that's the other half)
So assuming I expanded it correctly, it's a law.
Very good. And so
((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE))
Is true in all worlds!
Of course it is non sense if you interpret the arrow in p - q as
a causal implication. I use that formula to explain that - is not
a causal relation.
I will try more later...
OK.
Oh, it looks we are later:
Actually, you will have to remind me what [] and mean before I
go any further.
OK.
Let us take only 3 propositional variable, or letters, in our
language; p, q, and r, say.
A world (in that context) is given when we say which (atomic)
proposition is true, and which is false.
So, with three propositional variable we get 8 worlds, in the
multiverse associated to {p, q, r} (our language).
They are the one in which p, q, and r are all true, the one in which
p, and q are true, but r is false, ..., the one in which p, q and r
are all false.
OK?
If we fix the order p, q, r on {p, q, r}, we can represent a world
by a sequence of 0 and 1 (which by the way are often used to
represent true and false)? The 8 worlds of the multiverse are given by
000
100
010
110
101
001
111
011
OK?
Let A, B, C range over arbitrary formula. I recall that there are
two kind of formula. the atomic formula and the compound formula.
(A, B, C are metavariable. A-B is NOT a formula, unless A and B
designate some formula (which can contains only the formal p, q, r
(and the logical symbols, parentheses, etc.)
It is the same in algebra. x is not number, unless x designate some
number, like when x = 42. OK?
An atomic formula is just a letter from our set of propositional
letter. We call it an atomic proposition when we think about it in
the company of some truth or false assignment (a proposition can be
said true, or false, not a letter!). OK?
CPL is truth-functional. It means that the truth value (in some
world, thus) of a compound formula is determined by the the truth
value of its subformula, that is eventually by its atomic components.
So the semantic here is very easy, and can be described by the truth
tables, where the truth value of a compound formula is put under the
main connector of the formula :
~ p
0 1
1 0
p q
1 1 1
0 0 1
1 0 0
0 0 0
p V q
1 1 1
0 1 1
1 1 0
0 0 0
p - q (~p V q)
1 1 1
0 1 1
1 0 0
0 1 0
OK?
Now, your question, what does mean []A, for A some formula. And what
does mean A
Well, for Leibniz and Aristotle, it means that A has the value true
in all worlds, or A is true in all worlds, or that all worlds (in
the multiverse) satisfy A.
In particular, given that you have shown that (p - p) is a law,
true in all worlds, we have that
[] (p - p)
OK? For each CPL laws A, sometime called tautology, we will have
that []A.
examples are easly derived from you work. We will have
[] (p - p)
[] (p - (q - p))
[] ((p q) - r) - ((p - r) V (q - r))
etc.
OK? That is easy. tautologies, that is laws, are universal, so
they are verified or satisfied in all worlds, and so the fact that
they are laws is true in all worlds.
This will remain valid for the more general Kripke semantics so you
might try to remind this:
If A is true in all worlds, then []A is true in all worlds.
OK?
from this, and the semantic of the not ('~'), can you find the
semantics for the diamond ?
p is true if ~[]~p is true, if []~p is false, which means that
there is a world in which p is true.
Unfortunately all this
Re: Modal Logic (Part 1: Leibniz)
I think after looking at your next post that I have messed up []p - p and therefore, no doubt, everything else. I need to do the truth table business ... later! -- 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.
Re: Modal Logic (Part 1: Leibniz)
On 23 January 2014 08:18, Bruno Marchal marc...@ulb.ac.be wrote: OK. A last little exercise in the same vein, for the night. (coming from a book by Jeffrey): Alicia was singing this: Everybody loves my baby. My baby loves nobody but me. Can we deduce from this that everybody loves Alicia? Surely we can't deduce anything about A and her baby, unless we know that the song is true! :-) But *if* it is... Everybody loves my baby. Therefore my baby loves my baby. But my baby loves nobody but me. Therefore - the only way this can be true - is if Alicia *is*her baby. So the answer is yes! -- 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.
Re: Modal Logic (Part 1: Leibniz)
On 20 Jan 2014, at 23:47, LizR wrote: On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: If you remember Cantor, you see that if we take all variables into account, the multiverse is already a continuum. OK? A world is defined by a infinite sequence like true, false, false, true, true, true, ... corresponding to p, q, r, p1, q1, r1, p2, q2, ... I assume it's a continuum, rather than a countable infinity because if it was countable we could list all the worlds, but of course we can diagonalise the list by changing each truth value. Very good. (Those who does not get this can ask for more explanations). On 21 Jan 2014, at 01:32, LizR wrote: On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: Are the following laws? I don't put the last outer parenthesis for reason of readability. p - p This is a law because p - q is equivalent to (~p V q) and (p V ~p) must be (true OR false), or (false OR true) which are both true (p q) - p using (~p V q) gives (~(p q) V q) ... using 0 and 1 for false and true ... (0,0), (0,1) and (1,0) give 1, (1,1) gives 1 ... so this is true. So it is a law. I think. (p q) - q Hmm. (~(p q) V q) is ... the same as above. p - (p V q) (~p V (p V q)) must be true because of the p V ~p that's in there (as per the first one) q - (p V q) Is the same...hm, these are all laws (apparently). I feel as though I'm probably missing something and getting this all wrong. Have I misunderstood something ? No, it is all good, Liz! What about: (p V q) - p and p - (p q) What about (still in CPL) the question: is (p q) - r equivalent with p - (q - r) Oh! You did not answer: ((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE)) So what? Afraid of the logician's trick? Or of the logician's madness? Try this one if you are afraid to be influenced by your intuition aboutCOLD, WET and ICE: ((p q) - r) - ((p - r) V (q - r)) Is that a law? And what about the modal []p - p ? What about the []p - [][]p, and p - []p ? Is that true in all worlds? Let me an answer the first one: []p - p. The difficulty is that we can't use the truth table, (can you see why) but we have the meaning of []p. Indeed it means that p is true in all world. Now, p itself is either true in all worlds, or it is not true in all worlds. Note that p - p is true in all world (as you have shown above, it is (~p V p), so in each world each p is either true or false. If p is true in all worlds, then p is a law. But if p is true in all world, any A - p will be true too, given that for making A - p false, you need p false (truth is implied by anything, in CPL). So if p itself is a law, []p - is a law. For example (p-p) is a law, so [] (p-p) - (p-p) for example. But what if p is not a law? then ~[]p is true, and has to be true in all worlds. With this simple semantic of Leibniz, []p really simply means that p is true in all world, that is automatically true in all world. If p is not a law, ~[]p is true, and, as I said, this has to be true in all world (in all world we have that p is not a law). So []p is false in all worlds. But false - anything in CPL. OK? So []p - p is always satisfied in that case too. So, no matter what, p being a law or not, in that Leibnizian universe: []p - p *is* a law. In Leibniz semantic, you have just a collection of worlds. If []p is true, it entails that p is true in all worlds, so []p is true in all world too. Can you try to reason for []p - [][]p, and p - []p ? What about p - []p ? What about this one: ([]p [](p - q)) - []q ? Ask any question, up to be able to find the solution. tell me where you are stuck, in case you are stuck. I might go too much quickly, you have to speed me down, by questioning! It may seem astonishing, but with the simple Leibnizian semantic, we can answer all those questions. With Kripke semantics, the multiverse will get more structure. But in Leibniz, all worlds are completely independent, so if p is a law, the fact that it is a law is itself a law, and []p will be true in all worlds, and be a law itself. Indeed if []p was not a law, there would be a world where ~p is true, and p would not be a law, OK? Take those questions as puzzle, or delicious torture :) Bruno http://iridia.ulb.ac.be/~marchal/ -- 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.
Re: Modal Logic (Part 1: Leibniz)
Thanks for the info. It is very interesting and It helps in many ways. The problem with mathematical notation is that it is good to store and systematize knowledge, not to make it understandable. The transmission of knowledge can only be done by replaying the historical process that produces the discovery of that knowledge, as Feyerabend said. And this historical process of discovery-learning-transmission can never have the form of some formalism, but the form of concrete problems and partial steps to a solution in a narrative in which the formalism is nothing but the conclussion of the history, not the starting point. Doing it in the reverse order is one of the greatest mistake of education at all levels that the positivist rationalsim has perpetrated and it is a product of a complete misunderstanding that the modern rationalism has about the human mind since it rejected the greek philosophy. Another problem of mathematical notation, like any other language, is that it tries to be formal, but part of the definitions necessary for his understanding are necessarily outside of itself. Mathematics may be a context-free language, but philosophy is not, as well as mathematics when it is applied to something outside of itself. but that is only an intuition that I have not entirely formalized. 2014/1/21, Bruno Marchal marc...@ulb.ac.be: On 20 Jan 2014, at 23:47, LizR wrote: On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: If you remember Cantor, you see that if we take all variables into account, the multiverse is already a continuum. OK? A world is defined by a infinite sequence like true, false, false, true, true, true, ... corresponding to p, q, r, p1, q1, r1, p2, q2, ... I assume it's a continuum, rather than a countable infinity because if it was countable we could list all the worlds, but of course we can diagonalise the list by changing each truth value. Very good. (Those who does not get this can ask for more explanations). On 21 Jan 2014, at 01:32, LizR wrote: On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: Are the following laws? I don't put the last outer parenthesis for reason of readability. p - p This is a law because p - q is equivalent to (~p V q) and (p V ~p) must be (true OR false), or (false OR true) which are both true (p q) - p using (~p V q) gives (~(p q) V q) ... using 0 and 1 for false and true ... (0,0), (0,1) and (1,0) give 1, (1,1) gives 1 ... so this is true. So it is a law. I think. (p q) - q Hmm. (~(p q) V q) is ... the same as above. p - (p V q) (~p V (p V q)) must be true because of the p V ~p that's in there (as per the first one) q - (p V q) Is the same...hm, these are all laws (apparently). I feel as though I'm probably missing something and getting this all wrong. Have I misunderstood something ? No, it is all good, Liz! What about: (p V q) - p and p - (p q) What about (still in CPL) the question: is (p q) - r equivalent with p - (q - r) Oh! You did not answer: ((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE)) So what? Afraid of the logician's trick? Or of the logician's madness? Try this one if you are afraid to be influenced by your intuition aboutCOLD, WET and ICE: ((p q) - r) - ((p - r) V (q - r)) Is that a law? And what about the modal []p - p ? What about the []p - [][]p, and p - []p ? Is that true in all worlds? Let me an answer the first one: []p - p. The difficulty is that we can't use the truth table, (can you see why) but we have the meaning of []p. Indeed it means that p is true in all world. Now, p itself is either true in all worlds, or it is not true in all worlds. Note that p - p is true in all world (as you have shown above, it is (~p V p), so in each world each p is either true or false. If p is true in all worlds, then p is a law. But if p is true in all world, any A - p will be true too, given that for making A - p false, you need p false (truth is implied by anything, in CPL). So if p itself is a law, []p - is a law. For example (p-p) is a law, so [] (p-p) - (p-p) for example. But what if p is not a law? then ~[]p is true, and has to be true in all worlds. With this simple semantic of Leibniz, []p really simply means that p is true in all world, that is automatically true in all world. If p is not a law, ~[]p is true, and, as I said, this has to be true in all world (in all world we have that p is not a law). So []p is false in all worlds. But false - anything in CPL. OK? So []p - p is always satisfied in that case too. So, no matter what, p being a law or not, in that Leibnizian universe: []p - p *is* a law. In Leibniz semantic, you have just a collection of worlds. If []p is true, it entails that p is true in all worlds, so []p is true in all world too. Can you try to reason for []p - [][]p, and p - []p ? What about p - []p ? What about this one: ([]p [](p - q)) - []q ? Ask any
Re: Modal Logic (Part 1: Leibniz)
On 1/21/2014 2:14 AM, Alberto G. Corona wrote: Thanks for the info. It is very interesting and It helps in many ways. The problem with mathematical notation is that it is good to store and systematize knowledge, not to make it understandable. The transmission of knowledge can only be done by replaying the historical process that produces the discovery of that knowledge, as Feyerabend said. And this historical process of discovery-learning-transmission can never have the form of some formalism, but the form of concrete problems and partial steps to a solution in a narrative in which the formalism is nothing but the conclussion of the history, not the starting point. Right, learning the formalism must ultimately be grounded in examples and ostensive definitions. Doing it in the reverse order is one of the greatest mistake of education at all levels that the positivist rationalsim has perpetrated and it is a product of a complete misunderstanding that the modern rationalism has about the human mind since it rejected the greek philosophy. On the contrary positivism denigrated formalism as mere talk connecting one observation to another. It over emphasized the role of observations and ostensive definition. Brent Another problem of mathematical notation, like any other language, is that it tries to be formal, but part of the definitions necessary for his understanding are necessarily outside of itself. Mathematics may be a context-free language, but philosophy is not, as well as mathematics when it is applied to something outside of itself. but that is only an intuition that I have not entirely formalized. 2014/1/21, Bruno Marchal marc...@ulb.ac.be: On 20 Jan 2014, at 23:47, LizR wrote: On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: If you remember Cantor, you see that if we take all variables into account, the multiverse is already a continuum. OK? A world is defined by a infinite sequence like true, false, false, true, true, true, ... corresponding to p, q, r, p1, q1, r1, p2, q2, ... I assume it's a continuum, rather than a countable infinity because if it was countable we could list all the worlds, but of course we can diagonalise the list by changing each truth value. Very good. (Those who does not get this can ask for more explanations). On 21 Jan 2014, at 01:32, LizR wrote: On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: Are the following laws? I don't put the last outer parenthesis for reason of readability. p - p This is a law because p - q is equivalent to (~p V q) and (p V ~p) must be (true OR false), or (false OR true) which are both true (p q) - p using (~p V q) gives (~(p q) V q) ... using 0 and 1 for false and true ... (0,0), (0,1) and (1,0) give 1, (1,1) gives 1 ... so this is true. So it is a law. I think. (p q) - q Hmm. (~(p q) V q) is ... the same as above. p - (p V q) (~p V (p V q)) must be true because of the p V ~p that's in there (as per the first one) q - (p V q) Is the same...hm, these are all laws (apparently). I feel as though I'm probably missing something and getting this all wrong. Have I misunderstood something ? No, it is all good, Liz! What about: (p V q) - p and p - (p q) What about (still in CPL) the question: is (p q) - r equivalent with p - (q - r) Oh! You did not answer: ((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE)) So what? Afraid of the logician's trick? Or of the logician's madness? Try this one if you are afraid to be influenced by your intuition aboutCOLD, WET and ICE: ((p q) - r) - ((p - r) V (q - r)) Is that a law? And what about the modal []p - p ? What about the []p - [][]p, and p - []p ? Is that true in all worlds? Let me an answer the first one: []p - p. The difficulty is that we can't use the truth table, (can you see why) but we have the meaning of []p. Indeed it means that p is true in all world. Now, p itself is either true in all worlds, or it is not true in all worlds. Note that p - p is true in all world (as you have shown above, it is (~p V p), so in each world each p is either true or false. If p is true in all worlds, then p is a law. But if p is true in all world, any A - p will be true too, given that for making A - p false, you need p false (truth is implied by anything, in CPL). So if p itself is a law, []p - is a law. For example (p-p) is a law, so [] (p-p) - (p-p) for example. But what if p is not a law? then ~[]p is true, and has to be true in all worlds. With this simple semantic of Leibniz, []p really simply means that p is true in all world, that is automatically true in all world. If p is not a law, ~[]p is true, and, as I said, this has to be true in all world (in all world we have that p is not a law). So []p is false in all worlds. But false - anything in CPL. OK? So []p - p is always satisfied in that case too. So, no matter what, p being a law or not, in that Leibnizian universe: []p - p *is* a law. In Leibniz
Re: Modal Logic (Part 1: Leibniz)
On 21 January 2014 22:29, Bruno Marchal marc...@ulb.ac.be wrote: Oh! You did not answer: ((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE)) So what? Afraid of the logician's trick? Or of the logician's madness? Try this one if you are afraid to be influenced by your intuition aboutCOLD, WET and ICE: No, I will get back to you on the rest when I have time. I ran out of time (plus I thought maybe I was going up the wrong path...you have reassured me about that 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.
Re: Modal Logic (Part 1: Leibniz)
On 21 January 2014 22:29, Bruno Marchal marc...@ulb.ac.be wrote: No, it is all good, Liz! What about: (p V q) - p Using the same formula this is equivalent to(~(p V q) V p), which for (0,1) is 0, hence not a law. and p - (p q) And this is (~p V (p q)) which is 0 for (1,0), hence also not a law :-) What about (still in CPL) the question: is (p q) - r equivalent with p - (q - r) is (~(p q) V r) equal to (~p V (~q V r)) ? or is ~((p q) ~r) equal to ~(p ~~(q ~r)) i.e. is ~((p q) ~r) equal to ~(p (q ~r)) 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? Oh! You did not answer: ((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE)) Eek! That is even more difficult. Luckily you provided something that didn't involve so much typing... ((p q) - r) - ((p - r) V (q - r)) Expanding furiously and trying not to make any mistakes... ~((p q) ~r) - ( ~(p ~r) V ~(q ~r)) ~(~((p q) ~r) ~(~(p ~r) V ~(q ~r))) Um! Assuming for a moment that's correct, we have 8 possible combinations of values for p,q,r r = 0 gives 1 (so that's half the values sorted) r = 1 also gives 1 (so that's the other half) So assuming I expanded it correctly, it's a law. I will try more later... -- 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.
Re: Modal Logic (Part 1: Leibniz)
Actually, you will have to remind me what [] and mean before I go any further. -- 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.
Modal Logic (Part 1: Leibniz)
Hi Liz, and others,
I explain the classical modal logic.
It extends classical propositional logic (CPL), that we have already
encounter.
I will recall it first, and present it in a way which will suit well
the modal extensions of CPL.
One big advantage of CPL on all other propositional logic, is the
extreme simplicity of its semantics, which is truth functional, but I
will explain this later.
And what is logic? Also.
Logicians are interested in reasoning, and they want the validity of
the reasoning guarantied by its formal structure. He want to present
the reasoning in a way which does not depend on the interpretation of
the formula involved. That will guarantied that the reasoning is
independent of the interpretation, or of the world which might satisfy
the formula, and the formula will be true in all worlds, or in all
situations, or in all interpretations, etc. making a formula true
independently of the interpretation makes it into a law. It makes it
into something necessary, universally true.
But what is a world? For logician, worlds are rather abstract things.
Here, let me give the simple definition which suits well the goal of
classical propositional logic (CPL).
We make this formal, so we have a list of so-called propositional
variable p, q, r, p1, q1, r1, p2, ...
It helps some people, but distracts others, to instantiate the
propositional variable by concrete propositions like Obama is
president of the US, there is a planet called Earth, Earth has two
natural satellites, 34 is prime, etc.
The idea is that we take *all* propositions (later represented by
sentences, and restricted to more particular language, like predicate
calculus, or arithmetic, or set theory, ...)
Having those propositional variables we can define now a world, or an
interpretation, by an assignment of truth value (true, false) to each
proposition.
To simplify, let us use a simple finite finite set of propositions {p,
q, r}.
With just that set we get 8 worlds (2^3):
- the world where p is true, q is true, and r is true
- the world where p is true, q is true, and r is false
- the world where p is true, q is false, and r is true
- the world where p is true, q is false, and r is false
- the world where p is false, q is true, and r is true
- the world where p is false, q is true, and r is false
- the world where p is false, q is false, and r is true
- the world where p is false, q is false, and r is false
OK?
If you want the set of all possible worlds, having three propositional
variables, can be said to be the mutiverse of that logic.
If you remember Cantor, you see that if we take all variables into
account, the multiverse is already a continuum. OK? A world is defined
by a infinite sequence like true, false, false, true, true,
true, ... corresponding to p, q, r, p1, q1, r1, p2, q2, ...
If p is true in a word, I will often express this by saying that the
world satisfies p, or obeys p.
Now, what makes CPL much more easy than any other propositional logic,
is that the truth of the compound non atomic formula, those build with
the usual symbol , V, ~, -, -, ... is entirely determined by the
worlds, that is by the assignment of truth value to the components of
the formula (eventually the atomic formula, that is the propositional
variables).
For example, if a world satisfies p, and if it satisfies q, then it
satisfy the compound formula (p q).
And in fact, the truth value (true or false) of a conjunction (p q)
is entirely determined in all 8 worlds above.
(And also in richer multiverse with assignment to all variables)
Is the formula (p q) a law? is it true in all possible worlds (in
the multiverse, which here contains 8 worlds).
Obviously not. OK? (p q) is true only in the two first world
described above. All others does not satisfy (p q), they contradicts
it. They constitute counter-examples.
Do you remember the semantic (the association of truth value) for (p V
q)? for ~p?
I recall that the semantic of (p - q) is the one for (~p V q), or
similarly ~(p ~q). OK? (p - q) is false only in the worlds where p
is true and q is false. If p is false in some world, then p - q is
true (trivially true, said some logicians).
Can you find laws? It means a formula true in all worlds (in our
little multiverse, or bigger one).
Are the following laws? I don't put the last outer parenthesis for
reason of readability.
p - p
(p q) - p
(p q) - q
p - (p V q)
q - (p V q)
p - (q - p)
(p - (q - r)) - ((p - q) - (p - r))
What about, with the propositional variable WET, COLD, ICE:
((COLD WET) - ICE) - ((COLD - ICE) V (WET - ICE))
Is that a law? (hint: beware the logician's trick!)
Anyone asks any question if something is not understood. I use the
fact that we have already done a bit of CPL on FOAR more or less
recently, and on this list too, but much less recently.
Let us begin the modal
Re: Modal Logic (Part 1: Leibniz)
On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: If you remember Cantor, you see that if we take all variables into account, the multiverse is already a continuum. OK? A world is defined by a infinite sequence like true, false, false, true, true, true, ... corresponding to p, q, r, p1, q1, r1, p2, q2, ... I assume it's a continuum, rather than a countable infinity because if it was countable we could list all the worlds, but of course we can diagonalise the list by changing each truth value. -- 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.
Re: Modal Logic (Part 1: Leibniz)
On 21 January 2014 08:38, Bruno Marchal marc...@ulb.ac.be wrote: Are the following laws? I don't put the last outer parenthesis for reason of readability. p - p This is a law because p - q is equivalent to (~p V q) and (p V ~p) must be (true OR false), or (false OR true) which are both true (p q) - p using (~p V q) gives (~(p q) V q) ... using 0 and 1 for false and true ... (0,0), (0,1) and (1,0) give 1, (1,1) gives 1 ... so this is true. So it is a law. I think. (p q) - q Hmm. (~(p q) V q) is ... the same as above. p - (p V q) (~p V (p V q)) must be true because of the p V ~p that's in there (as per the first one) q - (p V q) Is the same...hm, these are all laws (apparently). I feel as though I'm probably missing something and getting this all wrong. Have I misunderstood something ? -- 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.
