Re: MODAL Last exercise
On 08 Mar 2014, at 06:32, LizR wrote: On 6 March 2014 22:06, Bruno Marchal marc...@ulb.ac.be wrote: Liz, meanwhile you might try this one, which is a bit more easy than the transitivity case: Show that (W,R) respects []A - A if and only if R is ideal. (I remind you that R is ideal means that there is no cul-de-sac world at all in (W,R)). OK, I consult my diary and... Ideal is as you say, yes! :-) Excellent :) So []A - A means that A is some proposition universally true in an illuminated, accessible multiverse, and this implies that A is possible in that multiverse. Not at all, and you know that, as you show below. Hang on I must be missing something. OK, I hang on. That seems trivially obvious! Maybe you could point out what I've misunderstood here... It is not misunderstanding, it is precipitation. Let me try again. OK. []p means that for any world alpha, p is true in all worlds accessible from alpha. That is much better. (Doesn't it? Well if p is a proposition, which might be 'x is false' then that seems reasonable). You lost me here, but it looks like non relevant, even for you. And p means that, ah, ~[]~p iirc. Which is to say it isn't true that there is a world accessible from alpha in which ~p. ~[]~p means that it is not true that ~p is true in all worlds accessible from alpha. That is, it means that there is a world beta, with alpha R beta, such that beta verifies p. p true in alpha = I can access, from alpha, to a world where p is true. But isn't that implied by []p? You fall again back in Leibniz. May be I should have started from Kripke immediately. Have you hang on, in your toilet, the fundamentals two Kripke principles? * * * * []p is true in alpha = For all beta such that (alpha R beta) we have beta verifies p. * * * * p is true in alpha = There is a beta verifying p such that (alpha R beta) * * * * I must have a definition wrong somewhere. Correct. Do you see that (W, R) is reflexive entails that (W,R) is ideal? If all worlds access to themselves, no world can be a cul-de-sac world, as a cul-de-sac world don't access to any world, including themselves. Reflexive is alpha R alpha for all alpha, so no cul de sac is possible. Correct. More precision later, notably for the transitive case. 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/d/optout.
Re: MODAL Last exercise
On 08 Mar 2014, at 06:20, LizR wrote:
On 6 March 2014 22:06, Bruno Marchal marc...@ulb.ac.be wrote:
On 05 Mar 2014, at 23:31, LizR wrote:
Let's take 3 worlds A B C making a minimal transitive multiverse.
ARB and BRC implies ARC. So if we assume ARB and BRC we also get ARC
Right.
(if we don't assume this we don't have a multiverse or at least not
one we can say anything about.
This, or something like this ...
[]p in this case means the value of p in A is the same as its value
in B and C (t or f).
What if p is false in A, and true in all worlds accessible from A?
Well that means ~[]p, doesn't it?
That would contradict Kripke semantics. it says that []p is true in
alpha IFF p is true in all worlds accessible from alpha (and in this
case alpha does not access to itself, and so the falsity of p in alpha
does not entails ~[]p is true in alpha. Indeed, as p is true in all
beta such that alpha R beta, we do have []p true in alpha.
I think you were just thinking with Leibniz semantics.
This also means that in A B and C, []p is true, hence we can also
say that in all worlds [][]p.
Correct.
(And indeed [][][]p and so on?)
Sure. at least in a multiverse where []A - [][]A is a law. In that
case it is true for any A, and so it is true if A is substituted
with []A, and so [][]A - [][][]A, and so []A - [][][]A, and so on.
So it's true for the minimal case that []p - [][]p
But then adding more worlds will just give the same result in each
set of 3... so does that prove it?
Not sure.
Me neither, as will now be demonstrated.
OK :)
No, hang on. Take { A B C } with p having values { t t f }. []p is
true in C, because C is not connected to anywhere else, which makes
it trivially true if I remember correctly. But []p is false in A
and B. So [][]p is false, even though []p is true in C. So []p
being true in C doesn't imply [][]p.
I might need to see your drawing. If C is not connected to anywhere
else, C is a cul-de-sac world, and so we have certainly that [][]p
is true in C (as []#anything# is true in all cul-de-sac worlds).
A --- B --- C
and
A --- C
where --- means 'can access' - so C is a cul-de-sac and { A B C }
is transitive.
OK.
OK, []X is true in C where X is anything.
So if []p isn't true in A, then [][]p isn't true for { A,B,C }
(though it's true in C treated as a multiverse)
You lost me, here. You suppose R transitive, and I guess you are
trying to prove that []p - [][]p has to be true in all the worlds A,
B and C, and this for any valuations V.
It is simpler to assume that you have a counter-example (a world in
which []p - [][]p is false), and get a contradiction from that (by
absurdum).
But for []p to be true in A, that means p is true (or false) in all
worlds accessible from A, including C. That is, p has the same value
in A B and C. So does that imply []p is true in all worlds
accessible from A? Yes, I think so.
In your little structure, but is it clear if that is preserved in all
transitive multiverses?
And that implies [][]p for all worlds accessible from A, including C
(trivially).
Isn't that what I was trying to prove?
Not sure. A bit fuzzy. The question is more are you convinced
yourself by your reasoning?.
Or have I just wandered off into a cul-de-sac myself?
No worry. It is very good that you seem aware you have not yet make a
proof.
More on this in the sequel.
Bruno
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Re: MODAL Last exercise
I think I need the sequel. -- 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/d/optout.
Re: MODAL Last exercise
On 09 Mar 2014, at 23:01, LizR wrote: I think I need the sequel. Nice. OK. (asap). 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/d/optout. 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/d/optout.
Re: MODAL Last exercise
On 08 Mar 2014, at 06:12, LizR wrote: On 6 March 2014 21:44, Bruno Marchal marc...@ulb.ac.be wrote: On 05 Mar 2014, at 23:06, LizR wrote: On 5 March 2014 20:59, Bruno Marchal marc...@ulb.ac.be wrote: You have to show two things: 1) R is transitive - (W,R) respects []A - [][]A and 2) (W,R) respects []A - [][]A- R is transitive Let us look at 1). To show that R is transitive - (W,R) respects []A - [][]A, you might try to derive a contradiction from R is transitive, and (W,R) does not respect []A - [][]A. What does it mean that (W,R) does not respect a formula? It can only mean that in some (W,R,V) there is world alpha where that formula is false. To say that []A-[][]A is false in alpha means only that []A is true in that world and that [][]A is false in that world. OK. I'm not sure where V came from, but anyway... W = the set of worlds R = the binary relation (of accessibility) (W, R) = the multiverse, or the frame (W, R, V) is the same as the multiverse, except that now, in each worlds of W, the sentence letters p, q, r, ... got a value 1, or 0. And so, all formula can be said to be true or false in each world, by the use of classical logic and of the semantic of Kripke (the fact that []A is determined in alpha by the value of A in its accessible worlds). So V is an illumination? Yes. Also called valuation. That is the function which gives, for each world, the true values of the atomic letters (p, q, r, ...). A formula A is a law in a multiverse (W,R), or (W,R) respects A, if and only if for all V, (W,R,V) satisfies A. And this means that for all V, A is true in all worlds in that multiverse. In modal logic the semantics is three-leveled: a proposition can be true in a world (in some (W,R,V)), it can be true in all worlds, and it can be true in all worlds whatever V is, that is true in all (W,R,V), with some W and and R. So as you say a contradiction is t - f (because f - x is always true, as it t - t) Like a tautology is true in all worlds, a contradiction is a proposition false in all worlds, like f, or (A ~A), or 0 = 1 (in arithmetic). f - x is a tautology, yes, and x - t also. So []A is true in a world alpha. I guess you assume []A - [][]A is false in alpha, which belong to a transitive multiverse, and you want to show that we will arrive at a contradiction. Hence if alpha is transitive, I understand what you mean. But of course it is R which is transitive. and if []A is true in all worlds reachable from alpha, let's call one beta, then []A is also true in all worlds reachable from beta. It looks like now you suppose []A - [][]A is true in alpha. So I am no more sure of what you try to prove. We don't know if alpha is reachable from beta, but we do know that if []A is true in beta then it's true in all worlds reachable from beta. I let you or Brent continue, or anyone else. I don't want to spoil the pleasure of finding the contradiction. Then we can discuss the 2). Surely the pleasure of NOT finding a contradiction? No, the pleasure of finding a contradiction from []A - [][]A is false in alpha and R is transitive I was suggesting you to prove P - Q, by showing that P ~Q implies a contradiction. it is the easiest way, although there are (infinitely many) other ways to proceed. Oh dear I don't think my brain can take this! Maybe a diagram would help. Anyway I have to go now :) Diagram would help a lot. I teach this basically every years since a long time, and I only draw diagrams on the black board, not one symbols, except for the sentence letters true or false in the worlds. Take it easy, we have all the time. My feeling is that you are impatient with yourself. Just calm down. It seems like I have no time! (But you sound like Charles :-) And you sound like the White Rabbit: No time, no time :) You will eventually NOT understand why you ever did find this difficult. But this takes times and work, that's normal. I hope you're right. No problem, unless you try hard to convince yourself that you will never learn, which is a good technic to not learn. But as long as you find the courage to be mistaken, you will learn, and eventually get the familiarity. But you need to be indulgent and patient with yourself. I am more anxious about Brent, who solved well the reflexive multiverse case, but seems mute on the transitive multiverse case. Well, I know what it is, all those mails, Bruno PS I have to go now. Will answer other mails later. Next week is also busy, and I expect some black hole in the basement. Thanks for your patience. 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
Re: MODAL Last exercise
On 6 March 2014 21:44, Bruno Marchal marc...@ulb.ac.be wrote: On 05 Mar 2014, at 23:06, LizR wrote: On 5 March 2014 20:59, Bruno Marchal marc...@ulb.ac.be wrote: You have to show two things: 1) R is transitive - (W,R) respects []A - [][]A and 2) (W,R) respects []A - [][]A- R is transitive Let us look at 1). To show that R is transitive - (W,R) respects []A - [][]A, you might try to derive a contradiction from R is transitive, and (W,R) does not respect []A - [][]A. What does it mean that (W,R) does not respect a formula? It can only mean that in some (W,R,V) there is world alpha where that formula is false. To say that []A-[][]A is false in alpha means only that []A is true in that world and that [][]A is false in that world. OK. I'm not sure where V came from, but anyway... W = the set of worlds R = the binary relation (of accessibility) (W, R) = the multiverse, or the frame (W, R, V) is the same as the multiverse, except that now, in each worlds of W, the sentence letters p, q, r, ... got a value 1, or 0. And so, all formula can be said to be true or false in each world, by the use of classical logic and of the semantic of Kripke (the fact that []A is determined in alpha by the value of A in its accessible worlds). So V is an illumination? So as you say a contradiction is t - f (because f - x is always true, as it t - t) Like a tautology is true in all worlds, a contradiction is a proposition false in all worlds, like f, or (A ~A), or 0 = 1 (in arithmetic). f - x is a tautology, yes, and x - t also. So []A is true in a world alpha. I guess you assume []A - [][]A is false in alpha, which belong to a transitive multiverse, and you want to show that we will arrive at a contradiction. Hence if alpha is transitive, I understand what you mean. But of course it is R which is transitive. and if []A is true in all worlds reachable from alpha, let's call one beta, then []A is also true in all worlds reachable from beta. It looks like now you suppose []A - [][]A is true in alpha. So I am no more sure of what you try to prove. We don't know if alpha is reachable from beta, but we do know that if []A is true in beta then it's true in all worlds reachable from beta. I let you or Brent continue, or anyone else. I don't want to spoil the pleasure of finding the contradiction. Then we can discuss the 2). Surely the pleasure of NOT finding a contradiction? No, the pleasure of finding a contradiction from []A - [][]A is false in alpha and R is transitive I was suggesting you to prove P - Q, by showing that P ~Q implies a contradiction. it is the easiest way, although there are (infinitely many) other ways to proceed. Oh dear I don't think my brain can take this! Maybe a diagram would help. Anyway I have to go now :) Diagram would help a lot. I teach this basically every years since a long time, and I only draw diagrams on the black board, not one symbols, except for the sentence letters true or false in the worlds. Take it easy, we have all the time. My feeling is that you are impatient with yourself. Just calm down. It seems like I have no time! (But you sound like Charles :-) You will eventually NOT understand why you ever did find this difficult. But this takes times and work, that's normal. I hope you're right. -- 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/d/optout.
Re: MODAL Last exercise
On 6 March 2014 22:06, Bruno Marchal marc...@ulb.ac.be wrote:
On 05 Mar 2014, at 23:31, LizR wrote:
Let's take 3 worlds A B C making a minimal transitive multiverse. ARB and
BRC implies ARC. So if we assume ARB and BRC we also get ARC
Right.
(if we don't assume this we don't have a multiverse or at least not one we
can say anything about.
This, or something like this ...
[]p in this case means the value of p in A is the same as its value in B
and C (t or f).
What if p is false in A, and true in all worlds accessible from A?
Well that means ~[]p, doesn't it?
This also means that in A B and C, []p is true, hence we can also say that
in all worlds [][]p.
Correct.
(And indeed [][][]p and so on?)
Sure. at least in a multiverse where []A - [][]A is a law. In that case
it is true for any A, and so it is true if A is substituted with []A, and
so [][]A - [][][]A, and so []A - [][][]A, and so on.
So it's true for the minimal case that []p - [][]p
But then adding more worlds will just give the same result in each set of
3... so does that prove it?
Not sure.
Me neither, as will now be demonstrated.
No, hang on. Take { A B C } with p having values { t t f }. []p is true in
C, because C is not connected to anywhere else, which makes it trivially
true if I remember correctly. But []p is false in A and B. So [][]p is
false, even though []p is true in C. So []p being true in C doesn't imply
[][]p.
I might need to see your drawing. If C is not connected to anywhere else,
C is a cul-de-sac world, and so we have certainly that [][]p is true in C
(as []#anything# is true in all cul-de-sac worlds).
A --- B --- C
and
A --- C
where --- means 'can access' - so C is a cul-de-sac and { A B C } is
transitive.
OK, []X is true in C where X is anything.
So if []p isn't true in A, then [][]p isn't true for { A,B,C } (though it's
true in C treated as a multiverse)
But for []p to be true in A, that means p is true (or false) in all worlds
accessible from A, including C. That is, p has the same value in A B and C.
So does that imply []p is true in all worlds accessible from A? Yes, I
think so. And that implies [][]p for all worlds accessible from A,
including C (trivially).
Isn't that what I was trying to prove? Or have I just wandered off into a
cul-de-sac myself?
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Re: MODAL Last exercise
On 6 March 2014 22:06, Bruno Marchal marc...@ulb.ac.be wrote: Liz, meanwhile you might try this one, which is a bit more easy than the transitivity case: Show that (W,R) respects []A - A if and only if R is ideal. (I remind you that R is ideal means that there is no cul-de-sac world at all in (W,R)). OK, I consult my diary and... Ideal is as you say, yes! :-) So []A - A means that A is some proposition universally true in an illuminated, accessible multiverse, and this implies that A is possible in that multiverse. Hang on I must be missing something. That seems trivially obvious! Maybe you could point out what I've misunderstood here... Let me try again. []p means that for any world alpha, p is true in all worlds accessible from alpha. (Doesn't it? Well if p is a proposition, which might be 'x is false' then that seems reasonable). And p means that, ah, ~[]~p iirc. Which is to say it isn't true that there is a world accessible from alpha in which ~p. But isn't that implied by []p? I must have a definition wrong somewhere. Do you see that (W, R) is reflexive entails that (W,R) is ideal? If all worlds access to themselves, no world can be a cul-de-sac world, as a cul-de-sac world don't access to any world, including themselves. Reflexive is alpha R alpha for all alpha, so no cul de sac is possible. 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. -- 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/d/optout.
Re: MODAL Last exercise
On 05 Mar 2014, at 23:06, LizR wrote: On 5 March 2014 20:59, Bruno Marchal marc...@ulb.ac.be wrote: You have to show two things: 1) R is transitive - (W,R) respects []A - [][]A and 2) (W,R) respects []A - [][]A- R is transitive Let us look at 1). To show that R is transitive - (W,R) respects []A - [][]A, you might try to derive a contradiction from R is transitive, and (W,R) does not respect []A - [][]A. What does it mean that (W,R) does not respect a formula? It can only mean that in some (W,R,V) there is world alpha where that formula is false. To say that []A-[][]A is false in alpha means only that []A is true in that world and that [][]A is false in that world. OK. I'm not sure where V came from, but anyway... W = the set of worlds R = the binary relation (of accessibility) (W, R) = the multiverse, or the frame (W, R, V) is the same as the multiverse, except that now, in each worlds of W, the sentence letters p, q, r, ... got a value 1, or 0. And so, all formula can be said to be true or false in each world, by the use of classical logic and of the semantic of Kripke (the fact that []A is determined in alpha by the value of A in its accessible worlds). So as you say a contradiction is t - f (because f - x is always true, as it t - t) Like a tautology is true in all worlds, a contradiction is a proposition false in all worlds, like f, or (A ~A), or 0 = 1 (in arithmetic). f - x is a tautology, yes, and x - t also. So []A is true in a world alpha. I guess you assume []A - [][]A is false in alpha, which belong to a transitive multiverse, and you want to show that we will arrive at a contradiction. Hence if alpha is transitive, I understand what you mean. But of course it is R which is transitive. and if []A is true in all worlds reachable from alpha, let's call one beta, then []A is also true in all worlds reachable from beta. It looks like now you suppose []A - [][]A is true in alpha. So I am no more sure of what you try to prove. We don't know if alpha is reachable from beta, but we do know that if []A is true in beta then it's true in all worlds reachable from beta. I let you or Brent continue, or anyone else. I don't want to spoil the pleasure of finding the contradiction. Then we can discuss the 2). Surely the pleasure of NOT finding a contradiction? No, the pleasure of finding a contradiction from []A - [][]A is false in alpha and R is transitive I was suggesting you to prove P - Q, by showing that P ~Q implies a contradiction. it is the easiest way, although there are (infinitely many) other ways to proceed. Oh dear I don't think my brain can take this! Maybe a diagram would help. Anyway I have to go now :) Diagram would help a lot. I teach this basically every years since a long time, and I only draw diagrams on the black board, not one symbols, except for the sentence letters true or false in the worlds. Take it easy, we have all the time. My feeling is that you are impatient with yourself. Just calm down. You will eventually NOT understand why you ever did find this difficult. But this takes times and work, that's normal. Bruno It is almost more easy to find this by yourself than reading the solution, and then searching the solution is part of the needed training to be sure you put the right sense on the matter. Keep in mind the semantic definitions. We assume some illuminated (W,R,V) Atomic proposition (like the initial p, q, r, ...) is true in a world alpha , iff V(p) = 1 for that word alpha. Classical propositional tautologies are true in all worlds. []A is true at world alpha iff A is true in all worlds accessible from alpha. (W,R,V) satisfies a formula if that formula is true in all worlds in W (with its R and V, of course). (W,R) respects a formula if that formula is satisfied for all V. So the formula is true in all worlds of W, whatever the valuation V is. Courage! 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 Last exercise
On 05 Mar 2014, at 23:31, LizR wrote:
Let's take 3 worlds A B C making a minimal transitive multiverse.
ARB and BRC implies ARC. So if we assume ARB and BRC we also get ARC
Right.
(if we don't assume this we don't have a multiverse or at least not
one we can say anything about.
This, or something like this ...
[]p in this case means the value of p in A is the same as its value
in B and C (t or f).
What if p is false in A, and true in all worlds accessible from A?
This also means that in A B and C, []p is true, hence we can also
say that in all worlds [][]p.
Correct.
(And indeed [][][]p and so on?)
Sure. at least in a multiverse where []A - [][]A is a law. In that
case it is true for any A, and so it is true if A is substituted with
[]A, and so [][]A - [][][]A, and so []A - [][][]A, and so on.
So it's true for the minimal case that []p - [][]p
But then adding more worlds will just give the same result in each
set of 3... so does that prove it?
Not sure.
No, hang on. Take { A B C } with p having values { t t f }. []p is
true in C, because C is not connected to anywhere else, which makes
it trivially true if I remember correctly. But []p is false in A and
B. So [][]p is false, even though []p is true in C. So []p being
true in C doesn't imply [][]p.
I might need to see your drawing. If C is not connected to anywhere
else, C is a cul-de-sac world, and so we have certainly that [][]p is
true in C (as []#anything# is true in all cul-de-sac worlds).
So that seems to disprove it, because C is in its own little
multiverse. There's nothing in the definition that says ARB and BRC
entails CRA or CRB, is there?
No, indeed.
Unless I have the trivially true thing wrong...
Yes, in the cul-de-sac world, [][]p is automatically vacuously true.
Good work Liz. I will provide clean solutions, but I wait a bit for
Brent. Brent?
Liz, meanwhile you might try this one, which is a bit more easy than
the transitivity case:
Show that (W,R) respects []A - A if and only if R is ideal.
(I remind you that R is ideal means that there is no cul-de-sac world
at all in (W,R)).
Do you see that (W, R) is reflexive entails that (W,R) is ideal? If
all worlds access to themselves, no world can be a cul-de-sac world,
as a cul-de-sac world don't access to any world, including themselves.
Bruno
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Re: MODAL Last exercise
On 05 Mar 2014, at 01:36, LizR wrote: On 5 March 2014 04:18, Bruno Marchal marc...@ulb.ac.be wrote: Good. To prove that P - Q, you can prove that P ~Q leads to a contradiction, or you can prove that ~Q leads to ~P. But it helps a lot if you start from what you want to prove, up to the conclusion, so that not only you prove it, but you know exactly what you discovered. In this case a necessary link, in Kripke semantics, between a binary relation (reflexivity) and a modal formula []A-A. I had to get my head around ... well, everything ... again. So I may have sneaked up on the result. You learned that the fact that (W, R) respects []A - A is equivalent with the fact that R is reflexive. OK? OK. OK. So, the next question was A Kripke multiverse (W, R) is said transitive if R is transitive. That is alpha R beta, and beta R gamma entails alpha R gamma, for all alpha beta and gamma in W. Show that (W, R) respects []A - [][]A if and only R is transitive, Damn. This looks too complicated for me to fake it! Hmm. You have to show two things: 1) R is transitive - (W,R) respects []A - [][]A and 2) (W,R) respects []A - [][]A- R is transitive Let us look at 1). To show that R is transitive - (W,R) respects []A - [][]A, you might try to derive a contradiction from R is transitive, and (W,R) does not respect []A - [][]A. What does it mean that (W,R) does not respect a formula? It can only mean that in some (W,R,V) there is world alpha where that formula is false. To say that []A-[][]A is false in alpha means only that []A is true in that world and that [][]A is false in that world. I let you or Brent continue, or anyone else. I don't want to spoil the pleasure of finding the contradiction. Then we can discuss the 2). It is almost more easy to find this by yourself than reading the solution, and then searching the solution is part of the needed training to be sure you put the right sense on the matter. Keep in mind the semantic definitions. We assume some illuminated (W,R,V) Atomic proposition (like the initial p, q, r, ...) is true in a world alpha , iff V(p) = 1 for that word alpha. Classical propositional tautologies are true in all worlds. []A is true at world alpha iff A is true in all worlds accessible from alpha. (W,R,V) satisfies a formula if that formula is true in all worlds in W (with its R and V, of course). (W,R) respects a formula if that formula is satisfied for all V. So the formula is true in all worlds of W, whatever the valuation V is. Courage! 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 Last exercise
Let's take 3 worlds A B C making a minimal transitive multiverse. ARB and
BRC implies ARC. So if we assume ARB and BRC we also get ARC (if we don't
assume this we don't have a multiverse or at least not one we can say
anything about. []p in this case means the value of p in A is the same as
its value in B and C (t or f). This also means that in A B and C, []p is
true, hence we can also say that in all worlds [][]p. (And indeed [][][]p
and so on?)
So it's true for the minimal case that []p - [][]p
But then adding more worlds will just give the same result in each set of
3... so does that prove it?
No, hang on. Take { A B C } with p having values { t t f }. []p is true in
C, because C is not connected to anywhere else, which makes it trivially
true if I remember correctly. But []p is false in A and B. So [][]p is
false, even though []p is true in C. So []p being true in C doesn't imply
[][]p.
So that seems to disprove it, because C is in its own little multiverse.
There's nothing in the definition that says ARB and BRC entails CRA or CRB,
is there?
Unless I have the trivially true thing wrong...
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Re: MODAL Last exercise
On 04 Mar 2014, at 01:18, LizR wrote:
OK, so ignoring Brent who I'm sure is way ahead of me...
The problem is to show that
(W, R) respects []A - A if and only if R is reflexive,
Where reflexive means for all alpha, { alpha R alpha } (and nothing
more is implied!)
OK.
(better not to use the accolade though, as you are just saying that
for all alpha, alpha R alpha (and then if you represent R by a set, R
will contain all couples like (alpha, alpha), so R = {(alpha, alpha),
(alpha beta), (beta beta), ... }.
And []p means that p is true in all worlds reachable from the world
being considered.
OK.
You could have said more precisely that
[]p means, in a world alpha, that p is true in all worlds reachable
from the world alpha.
(...I think. I just checked my diary and was told that []p means
that p is a law. Maybe that was the wrong page...)
Yes, that's was in the Leibniz semantics. Something similar will
happen with kripke, but if I explain now, it can be confusing.
OK, so anyway, before I get too confused
You should never allow this to happen. It happens because you allow
slight confusion, and then they add up. I know it is not easy.
let's consider world alpha which is part of W.
A part of W means usually a subset of W.
A world is an element of W.
If W = {a, b}, a and b are elements of W. The parts of W are { }, {a},
{b}, {a, b}. If W has n elements, we have seen that W has 2^n parts.
I hope you don't mind I help you to use the standard terminology, as
it will help us a lot later.
We know { alpha R alpha }.
?
At this stage I am not sure if you try to prove:
(W, R) respects []A - A - R is reflexive,
or
R is reflexive - (W, R) respects []A - A
I will have to guess. And here I guess you assume R is reflexive, and
so you intent to deduce from this that (W, R) respects []A - A.
[]p means p is true in all worlds reachable from alpha (I think)
which includes alpha itself, hence it means that p has to be true in
alpha, hence it means []p - p.
That's correct, but still a bit fuzzy. To say that (W, R) respects a
formula, like []A - A, means that (W,R, V) satisfies the formula, for
all valuations V.
May be I am just nitpicking, but what if p is not true in alpha. Do we
still have []p - p?
Your fuzziness, or perhaps my own imperfect brain, makes consistent
that you did treat that case, or that you did not.
(Conversely, if alpha wasn't reachable from itself, then p being
true in all worlds reachable from alpha wouldn't entail that p is
true in alpha.)
Very good. Just a bit lazy. When you say that p being true in all
worlds reachable from alpha wouldn't entail that p is true in alpha,
you might give us the counterexample, like chosing a valuation
(illumination) V with p false in alpha and true in all worlds
accessible from alpha.
We love counter-examples, you know.
QED, perhaps? Did I just prove something?
Yes. You did.
You have proved that
1) if R is reflexive, []A - A is automatically true in all worlds in
any reflexive illuminated multiverse.
And then, three lines above, beginning by Conversely, ... you have
proved that, conversely indeed:
2) ~ (R is reflexive) -~(For all V (W,R,V) respects []A -
A)) (by showing one V with a world in which []A - A is false,
when R is not reflexive).
If so, I'still m not sure that proves if any only if...
Oh! You might say that once you proved 2) you did prove that
2') (For all V (W,R,V) respects []A - A)) - (R is reflexive)
But you can derived P - Q from a derivation of ~Q - ~P. All
right.
You did it, and it would have been simpler for you, and for me, if you
just started from what you were asked to prove.
Although... maybe it does.
Sure it does.
For []p to imply p in a world alpha, where []p means p is true in
all worlds reachable from alpha, it can only imply p is true if
alpha is reachable from alpha.
This applies to all worlds in (W, R) hence it must be reflexive.
I think.
Good.
To prove that P - Q, you can prove that P ~Q leads to a
contradiction, or you can prove that ~Q leads to ~P.
But it helps a lot if you start from what you want to prove, up to the
conclusion, so that not only you prove it, but you know exactly what
you discovered. In this case a necessary link, in Kripke semantics,
between a binary relation (reflexivity) and a modal formula []A-A.
You learned that the fact that (W, R) respects []A - A is equivalent
with the fact that R is reflexive.
OK?
Bruno
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Re: MODAL Last exercise (+ a zest of the real thing)
On 04 Mar 2014, at 03:00, LizR wrote: Hm. I don't know if the first one was OK but anyway let's look at the second one. A Kripke multiverse (W, R) is said transitive if R is transitive. That is alpha R beta, and beta R gamma entails alpha R gamma, for all alpha beta and gamma in W. Show that (W, R) respects []A - [][]A if and only R is transitive, I think []A - [][]A means (for a world alpha in (W,R)) that if A is true in all worlds accessible from alpha, then it's true in all worlds reachable from alpha that A is true in all worlds reachable from alpha. I am not sure. []A - [][]A means, in a world alpha, in W, from (W,R), indeed, that []A - [][]A is true in alpha. So if []A is true in alpha, you know that [][]A is true in alpha, so that means that if A is true in all accessible worlds, then []A is true in all the accessible worlds. That's a bit - I don't know - recursive? I can feel a bit of boggling starting in my mind. Let's try to keep things (very, very) simple. No problem. Consider a world alpha in which p is true. I assume I can use p since I'm used to typing []p by now! And suppose we have beta and gamma as above. So []p implies that p is true in beta because alpha R beta... OK so far... Hang on, does transitive imply reflexive? This is hard to think about, having 3 things! For ALL a,b,c, in (W,R) we have (aRb bRc) - aRc. Specifically if a,b,c are the same (aRa aRa) - aRa, so we (kind of redundantly) get reflexivity too. I think. Well tried, but if (a Ra) is false, that is just f - f. Take a strict order relation like strictly less than, on N, or R, that relation is transitive, but not reflexive. Take less or equal, that relation is both reflexive and transitive. Strictly less than is even worse than not reflexive, it is irreflexive. For all a ~(aRa), or if you prefer ~ Exist a such that a R A. By the way, I suspect that the 3-fold nature of the transitivity rule somehow connects with the 3 []s in the thing I'm trying to prove! But I have no idea why or how that works, if it does. Maybe I should stop for a coffee break and let this percolate around my brain for a bit. Take the time. And don't worry, at some point I will have to re-explained all this, to what some people might take as a very dumb machine, which indeed believes only few axioms of elementary arithmetic. That will be the real things, some modal logics will impose themselves there, including the one corresponding to alternating consistent extensions. The theory of everything, here, is classical first order logic + the following formula: 0 ≠ s(x) s(x) = s(y) - x = y x+0 = x x+s(y) = s(x+y) x*0=0 x*s(y)=(x*y)+x An observer will be defined, in the theory above, by a sound extension believer of the axioms above, + some amount of induction axioms, of the type: (F(0) Ax(F(x) - F(s(x))) - AxF(x), with F(x) being a formula in the arithmetical language (with 0, s, +, *). We have to explain to a dumb machine, which understands only 0, s(0), s(s(0)), ... and can only add and multiply, but yet can reason in classical logic, the very functioning of such a dumb machine. There is no miracle. To define the variables, we can use the letter x, y, ..., it works well for many human people, but the dumb machine understands only 0, s(0), s(s(0)), so we will have to decide to say something like let the variable be defined by 0, s(s(0)), s(s(s(s(0, that is, the even number, so we will defined in arithmetic, the variable by the even numbers. Variable(x) - even(x) - Ey(2*y = x) And about , - t, and even what about (, and ) ? Well, again, there is no magic, you have to chose particular odd numbers (to not confuse them from variable) to represent them. That is both logic and polite. And then, how about finite sequences of symbols like 0≠s(x)? There too must be defined in terms of number relations, and in this case a simple way, if we allow ourselves the use of exponentiation, is given by the uniqueness of prime decomposition. If g(0), g(≠), ... represents the particular odd number symbol for 0, ≠, etc. then you can represent 0≠s(x) by 2^g(0)*3^g(≠)*5^g(s)*7^g(()*11^g(x)*13^g()). Then the theory itself can be defined or represented, as a number, being a finite sequences of the number corresponding to the axioms above. We will have to defined in arithmetic what we mean by a valid proof. A proof is itself a finite (or infinite) sequences of application of inference rules, making proof easy to check (and hard to find in many domains). So we can define in arithmetic a predicate b(x, y) true when y is a proof (in the dumb number language) of x. Then provable(x) can be defined by EyB(x, y). It is a Turing complete sigma_1 arithmetical predicate, a Löbian once it get few induction axioms. That provable(x), or believable(x), or assertable(x) by the modest believer in the axiom above,
Re: MODAL Last exercise
On 5 March 2014 04:18, Bruno Marchal marc...@ulb.ac.be wrote: Good. To prove that P - Q, you can prove that P ~Q leads to a contradiction, or you can prove that ~Q leads to ~P. But it helps a lot if you start from what you want to prove, up to the conclusion, so that not only you prove it, but you know exactly what you discovered. In this case a necessary link, in Kripke semantics, between a binary relation (reflexivity) and a modal formula []A-A. I had to get my head around ... well, everything ... again. So I may have sneaked up on the result. You learned that the fact that (W, R) respects []A - A is equivalent with the fact that R is reflexive. OK? OK. So, the next question was A Kripke multiverse (W, R) is said transitive if R is transitive. That is alpha R beta, and beta R gamma entails alpha R gamma, for all alpha beta and gamma in W. Show that (W, R) respects []A - [][]A if and only R is transitive, Damn. This looks too complicated for me to fake it! -- 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 Last exercise
On 03 Mar 2014, at 05:40, meekerdb wrote:
On 3/2/2014 9:57 AM, Bruno Marchal wrote:
Brent, Liz, others,
I sum up the main things, and give a lot of exercises, or
meditation subject.
Liz we can do them one at a time, even one halve. Ask questions if
the question asked seems unclear.
***
A Kripke frame, or multiverse, is a couple (W, R) with W a non
empty set of worlds, and R a binary relation of accessibility.
An illuminated, or valued, multiverse (W,R, V), is a Kripke
multiverse together with an assignment V of a truth value (0, or 1)
to each propositional letter for each world. We say that p is true
in that world, when V(p) = 1, for that world. If you want V is a
collection of functions V_alpha in {0, 1}, one for each world alpha.
***
Some class of multiverses will play some role.
A Kripke multiverse (W, R) is said reflexive if R is reflexive.
alpha R alpha, for all alpha in W.
A Kripke multiverse (W, R) is said transitive if R is transitive.
That is
alpha R beta, and beta R gamma entails alpha R gamma, for all alpha
beta and gamma in W.
A Kripke multiverse (W, R) is said symmetric if R is symmetric.
alpha R beta entails beta R alpha, for all alpha in W.
A Kripke multiverse (W, R) is said ideal if there are no cul-de-sac
worlds. For all alpha, there is beta such that alpha R beta.
A Kripke multiverse (W, R) is said realist if all non cul-de-sac
worlds can access to a cul-de-sac world.
***
Finally: (The key thing)
I say that a Kripke multiverse (W,R) respects a modal formula if
that formula is true in all worlds in W, and this for any valuation
V.
***
Show that
(W, R) respects []A - A if and only if R is reflexive,
R is reflexive implies (alpha R alpha) for all alpha. []A in alpha
implies A is true in all beta where (alpha R beta), which includes
the case beta=alpha. So R is reflexive implies (W,R) respects []A-A.
OK.
Assume (W,R) respects []A-A,
OK.
so that []A-A is true in all W.
Precisely: []A-A is true in all world in W, and this whatever the
valuation or illumination of the atomic sentences (p, q, r, ...), and
thus whatever is the truth value of A.
That means that every world has another R-accessible world and
whatever valuation any formula A has in the world, it has the same
valuation in the R-accessible world.
Hmm? I don't see how that implies R is reflexive, unless I can say
that any two worlds whose valuation is the same for every formula
are just the same world.
I let you think a bit more. May be another will have an insight. You
might try to reason by absurdum.
You must show that: [(W,R) respects []A-A] - R is reflexive.
(and I recall that respects means that []A - A is true, whatever A
is, and whatever his truth value is in any (W, R, V)).
Suppose that this is false.That means (definition of - actually)
that we have both (W, R) respects []A - A, *and* R is not reflexive.
But if R is not reflexive, there is some world alpha such that ~(alpha
R alpha). OK?
Consider all worlds beta so that alpha R beta, and consider the
valuation V where V(p) = true in all those worlds beta, and V(p) =
false in alpha.
In that case, []A is true in alpha, and A is false in alpha. OK?
That gives the counter-example, and we are done. The fact that R is
not reflexive makes it possible to find a valuation and a world such
that []A - A is false in that world, and so []A - A cannot be
respected by a non reflexive frame (multiverse).
Note that if the set of beta is empty, then []p is automatically true,
and V(p) = true will do. It is in fact taken into account above.
Oops, I have solved it. Are you OK? Liz?
If this is OK, try the next one, again in both direction, perhaps.
(W, R) respects []A - [][]A if and only R is transitive,
Bruno
Brent
(W, R) respects []A - [][]A if and only R is transitive,
(W, R) respects A - []A if and only R is symmetrical,
(W,R) respects []A - A if and only if R is ideal,
(W, R) respects A - ~[]A if and only if R is realist.
You can try to find small counter-examples, and guess the pattern
of what happens when you fail.
Of course proving that (W, R) respects []A - A if and only if R is
reflexive, consists in proving both
(W, R) respects []A - A if R is reflexive,
and
(W, R) respects []A - A only if R is reflexive, that is
R is reflexive if (W, R) respects []A - A
That's a lot of exercises. 10 exercises.
We can do them one at a time. Who propose a proof for
(W, R) respects []A - A if R is reflexive, That is:
R reflexive - (W, R) respects []A - A
?
Bruno
Oh! I forget this one:
Show that all the Kripke multiverses (W, R), whatever R is, respect
[](A - B) - ([]A - []B).
http://iridia.ulb.ac.be/~marchal/
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Re: MODAL Last exercise
OK, so ignoring Brent who I'm sure is way ahead of me...
The problem is to show that
(W, R) respects []A - A if and only if R is reflexive,
Where reflexive means for all alpha, { alpha R alpha } (and *nothing more *is
implied!)
And []p means that p is true in all worlds reachable from the world being
considered.
(...I think. I just checked my diary and was told that []p means that p is
a law. Maybe that was the wrong page...)
OK, so anyway, before I get too confused let's consider world alpha which
is part of W. We know { alpha R alpha }. []p means p is true in all worlds
reachable from alpha (I think) which includes alpha itself, hence it means
that p has to be true in alpha, hence it means []p - p. (Conversely, if
alpha wasn't reachable from itself, then p being true in all worlds
reachable from alpha *wouldn't* entail that p is true in alpha.)
QED, perhaps? Did I just prove something?
If so, I'still m not sure that proves if any only if...
Although... maybe it does. For []p to imply p in a world alpha, where []p
means p is true in all worlds reachable from alpha, it can *only* imply p
is true if alpha is reachable from alpha.
This applies to all worlds in (W, R) hence it must be reflexive.
I think.
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Re: MODAL Last exercise
Hm. I don't know if the first one was OK but anyway let's look at the second one. A Kripke multiverse (W, R) is said transitive if R is transitive. That is alpha R beta, and beta R gamma entails alpha R gamma, for all alpha beta and gamma in W. Show that (W, R) respects []A - [][]A if and only R is transitive, I think []A - [][]A means (for a world alpha in (W,R)) that if A is true in all worlds accessible from alpha, then it's true in all worlds reachable from alpha that A is true in all worlds reachable from alpha. That's a bit - I don't know - recursive? I can feel a bit of boggling starting in my mind. Let's try to keep things (very, very) simple. Consider a world alpha in which p is true. I assume I can use p since I'm used to typing []p by now! And suppose we have beta and gamma as above. So []p implies that p is true in beta because alpha R beta... OK so far... Hang on, does transitive imply reflexive? This is hard to think about, having 3 things! For ALL a,b,c, in (W,R) we have (aRb bRc) - aRc. Specifically if a,b,c are the same (aRa aRa) - aRa, so we (kind of redundantly) get reflexivity too. I think. By the way, I suspect that the 3-fold nature of the transitivity rule somehow connects with the 3 []s in the thing I'm trying to prove! But I have no idea why or how that works, if it does. Maybe I should stop for a coffee break and let this percolate around my brain for a bit. -- 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 Last exercise
Brent, Liz, others,
I sum up the main things, and give a lot of exercises, or meditation
subject.
Liz we can do them one at a time, even one halve. Ask questions if the
question asked seems unclear.
***
A Kripke frame, or multiverse, is a couple (W, R) with W a non empty
set of worlds, and R a binary relation of accessibility.
An illuminated, or valued, multiverse (W,R, V), is a Kripke multiverse
together with an assignment V of a truth value (0, or 1) to each
propositional letter for each world. We say that p is true in that
world, when V(p) = 1, for that world. If you want V is a collection of
functions V_alpha in {0, 1}, one for each world alpha.
***
Some class of multiverses will play some role.
A Kripke multiverse (W, R) is said reflexive if R is reflexive. alpha
R alpha, for all alpha in W.
A Kripke multiverse (W, R) is said transitive if R is transitive. That
is
alpha R beta, and beta R gamma entails alpha R gamma, for all alpha
beta and gamma in W.
A Kripke multiverse (W, R) is said symmetric if R is symmetric. alpha
R beta entails beta R alpha, for all alpha in W.
A Kripke multiverse (W, R) is said ideal if there are no cul-de-sac
worlds. For all alpha, there is beta such that alpha R beta.
A Kripke multiverse (W, R) is said realist if all non cul-de-sac
worlds can access to a cul-de-sac world.
***
Finally: (The key thing)
I say that a Kripke multiverse (W,R) respects a modal formula if that
formula is true in all worlds in W, and this for any valuation V.
***
Show that
(W, R) respects []A - A if and only if R is reflexive,
(W, R) respects []A - [][]A if and only R is transitive,
(W, R) respects A - []A if and only R is symmetrical,
(W,R) respects []A - A if and only if R is ideal,
(W, R) respects A - ~[]A if and only if R is realist.
You can try to find small counter-examples, and guess the pattern of
what happens when you fail.
Of course proving that (W, R) respects []A - A if and only if R is
reflexive, consists in proving both
(W, R) respects []A - A if R is reflexive,
and
(W, R) respects []A - A only if R is reflexive, that is
R is reflexive if (W, R) respects []A - A
That's a lot of exercises. 10 exercises.
We can do them one at a time. Who propose a proof for
(W, R) respects []A - A if R is reflexive, That is:
R reflexive - (W, R) respects []A - A
?
Bruno
Oh! I forget this one:
Show that all the Kripke multiverses (W, R), whatever R is, respect []
(A - B) - ([]A - []B).
http://iridia.ulb.ac.be/~marchal/
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Re: MODAL Last exercise
So does reflexive (alpha R alpha) mean that all universes are *only*accessible to themselves, or does it mean that all universes are accessible to themselves and possibly, but not necessarily, to each other? -- 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 Last exercise
On 3/2/2014 9:57 AM, Bruno Marchal wrote:
Brent, Liz, others,
I sum up the main things, and give a lot of exercises, or meditation subject.
Liz we can do them one at a time, even one halve. Ask questions if the question asked
seems unclear.
***
A Kripke frame, or multiverse, is a couple (W, R) with W a non empty set of worlds, and
R a binary relation of accessibility.
An illuminated, or valued, multiverse (W,R, V), is a Kripke multiverse together with an
assignment V of a truth value (0, or 1) to each propositional letter for each world. We
say that p is true in that world, when V(p) = 1, for that world. If you want V is a
collection of functions V_alpha in {0, 1}, one for each world alpha.
***
Some class of multiverses will play some role.
A Kripke multiverse (W, R) is said reflexive if R is reflexive. alpha R alpha, for all
alpha in W.
A Kripke multiverse (W, R) is said transitive if R is transitive. That is
alpha R beta, and beta R gamma entails alpha R gamma, for all alpha beta and
gamma in W.
A Kripke multiverse (W, R) is said symmetric if R is symmetric. alpha R beta entails
beta R alpha, for all alpha in W.
A Kripke multiverse (W, R) is said ideal if there are no cul-de-sac worlds. For all
alpha, there is beta such that alpha R beta.
A Kripke multiverse (W, R) is said realist if all non cul-de-sac worlds can access to a
cul-de-sac world.
***
Finally: (The key thing)
*I say that a Kripke multiverse (W,R) respects a modal formula if that formula is true
in all worlds in W, and this for any valuation V.*
***
Show that
(W, R) respects []A - A if and only if R is reflexive,
R is reflexive implies (alpha R alpha) for all alpha. []A in alpha implies A is true in
all beta where (alpha R beta), which includes the case beta=alpha. So R is reflexive
implies (W,R) respects []A-A.
Assume (W,R) respects []A-A, so that []A-A is true in all W. That means that every
world has another R-accessible world and whatever valuation any formula A has in the
world, it has the same valuation in the R-accessible world. Hmm? I don't see how that
implies R is reflexive, unless I can say that any two worlds whose valuation is the same
for every formula are just the same world.
Brent
(W, R) respects []A - [][]A if and only R is transitive,
(W, R) respects A - []A if and only R is symmetrical,
(W,R) respects []A - A if and only if R is ideal,
(W, R) respects A - ~[]A if and only if R is realist.
You can try to find small counter-examples, and guess the pattern of what happens when
you fail.
Of course proving that (W, R) respects []A - A if and only if R is reflexive, consists
in proving both
(W, R) respects []A - A if R is reflexive,
and
(W, R) respects []A - A only if R is reflexive, that is
R is reflexive if (W, R) respects []A - A
That's a lot of exercises. 10 exercises.
We can do them one at a time. Who propose a proof for
(W, R) respects []A - A if R is reflexive, That is:
R reflexive - (W, R) respects []A - A
?
Bruno
Oh! I forget this one:
Show that all the Kripke multiverses (W, R), whatever R is, respect [](A - B) - ([]A
- []B).
http://iridia.ulb.ac.be/~marchal/ http://iridia.ulb.ac.be/%7Emarchal/
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Re: MODAL Last exercise
On 03 Mar 2014, at 04:55, LizR wrote: So does reflexive (alpha R alpha) mean that all universes are only accessible to themselves, or does it mean that all universes are accessible to themselves and possibly, but not necessarily, to each other? Good question. Mathematician are literalist, so when we say: A Kripke multiverse (W, R) is said reflexive if R is reflexive. alpha R alpha, for all alpha in W. It means that for all alpha, we have alpha R alpha. And you cannot deduce from this, that ~(alpha R beta) when alpha R beta. For example, a (W, R) can be both reflexive and transitive. You can also derive that if R is reflexive then R is ideal. If R is reflexive, there is no cul-de-sac, as all worlds can access *at least one world*, itself. OK? 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.
