On 19 February 2014 23:00, Bruno Marchal <[email protected]> wrote:
> Liz, Others,
>
> I was waiting for you to answer the last questions to proceed. Any problem?
>
> I give the correction of the last exercise.
>
>
> On 14 Feb 2014, at 19:18, Bruno Marchal wrote:
>
> <snip>
>
>
> On 13 Feb 2014, at 22:23, LizR wrote:
>
> On 14 February 2014 07:49, Bruno Marchal <[email protected]> wrote:
>
>> Liz, and others,
>>
>>
>> On 13 Feb 2014, at 10:04, LizR wrote:
>>
>
>>
> Well, we get { p=t } and { p=f } regardless of the accessibility
> relations. (If that's how you write it)
>
>
> Well ... OK.
>
> More precisely we get
>
> 1) {alpha}, with R = { } with p=t in alpha
> 2) {alpha}, with R = { } with p=f in alpha
>
> and
>
> 3) {alpha}, with R = {(alpha R alpha) } with p=t in alpha
> 4) {alpha}, with R = {(alpha R alpha) } with p=f in alpha
>
>
>
>
>
>
>> Which of those propositions are true of false in alpha, in the
>> illuminated simplest multiverses.
>> And which one are law (meaning true in all worlds, but true for all
>> valuation of p, that is valid with A = p, but also with A = ~p)
>>
>> 1) []A -> A
>>
>
> This is true in alpha R alpha (because it's just a Leibniz type world)
>
>
> Very good. It is a law there.
>
>
> . []p is "vacuously true" in "alpha" (the disconnected multiverse) - as
> you said above - so []p -> p is false, because []p is true regardless of p.
>
>
> OK. It is not a law, but it might still be true in some circumstances.
> Give me which among 1), 2), 3), 4), above.
>
> Any problem with this?
>
OK, let me see. []A -> A is true in 3 and 4, I think we established that.
So it depends on []p in the disconnected multiverses...
[]p is true if p is true in all worlds connected to alpha. Or to put it
another way, it isn't true that there are worlds connected to alpha in
which p is false, which I think would be ~<>~p. Oh dear. I'm afraid I can't
get my head around this again. If no worlds are accessible, then it seems
that p can't be true or false in all worlds accessible from alpha! I think
I have missed the point somehow. So I think yes there is a problem with
this.
> 2) []A -> [][]A
>>
>
> This is true in alpha R alpha, and in alpha I guess it's true too, because
> vacuously true implies vacuously true?
>
> Exact.
>
> 3) <>A -> []<>A
>>
>
> true in alpha R alpha again, because there's only one world to consider so
> <>A is equivalent to []A in this case (isn't it?)
>
> Well seen!
>
> not true in alpha because []<>A is vacuously true regardless of <>A - I
> think
>
> Not correct. You jump to hastily.
>
> in your language the answer is:
>
> true in alpha, because []<>A is vacuously true, so that <>A -> []<>A is
> vacuously true too (as "p -> q" is false only if q is false and p is true).
> The type of <>A -> []<>A is really f -> t, which is as much tautological
> than f -> A, and A -> f, for any A.
>
> Are you OK with this?
>
> OK, let me see. <>A -> []<>A in alpha. You say []<>A is vacuously true, I
think we established that []X is true for all X in {alpha}, so ... I think
whatever the left hand side is, it can imply true. If left side is false,
then right side is true - if I have a pet unicorn then I am the Queen of
France. And if left side is true that can also imply true...isn't it???
>
>
>> 4) []A -> <>A
>>
>
> Well I think this is true for reason given above.
>
>
> You begin to try to go to quickly. I have some doubt that []A -> <>A can
> be a law in a cul-de-sac world, like poor alpha, with R = { }.
>
> OK?
>
Yes I probably go too quickly. I find that I have no time and so I rush
through, and it turns out there are a lot more questions than I thought...
(Then at the end you say "one at a time!" :)
So []A is always true in alpha, but t -> f isn't true, so is <>A false in
alpha for some A? I'm not sure, it's quite hard to think about this for a
disconnected world. <>A is "there is some world connected to alpha in which
A is true" but there are no worlds connected to alpha, so...help! How does
one think about this?
5)A -> []<>A
>>
>
> True in alpha R alpha.
>
> OK.
>
> In alpha not true because []<>A is always true and A isn't
>
> Not a law. OK. Again there are case where it is true, like when A is true.
>
> Ah, so we're back to X -> t being true in all cases of X.
So in fact X -> []A is always true in {alpha}
>
>
>> 6) <>A -> ~[]<>A
>>
>
> False in alpha R alpha, surely? With one world, <>A -> []<>A (above)
>
> Correct.
>
> Phew :-)
> Not true in alpha because ~{}<>A is vacuously false regardless of <>A
>
> OK, this seems wrong, looking ahead so...
~[]X is f is {alpha}
So if <>A can be t we get t -> f which is itself false.
This comes back to how <>A works in {alpha} again, which I'm not sure
about. If say <>A is the same as A this would give t -> f in one case, and
that's false.
> Unfortunately as much as ~{}<>A is vacuously false regardless of <>A, as
> you say, we are interested in
> <>A -> ~[]<>A and in poor alpha (case 1) 2)) <>A is *also vacuously
> false, so that we are in the f -> f, case, which is, vacuously or not,
> always true.
>
> Are you OK with this?
>
> OK, if <>A is vacuously false, that is the key thing I've been missing.
Maybe you can explain that again.
> Keep in mind that in CPL both
>
> f -> A
>
> and
>
> A -> t
>
> are always tautologies. They are true in all worlds, whatever A is.
>
> I didn't realise the second one was true, or maybe I did, because of
course f -> A implies f -> t is true, which only leaves t -> t which is
also true!
OK, maybe I knew that.
>
>
>> 7) []([]A -> A) -> []A
>>
>
> True in alpha R alpha I think.
>
> A law? True in both 3), 4) ?
>
> And what about alpha (case 1 and 2)?
>
Well in {alpha} this is t -> t
>
> Let us look in alpha R alpha (case 3 and 4 above):
>
In {alpha R alpha} []A -> A is a law, so []([]A -> A) is true. But that
doesn't imply []A (I think) because A could be false.
>
> in W = {alpha}, with alpha R alpha, and with V(p) = t (V = the valuation
> or illumination):
>
> We have p is true in alpha, and p is true in all worlds accessed to alpha.
> OK?
> So, []p is true, and A -> []p is true, whatever A is, so []([]p -> p) ->
> []p is true.
>
> What if A = ~p, in []([]A -> A) -> []A? (that is really case 4)
>
> In that case the right hand side is []A = [] ~p, and is false. In the left
> hand side, A is false, but []A is false too, so "[]A -> A" is true (f -> f
> is true), and thus, it is also true in all worlds accessed from alpha, and
> thus we have that []([]A -> A) is true, and so []([]A -> A) -> []A is of
> the type t -> f, and so is false, and so []([]A -> A) -> []A is NOT a law.
> Same reasoning in the "4" case, with p exchanged with ~p.
>
> Conclusion: []([]A -> A) -> []A is NOT a law in the little reflexive
> (alpha R alpha) multiverse.
>
> OK?
>
Yes. I even managed to work that out, too :-)
> And vacuously true in alpha because both sides of the rightmost -> have to
> be true.
>
> Correct. It is law. True in 1) 2).
>
> Not sure if that means it's implied though...
>
> I am not sure what you are asking here.
>
I'm not sure either, now.
>
>
>> 8) []([](A -> []A) -> A) -> A
>>
>
> Not true in alpha because to the left of rightmost -> is vacuously true
> regarldess of A.
>
> OK. Not a law.
>
> Precisely, if A is false, []([](A -> []A) -> A) is still vacuously true,
> so we get T -> f, which is false.
>
OK.
>
> Don't know about alpha R alpha because my head exploded...
>
> Take a break. I said "one a at a time" !
>
> OK. I do it.
>
> Consider []([](A -> []A) -> A) -> A with A true (for example A = p in
> case p is true, or equivalently A = ~p with p false).
>
> In that case we have something like # -> t, but that is always true, so
> []([](A
> -> []A) -> A) -> A is true.
>
> So, the less easy case, consider []([](A -> []A) -> A) -> A with A false.
>
> So we have <something> -> f, and this can be true only if <something> is
> false.
>
> What about the inner A -> []A ? A is false, so A->[]A is true (f -> # is
> always true).
>
> But if A -> []A is true in "alpha R alpha", then A -> []A is true in all
> worlds accessible from alpha, and so [](A->[]A) is true too. We get
> something like [](t -> f) -> A; with A = f. (I have substituted [](A-> []A)
> by t, in the initial formula. But t -> f is false, and thus [](t->f) is
> false too. OK? So the initial Grz formula is of the type f -> f, and thus
> it is true! So []([](A -> []A) -> A) -> A is true, with A false and A
> true, and it is a law for the 3 and 4 case.
>
> OK?
>
> Yes.
>
>
>
>
>
>
>
>
>> Let me solve one case, to illustrate. Let us look at []A -> <>A, in the
>> illuminated multiverse {alpha}, with empty R, and with p true in alpha.
>> Well, what about []A?
>> Alpha is a cul-de-sac world, so we have seen that []A must be true (if
>> not <>~A has to be true), so []A is true and in particular []p is true
>> (whatever the value of p is). What about <>A ? well this say that there is
>> some beta accessible from alpha in which A is true. But alpha is a
>> cul-de-sac world, so there is no such world, and so <>A is false, whatever
>> A is (notably p or ~p). So []A is always true, and <>A is always false, so
>> []A -> <>A is always false (by CPL), whatever illumination is chosen (p or
>> ~p true at alpha).
>>
>> OK?
>>
>
> OK
>
>>
>> It is just CPL used in the world alpha, with the value of "[]A"
>> determined by the Kripke semantics.
>>
>> Can you see the truth value of the 7 other formula in those simplest
>> multiverses. I think it is a good training, and a good way to demystify the
>> difficulties. Tell me what. You can do one formula at a time, in 8 posts.
>>
>
> I did them all (or more likely I didn't) before I got to this
> point..........see above..........
>
>
> Yeah, I should have said "one at a time" earlier :)
>
> Very good work.
>
> I give you more time for Grzegorczyk []([](A -> []A) -> A) -> A
>
> And just one supplementary exercise. What can be said about
>
> [](A -> B) -> ([]A -> []B), in those simple multiworlds structure? This
> formula has two (meta) variables A and B.
>
> Is that a law?
>
>
> I let you think a bit more on this before providing the solution. Tell me
> if you are OK with the correction.
>
> You solved correctly 6 cases on 8. I think that your head go near
> explosion, because you still panic in front of "too much symbols", but I
> hope having shown here that it is just a matter at looking what happens,
> beginning by the sub-formula, in each world, for each valuation. You must
> keep in mind the truth table of the implication too. The tautologies f -> #
> and # -> t are very handy here.
>
> Too many symbols, and I turn into the kid in "The Waves" who saw them all
as weird marks with no meaning.
> OK?
>
> About [](A -> B) -> ([]A -> []B), let me ask you a more precise exercise.
>
> Convince yourself that this formula is true in all worlds, of all Kripke
> multiverses, with any illumination.
> Hint: you might try a reductio ad absurdum. try to build a multiverse in
> which that law would be violated.
>
[](A -> B) -> ([]A -> []B)
OK. For a disconnected universe this is t -> (t -> t) or t -> t which is
true.
And for a Leibniz universe, I'm fairly sure this is also true.
So that leaves {alpha R alpha} and {alpha R beta} and .... so on, for any
number of universes + relations.
Maybe I can come back on this one.
> Don't hesitate also to test your memory on exercises that you have already
> been able to solve, just to be sure you memorize correctly the definition.
>
> Hope you enjoy this a little bit and that it is not too much technical. We
> are at the half of the modal prerequisites for AUDA, except that the real
> thing must still be done, but it is not modal logic per se, it is the
> provability logic. I will not give all details (as that would be very long
> and highly technical, but i will give enough so that you get the idea).
> Then I will be able to explain Solovay, and how to extract the 8
> hypostases, including physics (and why) from there. (That was the question!)
>
> We might need to come back also on UDA step seven, if only to understand
> in which sense AUDA "translated" it in arithmetic, and to add some light on
> it, as it seems we have to do (given some recent posts).
>
> Bon courage!
>
> Bruno
>
>
>
> http://iridia.ulb.ac.be/~marchal/
>
>
>
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