On 8 February 2014 08:43, Bruno Marchal <[email protected]> wrote:
>
> On 07 Feb 2014, at 02:29, LizR wrote:
>
> On 7 February 2014 09:14, Bruno Marchal <[email protected]> wrote:
>
>>
>> On 06 Feb 2014, at 07:39, LizR wrote:
>>
>> On 6 February 2014 08:25, Bruno Marchal <[email protected]> wrote:
>>
>>>
>>>
>>> Which among the next symbolic expressions is the one being a well formed
>>> formula:
>>>
>>> ((p -> q) -> ((p& (p V r)) -> q))
>>>
>>> ))(p-)##à89-< a -> q)
>>>
>>> OK?
>>>
>>
>> I sure hope so.
>>
>> Well, I will pray a little bit.
>>
>>
>>> (to be sure the irst one might contain a typo, but I assure you there
>>> are no typo in the second one (and there is no cat walking on the keyboard).
>>>
>>> ***
>>>
>>> Then a set of worlds get alive when each proposition (p, q, r), in each
>>> world get some truth value, t, or f. I will say that the mutiverse is
>>> illuminated.
>>>
>>> And we can decide to put f and t is the propositional symbol for the
>>> boolean constant true and false.
>>> (meaning that "p -> f" is a proposition, or well formed formula).
>>>
>>> In modal logic it is often simpler to use only the connector "->" and
>>> that if possible if you have the constant f.
>>>
>>> For example you can define ~p as an abbreviation for (p -> f), as you
>>> should see by doing a truth table. OK?
>>>
>>
>> p -> f is (~p V f), for which the truth table is indeed the same as ~p
>>
>> OK.
>>
>>
>>> (Can you define "&", "V", with "->" and "f" in the same way? This is not
>>> an exercise, just a question!).
>>>
>>
>> I don't think I can define those *literally* with p, -> and f if that's
>> what you mean.
>>
>>
>> That is what I mean, indeed.
>>
>> OK, having had a look at what you say below, let's have another go. Start
> from p -> q being equivalent to (~p V q)
>
> That gives us ~p -> q equiv (p V q) and from the above ~p is (p -> f) so p
> V q is (p -> f) -> q which I seem to remember is what you got. OK so far.
>
> p & q --- well, p -> q is ~(p & ~q), so ~(p -> q) = (p & ~q) and ~(p ->
> ~q) = (p & q)
>
> so ~(p -> (q -> f)) which I guess is ((p -> ( q -> f)) -> f) = (p & q)
>
> Does it?!?! Looking below, I see that it does. Wow.
>
> I knew you can do that.
>
With hints.
>
> But that doesn't make sense, because & requires two arguments, so it would
>> have to be something like ... well, p -> q is (~p V q) and it's also ~(p &
>> ~q), which contain V and & ... I'm not sure I know what you mean.
>>
>>
>> Like for "~", to define "&" and "V" to a machine which knows only "->"
>> and "f". You can use the "~", as you have alredy see that you can define
>> it with "->" and "f".
>>
>> I reason aloud. Please tell me if you understand.
>>
>> First we know that "p -> q" is just "~p V q", OK?
>>
>> So the "V" looks already close to "->". Except that instead of "~p V q"
>> (which is p -> q) we want "p V q".
>>
>> May be we can substitute just p by ~p: and p V q might be then ~p -> q,
>>
>> Well, you can do the truth table of ~p -> q, and see that it is the same
>> as p V q.
>>
>> To finish it of course, we can eliminate the "~", and we have that p V q
>> is entirely defined by (p -> f) -> q.
>>
>> OK?
>>
>> And the "&":
>>
>> Well, we already know a relationship between the "&" and the "V", OK? The
>> De Morgan relations.
>>
>> So, applying the de Morgan relation, p & q is the same as ~(~p V ~q),
>> (the same "logically", not pragmatically, of course).
>>
>> That solves the problem.
>>
>> But we can verify, perhaps simplify. We can eliminate the "V" by the
>> definition above (A V B = ~A -> B),
>> ~(~p V ~q) becomes ~(~~p -> ~q), that is ~(p -> ~q). Or, to really settle
>> the things, and define & from -> and f:
>> p & q = ((p -> (q -> f)) -> f).
>>
>> OK?
>>
>
> Apparently, yes.
>
> OK. (Not sure what you mean by "apparently", though).
>
Well, even though I did it, the result still looks rather strange to me!
>
>>> Each world, once "illuminated" (that is once each proposition letter has
>>> a value f or t) inherits of the semantics of classical proposition logic.
>>>
>>> This means that if p and q are true in some world alpha, then (p & q) is
>>> true in that world alpha, etc.
>>> in particular all tautologies, or propositional laws, is true in all
>>> illuminated multiverse, and this for all illuminations (that for all
>>> possible assignment of truth value to the world).
>>>
>>> OK?
>>>
>>> Question: If the multiverse is the set {a, b}, how many illuminated
>>> multiverses can we get?
>>>
>>
>> I suppose 4, since we have a world with 2 propositions, and each can be t
>> or f?
>>
>>>
>>> Answer: there is three letters p, q, r, leading to eight valuations
>>> possible in a, and the same in b, making a total of 64 valuations, if I am
>>> not too much distracted. I go quick. This is just to test if you get the
>>> precise meanings.
>>>
>>
>> Oh, OK. So a and b are worlds, not ... sorry. I see.
>>
>>
>> Good.
>>
>>
>> So that is 2^3 x 2^3 because a has p,q,r = 3 values, all t or f, as does
>> b. OK now I see what you meant.
>>
>>
>> OK.
>>
>>
>>
>>> Of course with the infinite alphabet {p, q, r, p1, q1, r1, p2, ... } we
>>> already have a continuum of multiverses.
>>>
>>
>> I can't quite see why it's a continuum. Each world has a countable
>> infinity of letters, and the number of worlds is therefore 2 ^ countable
>> infinity! Is that a continuum?
>>
>>
>> Yes. We proved it, Liz.
>>
>
> Yes I had a sneaky suspicion we did. It seems familiar ... a bit.
>
> Understanding is good.
> Understanding and memorizing, even with the help of a well presented
> diary, is better, as it saves the future possible works.
>
I agree. I'm sure I started one, too, but I can't find it now. (So
sometimes I have to treat you as my diary...)
> Take a the infinite propositional symbol letters {p, q, r, p1, q1, r1, p2,
>> ... } . They are well ordered. So a sequence of 1 and 0 (other common name
>> for t and f) can be interpreted as being a valuation. The valuation are the
>> infinite sequences of 1 and 0. Or the function from N to {0, 1}.
>>
>> If such a set of function was in bijection with N, i -> f_i, the function
>> g defined by g(n) = f_n(n) + 1 would be a function f_i, let us sat f_k, and
>> f_k, applied on k, would gives both f_k(k) + 1 and f_k(k), and be well
>> defined, making 0 = 1.
>>
>> OK. I think.
>
> Hmm... OK. I think. For now. (That was quick).
>
> I meant it's clear once you assign values to them and make binary strings
that they can be diagonalised. And I remember the above proof, at least in
outline.
>
>> My transfinite maths may not be quite up to that one.
>>
>>
>> The infinite sequence of 0, and 1, if you put "0." at the front, you get
>>
>> 0.000011011111000011101101011111111000...
>>
>> for all sequences of 0 and 1, that is you get the real numbers, written
>> in binary, belonging to the interval (0, 1].
>>
>> That is the continuum. 2^aleph_0.
>>
>
> Ah yes because you can diagonalise it.
>
> I can diagonalize the bijection between N and 2^N, which send i on f_i.
>
>
>
>
>>> Well, that was Leibniz sort of multiverse, with all worlds quite
>>> independent of each other.
>>> With Kripke, we introduce a binary relation R on the set of world.
>>> That's all. We read alpha R beta, as beta is accessible from alpha.
>>>
>>
>> This seems like a way of getting subsets from the multiverse, but I'm not
>> completely sure what accessible means here.
>>
>>
>> I give you a concrete example. I take the set {a, b, c, d, e}, as the set
>> of worlds. That is already enough to get a Leibnizian multiverse, which for
>> all valuations (illuminations) of its propositional letters, in all worlds,
>> obeys classical logic, extended by the modal laws:
>>
>> [](A->B) -> ([]A -> []B)
>> []A -> [][]A
>> <>A -> []<>A
>> []A -> <>A
>> p -> []<>A
>>
>> As you have verified OK?
>>
>> It is nice, but we want modal logic in which those modal laws are
>> independent, semantically, and later deductively.
>>
>> So Kripke added a relation of accessibility between the worlds. And the
>> Kripke multiverse are any set (of "worlds") having a binary relation.
>>
>> So, to get a Kripke multiverse, I add, or not (!) a binary relation.
>> Written R. (but we could use another name).
>>
>> So the following are example of kripke multiverse:
>>
>> 1) the set {a, b, c, d, e}, together with the complete relation: all
>> worlds can access to all worlds. We will get xRy for all worlds x y in {a,
>> b, c, d, e}.
>>
>> 2) the set {a, b, c, d, e}, together with the empty relation, no world
>> can access to any other worlds, nor even themselves. A dust of cul-de-sac
>> world!
>>
>> 3) the set {a, b, c, d, e} with the relation aRb, cRc, and dRe. And that
>> all.
>>
>> 4) any relation you want about: well good exercise: how many binary
>> relations can we defined on {a, b, c, d, e} ?
>>
>
> I would guess 2 x 5^5 in this case? Because each one can have a relation
> with each other one (or not). Oops, I see that was way out. You can link
> each one to each other one, or not, so that's where the 2^ comes from. That
> makes sense, I made a similar argument about my hard drive in another
> thread. So we are looking for all the numbers that can be formed from aRa,
> aRb, aRc... hmm.
>
>>
>> Answer: A relation on {a, b, c, d, e} is really just a part, or a
>> subset, of the cartesian product
>>
>> {a, b, c, d, e} X {a, b, c, d, e},
>>
>> And a subset is really (modulo natural bijection) a function from that
>> set in the set {0, 1},
>> so we have 2^(5*5), that is 33554432 Kripke multiverse (non illuminated!).
>>
>> So there are 5*5 because ... each one can link to each other one and
> itself. I somehow put that as 5^5, which definitely isn't right, I almost
> got that bit right too, because I first thought 5! and then thought, no,
> the link to yourself makes it ... 5*5, of course, not 5^5. The world of
> maths kicks back!
>
>
> Yeah! That's its charm.
>
> Some would disagree...
>
>
>
>> A Leibniz multiverse is just a set.
>> A Kripke multiverse is just a set with a binary relations.
>>
>> The set of reals, with the relation x < y, is also an example of kripke
>> multiverse. Using R instead of {a, b, c, d, e}.
>>
>> The goal of the logician is not to explore the mutiverses, but to find
>> couterexamples to invalidate reasoning, notably here, modal reasoning.
>>
>>
>>
>>> OK. Time for the main recall:
>>>
>>> We add then new unary connector "[]", and define <> by ~[]~
>>>
>>> In Leibniz semantics, []A is true (*absolutely*) means that A is true
>>> in all worlds.
>>>
>>
>> The lines above are the important line.
>>
>> And even more important is the passage from the line above to the line
>> below
>>
>>
>>> In Kripke semantics []A is true *in a world alpha* means that A is true
>>> in all worlds *accessible from alpha*.
>>>
>>
>> If you remember this, there should be no problem.
>>
>>
>>
>>
>>
>>
>>> And the only one exercise:
>>>
>>> prove that "[]A -> A" is true in all worlds of a multiverse, for all
>>> illumination possible (choice of valuation for the letter)
>>>
>>
>> So []A means the proposition A is true in all worlds accessible from ...
>> somewhere.
>>
>>
>> Yes and no. []A has no more meaning at all.
>>
>> In Kripke semantic all statements are relativized to the world you are
>> in. []A can be true in some world and false in another. The meaning of "[]"
>> is restricted, for each world, to the world they can access (through the
>> accessibility relation available in the Kripke multiverse).
>>
>> []A still keep a meaning, but only in each world. So everything is said
>> when we define the new meaning of "[]" by the rule
>>
>> []A is true in alpha, by definition, means that A is true in all world
>> beta *accessible* from alpha.
>>
>> And
>>
>> <>A is true in alpha iff there is a world beta; where A is true,
>> accessible from alpha.
>>
>> OK. That makes sense, but I'm not sure I can use that fact to work things
> out...
>
>
>
> Understanding is good.
>
> But understanding + familiarity is better, and that comes with *some*
> practice.
>
Yes, I know. But you are trying to teach an old dog new tricks (as we say).
I have recently learned how cryptic crosswords work and (maybe not quite so
recently) the craft of novel writing. Plus my work requires me to learn new
things, at least if I want to live in New Zealand (because NZ is too small
to have the exact jobs I'm good at). And I have 2 kids and a husband to
look after. So I am trying to fit in some logic as well ... luckily I have
some spare time at work quite often...
>
>> Oh dear. I don't seem to be able to get my head around this.
>>
>>
>> That happens. Tell me if the explanations above help. Ask any question.
>>
>
> Well it does seem to make sense now. The binary relations help, I didn't
> really get that.
>
>
> I am a bit aware of that. But it might be because I go quick.
>
> But I might go quick because yo don't ask enough question, also.
>
That will be when I don't have time to really engage. I can do all this,
but only when I concentrate on it - it doesn't come naturally.
Now I have to go and make breakfast :)
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