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.
> 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.
>
>> 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.
>
> 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.
>
>
>
>
> 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.
>
>> 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!
>
> 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...
>
>
>
>
> 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.
> Maybe because I'm not sure what accessible means here...
>
> Just keep in mind the definition: a Kripke muliverse is just a set with a
> binary relation (any one you like a priori (the goal will be to find
> counter-example, notably to Leibinizian laws).
>
>
> I let you breath, and will come back, if you don't mind, I am aware I go
> quick.
>
> Do extremely simple exercise, build simple multiverse, with few worlds,
> and simple relations accessibility, then illuminate them by assignment of t
> or f (1 or 0) to p, and q, in those world, evaluate the formula, by using
> classical logic, and the "new" definition of the meaning of "[]".
>
OK so I suppose I have a multiverse of 2 worlds a and b { a b }
So that gives me ... I think 16 ways in which they can be accessible, from
completely disconnected through to [ aRb aRa bRb bRa ]
So then we have two propositions which are defined in a and b, p and q.
So suppose p is t in a
and we have bRa but not aRb
then []p -> p doesn't hold because p can be t or f in b.
????????????
Do that with your son!
>
> Can you find counter-example to the Leibnizian laws?
>
> This is not an exercise, just a remind of Kripke's goal. To contradict all
> (well not all) leibnizian quasi "obvious" modal tautologies, and refined
> considerably the field and the use of modal logical systems (something
> known by the deductive approach, and older more topological or algebraical
> semantics).
>
> Next post: *the discovery of a Kripke multiverse violating the Leibnizian
> law []A -> A*. (tataaaah... :)
>
> Bruno
>
>
>
>
>
>
>>> iff the relation is reflexive (that is: all world can access themselves).
>>
>> Hint: this should be easy. Any difficulty here is due to my probable
>> unclarity, or my excess of verbosity, or a lack of familiarity with math of
>> your part. I suggest you might search for counterexample.
>> And yes, this is truly two exercises, because to prove an iff, you have
>> to prove two if. You must prove:
>>
>> 1) if a multiverse is reflexive, then, *whatever the illumination* is,
>> each world satisfy []A -> A (for all formula A).
>> 2) If, *whatever the illumination* is, each world satisfy []A -> A (for
>> all formula A), then the multiverse is reflexive.
>>
>> "*whatever the illumination*" is important: for example in the simple
>> multiverse with one world: {alpha}, and the empty accessibility relation
>> (so that alpha does not access to itself, ~ (alpha R alpha), and with p
>> valuated to 1 in alpha, you have that []p is true, p is true, so []p -> p
>> is true in alpha, yet the mutiverse is not reflexive.
>>
>> OK?
>>
>> Please, ask any question to clarify. Note in passing the beauty: a modal
>> formula, made into a law, impose some structure on a Kripke multiverse, and
>> inversely, an accessibility structure on a multiverse impose a modal law.
>>
>> And now a free subject of reflexion :) (to prepare the sequel)
>>
>> If reflexivity in Kripke multiverse characterizes []A -> A
>>
>> Which relations can characterize the following formula?
>>
>> The Leibnizian one:
>>
>> []A -> [][]A
>> []A -> <>A
>> p -> []<>A
>> <>A -> []<>A
>> [](A->B) -> ([]A -> []B)
>>
>> And what about (more hard) the non Leibnizian one, which will play some
>> role (as scheme of some machines discourses)
>>
>> <>A -> ~[]<>A (related to Gödel)
>> []([]A -> A) -> []A (related to Löb)
>> []([](p -> []p) -> p) -> p (related to Grzegorczyk, the Grz of S4Grz).
>>
>>
>> Bruno
>>
>>
>>
> http://iridia.ulb.ac.be/~marchal/
>
>
>
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