Re: Modal summary and new exercises + motivation

[Bruno Marchal]

On 25 Mar 2014, at 03:02, LizR wrote:

Thank you for the above, for my diary!

On 24 March 2014 20:14, Bruno Marchal <mar...@ulb.ac.be> wrote: 

New exercise:

show

(W,R) respects A -> []<>A
iff
R is symmetrical.

OK, symmetrical means for all a and b, a R b implies b R a.

A -> []<>A can (I hope) be read as "the truth of A in one particular world (which I will call this world) implies that for all worlds accessible from this world, there exists at least one world in which A is true".

I was about to write that you did that error again, but looking twice, I saw you are entirely correct, OK. 

Well, there is indeed one world accessible from those other worlds, in which A is true - this one! Because all worlds accessible from this one can access this world (due to symmetry) and in this world A is true.

Excellent.

You proved that 

If R is symmetrical then  (W, R) respects A -> []<>A.

What about finishing the work and prove the reciprocal? Hmm... Please look at the iff in the quoted quote above.

You should still show that  

if (W, R) respects A -> []<>A then R is symmetrical.

The mere fact that A -> []<>A is true in all worlds, whatever the valuation is, imposes the symmetry for the binary accessibility. 

To show that, you can reason by absurdum. You imagine that (W, R) respects A -> []<>A, and you consider that R is not symmetrical. Then you have to find a valuation leading to a counterexample, a world in which A is true and []<>A is false.

Let me do it, so that you can rest after the good work :)

*

If (W, R) is not symmetrical, there is two worlds a and b so that a R b, and ~(b R a). OK?

Let us choose the valuation V which assign 0 to p in all the worlds accessible from beta.

Well, but then if p is true in alpha, []<>p is true in alpha (as we assume that (W, R) respects A -> []<>A). But then <>p must be true in beta, OK?

But beta accesses only to worlds with p false. Contradiction.

We say that the illuminated multiverse (W, R) 

with W = {a, b, c}, 

together with the non symmetrical relation explicitly defined by aRb, bRc. (so we have 'not bRa'), together  with the valuation: p true in a, and false in c, constitutes a counterexample, to the idea here that a (W,R) with R non symmetrical can respect A -> []<>A. Indeed, in that illuminated multiverse the assymmetry break makes it possible to break the law, and in the world a p is verified and  []<>p is not contradicting the "law" A -> []<>A.

All right?

Bruno

PS I send this to FOAR as this is part of an answer to his question, and a key (albeit tiny) part of the derivation of the "physical laws", notably giving clues on the reversibility on the bottom of the domain of indeterminacy ( the true sigma_1 sentences).

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[LizR]

On 26 March 2014 06:17, Bruno Marchal <mar...@ulb.ac.be> wrote:

On 25 Mar 2014, at 03:02, LizR wrote:

Thank you for the above, for my diary!

On 24 March 2014 20:14, Bruno Marchal <mar...@ulb.ac.be> wrote: 

New exercise:

show

(W,R) respects A -> []<>A
iff
R is symmetrical.

OK, symmetrical means for all a and b, a R b implies b R a.

A -> []<>A can (I hope) be read as "the truth of A in one particular world (which I will call this world) implies that for all worlds accessible from this world, there exists at least one world in which A is true".

I was about to write that you did that error again, but looking twice, I saw you are entirely correct, OK.

Phew a timely alteration to reality there! :-)

Well, there is indeed one world accessible from those other worlds, in which A is true - this one! Because all worlds accessible from this one can access this world (due to symmetry) and in this world A is true.

Excellent.

You proved that 

If R is symmetrical then  (W, R) respects A -> []<>A.

What about finishing the work and prove the reciprocal? Hmm... Please look at the iff in the quoted quote above.

Oh! Yes of course, I noticed the "iff" when I started, but had forgotten about it by the time I finished.

So I am attempting to show that if R isn't symmetrical, then it won't respect A -> []<>A

I assume a mulitverse is not symmetrical if there is at least one a b for which aRb but not bRa.

...so I can see that there COULD be a situation in which the truth of A in a particular world doesn't necessarily imply []<>A. I'm not sure if this is necessarily true, however.

For an example, a multiverse of two worlds a b with aRb only doesn't respect A -> []<>A because the truth of A in a doesn't imply that for all worlds accessible from a (i.e. b) there exists a world accessible from b where A is true, because b is a cul-de-sac with nowhere accessible.

But could there be a more complicated multiverse which had aRb and maybe aRc and cRa and bRc (say) which did respect this? Well, I can construct that multiverse with an illumination that doesn't respect this, namely A is true in a but not true in c, so in all worlds accessible from a, there is one in which there are none accessible in which A is true (namely b). And I can probably construct larger multiverses and show it isn't true in them either....and it seems like it shouldn't be, because there is no implication (without symmetry) that A has to be true in a world accessible from, in this example, b. So there will always be a world b which is accessible from another world a, but in which the truth of A in a doesn't imply that it's true in any world accessible from b, because one can always construct an illumination in which it isn't true in any world except a, and a isn't accessible from b.

Actually, did I just prove it???!

If so...

[Bruno Marchal]

On 26 Mar 2014, at 02:12, LizR wrote:

On 26 March 2014 06:17, Bruno Marchal <mar...@ulb.ac.be> wrote:

On 25 Mar 2014, at 03:02, LizR wrote:

Thank you for the above, for my diary!

On 24 March 2014 20:14, Bruno Marchal <mar...@ulb.ac.be> wrote: 

New exercise:

show

(W,R) respects A -> []<>A
iff
R is symmetrical.

OK, symmetrical means for all a and b, a R b implies b R a.

A -> []<>A can (I hope) be read as "the truth of A in one particular world (which I will call this world) implies that for all worlds accessible from this world, there exists at least one world in which A is true".

I was about to write that you did that error again, but looking twice, I saw you are entirely correct, OK.

Phew a timely alteration to reality there! :-)

Well, there is indeed one world accessible from those other worlds, in which A is true - this one! Because all worlds accessible from this one can access this world (due to symmetry) and in this world A is true.

Excellent.

You proved that 

If R is symmetrical then  (W, R) respects A -> []<>A.

What about finishing the work and prove the reciprocal? Hmm... Please look at the iff in the quoted quote above.

Oh! Yes of course, I noticed the "iff" when I started, but had forgotten about it by the time I finished.

... by the time you half-finished (grin).

So I am attempting to show that if R isn't symmetrical, then it won't respect A -> []<>A

I assume a mulitverse is not symmetrical if there is at least one a b for which aRb but not bRa.

...so I can see that there COULD be a situation in which the truth of A in a particular world doesn't necessarily imply []<>A. I'm not sure if this is necessarily true, however.

One counter-example is enough.

For an example, a multiverse of two worlds a b with aRb only doesn't respect A -> []<>A because the truth of A in a doesn't imply that for all worlds accessible from a (i.e. b) there exists a world accessible from b where A is true, because b is a cul-de-sac with nowhere accessible.

Very good Liz, you found a simpler counter-example than mine.

But could there be a more complicated multiverse which had aRb and maybe aRc and cRa and bRc (say) which did respect this? Well, I can construct that multiverse with an illumination that doesn't respect this, namely A is true in a but not true in c, so in all worlds accessible from a, there is one in which there are none accessible in which A is true (namely b). And I can probably construct larger multiverses and show it isn't true in them either....and it seems like it shouldn't be, because there is no implication (without symmetry) that A has to be true in a world accessible from, in this example, b. So there will always be a world b which is accessible from another world a, but in which the truth of A in a doesn't imply that it's true in any world accessible from b, because one can always construct an illumination in which it isn't true in any world except a, and a isn't accessible from b.

Actually, did I just prove it???!

You proved it already above.  What you show after is slightly stronger: you show how to build a counterexample from any two worlds having a non symmetrical accessibility relation between (and so no need to invoke a cul-de-sac world), but the example above (with the cul-de-sac world) was enough. 

Good.

Why is it that you solved the case []A -> A (reflexive R), []A -> <>A, (ideal (W,R)), A -> []<>A (symmetry), but be so long on the not more complex case of 

[]A -> [][]A (transitivity: aRb and bRc entails aRc)?  I think that Brent solved one half of it.

You can also train yourself with the following

1)

(W, R) respects <>A -> []<>A

iff

R is euclidian  (meaning that if aRb and aRc, then bRc)  (sub exercise: show that if R is both reflexive and euclidean, then R is an equivalence relation (reflexive, transitive and symmetric. The converse is trivial). 

2)

(W, R) respects <>A -> ~[]<>A

iff

R is realist (meaning that all transitory worlds access to a cul-de-sac world.  A transitory world is just a non-cul-de-sac world).

Then I will give you the case for the Löb formula and Grzegorczyk, which are slightly more difficult, and I will have to impose some more condition on the Kripke multiverse. Later.

We will have also to come back to logic proper, and define what is a modal "theory", and what is a proof in such theory, and eventually we will be able to enunciate the "fundamental" soundness and completeness theorem, linking the valid modal proof and the Kripke semantics.

Then we will have to come back on the definition in the arithmetical language of the beweisbar provability predicate, and so we get closer to Solovay, which is used to derive the arithmetical hypostases and eventually derive physics from (arithmetical) self-reference. 

A lot of work remains to be done, but that is good, as you need a lot of work to solidify your understanding and become familiar with the material. I hope you enjoy this, well enough to pursue the thread.

We have all the time ... There is no rush.

Bruno

If so...

<360.gif>

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[LizR]

On 27 March 2014 00:35, Bruno Marchal <mar...@ulb.ac.be> wrote:

On 26 Mar 2014, at 02:12, LizR wrote:

... by the time you half-finished (grin).

By the time I thought I'd finished.

So I am attempting to show that if R isn't symmetrical, then it won't respect A -> []<>A

I assume a mulitverse is not symmetrical if there is at least one a b for which aRb but not bRa.

...so I can see that there COULD be a situation in which the truth of A in a particular world doesn't necessarily imply []<>A. I'm not sure if this is necessarily true, however.

One counter-example is enough.

For an example, a multiverse of two worlds a b with aRb only doesn't respect A -> []<>A because the truth of A in a doesn't imply that for all worlds accessible from a (i.e. b) there exists a world accessible from b where A is true, because b is a cul-de-sac with nowhere accessible.

Very good Liz, you found a simpler counter-example than mine.

But could there be a more complicated multiverse which had aRb and maybe aRc and cRa and bRc (say) which did respect this? Well, I can construct that multiverse with an illumination that doesn't respect this, namely A is true in a but not true in c, so in all worlds accessible from a, there is one in which there are none accessible in which A is true (namely b). And I can probably construct larger multiverses and show it isn't true in them either....and it seems like it shouldn't be, because there is no implication (without symmetry) that A has to be true in a world accessible from, in this example, b. So there will always be a world b which is accessible from another world a, but in which the truth of A in a doesn't imply that it's true in any world accessible from b, because one can always construct an illumination in which it isn't true in any world except a, and a isn't accessible from b.

Actually, did I just prove it???!

You proved it already above.  What you show after is slightly stronger: you show how to build a counterexample from any two worlds having a non symmetrical accessibility relation between (and so no need to invoke a cul-de-sac world), but the example above (with the cul-de-sac world) was enough. 

Good.

Phew.

I will look at the next part later