On 22 Oct 2013, at 03:21, Platonist Guitar Cowboy wrote:
On Mon, Oct 21, 2013 at 4:15 PM, Bruno Marchal <marc...@ulb.ac.be>
wrote:
On 21 Oct 2013, at 04:48, Platonist Guitar Cowboy wrote:
Disclaimer: No idea if I am even on the same planet on which this
discussion is taking place. So pardon my questions and confusions:
On Sun, Oct 20, 2013 at 11:15 PM, Russell Standish <li...@hpcoders.com.au
> wrote:
On Sun, Oct 20, 2013 at 06:22:15PM +0200, Bruno Marchal wrote:
>
> On 20 Oct 2013, at 12:01, Russell Standish wrote:
>
> >On Sun, Oct 20, 2013 at 08:52:41AM +0200, Bruno Marchal wrote:
> >>
> >>We have always that [o]p -> [o][o]p (like we have also always
that
> >>[]p -> [][]p)
> >>
> >
> >
> >There may be things we can prove, but about which we are in fact
> >mistaken, ie
> >[]p & -p
>
> That is consistent. (Shit happens, we became unsound).
>
Consistency is []p & ~[]~p. I was saying []p & ~p, ie mistaken
belief.
Why Isn't "mistaken belief" here merely unsound because it's
propositional variable, assuming we're speaking generally about all
systems?
Yes, []p & —p, makes the machine on which [] applies, unsound. But
still consistent. Such machines will believe (prove) an arithmetical
false (but consistent) sentence.
OK.
>
> >
> >Obviously, one cannot prove []p & p, for very many statements, ie
> >
> >[]p & p does not entail []([o]p)
Isn't that a rule for any modal sentence though, independent of
system? [o]p is ([]p & p)
Isn't []p & p = []([o]p) the definition of fixed point theorem?
That, plus modalization conditionals to remove p from the G
sentence so that every sentence H is a fixed point of it?
This is a bit unclear.
A fixed point is when a proposition says something about herself
(like p <-> []p, p <-> [] ~p, etc.). The fixed point will be a
proposition in which p does no more occur. The main one are:
Ok, that is clearer. But this is a general rule in provability logic
of every modal system, right?
You can consider formula like "p <-> [] (... p ...)" in all modal
logic, but few will have solution.
This can be said: any K4 reasonner will have the fixed point in case
he visit Knave/ Knight island. And that K4 reasonner will becaome a G
reasonner (a Lôbian entity).
Any K4 consistent machines "rich enough" (like PA, ZF, but unlike RA)
will become Löbian too, and got the G-like fixed point. The Gödel
diagonalization lemma will somehow emulate the Knight-Knaves island.
[](p <-> ~[]p) -> [](p <-> ~[]f) Gödel fixed point
[](p <-> [] ¬p) <=> [](p<-> []⊥)
Yes, that's the kind of thing I think we're talking about.
OK. An important one is:
[](p <-> <>p) <-> [](p <-> f) (all machines/sentences asserting their
own consistency can prove 0=1)
And the one related to Löb's theorem:
[](p <-> []p) <-> [](p <-> t)
That is very amazing: all machines/sentences asserting their own
provability are true and provable.
That's a sort arithmetical "placebo".
So that you get that list of instances with (G sentence on left and
H on right) examples like the following:
[]p corresponds to T
¬[]p corresponds to ¬[]⊥
[]¬p corresponds to []⊥, which is pretty cool because you get a
provability statement that is arithmetically equivalent to its own
non-provability iff the statement is equivalent to the statement
that arithmetic is inconsistent. Because G proves in this fashion:
[o](p <=> [] ¬p) <=> [o](p<=> []⊥)
OK.
This is the kind of thing that clarifies the pronoun issue imho.
In arithmetic. But in UDA I think that the definition of first
person/third person in term of reconstitution/annihilation is clear
enough for the indeterminacy purpose, and the necessity of deriving
physics from arithmetic.
it is just not precise enough to get the actual technical beginning
of the derivation of physics.
>
> []p -> [][]p OK?
>
Why? This is not obvious. It translates as being able to prove that
you can prove stuff when you can prove it.
If this were a theorem of G, then it suggests G does not capture
the nature of proof.
Oh, I see that you are just restating axiom "4". But how can you
prove
that you've proven something? How does Boolos justify that?
But it nonetheless is everywhere in Boolos: []p -> [][]p IS a
theorem of G, and useful, unless Bruno shoots the cowboy, because
he cannot prove it or find his damned Boolos book.
What? You don't find your sacred manual? You will have to do some
penitences or something :)
Let me give you a difficult exercise: derive []p -> [][]p in *any*
normal modal logic satisfying Löb's formula (that is derive []p -> []
[]p from []([]p -p) -> []p (and [](p->q)->([]p->[]q).
Hmm, I'll try but feel free to ignore if there is too much bs in my
attempt to remember from Boolos, for anything to be saved; "find the
book, return to earlier chapters" will suffice). I choose GL from
fuzzy memory as normal modal logic for the exercise:
GL derives []p -> [][]p in a confused cowboy's dream from []([]p -p)
-> []p (and [](p->q)->([]p->[]q):
1) Löb formula: []([]p ->p) -> []p
Apply truth reflexivity ([]p->p) to bold box above to isolate p as
([]p->p), so:
[]([]p->p) -> ([]p->p)
?
I will read the rest later, but in the meantime you might try to
explain what you mean by "so".
The formula []([]p->p) -> ([]p->p) is not a theorem (for any p), as
you can see by substituting p by false.
[]([]f -> f) -> ([]f -> f) would entail []<>t -> "correctness", but
[]<>t -> []f "inconsistency! (by Gödel's second incompleteness).
You cannot apply "[]p -> p" in G, as self-correctness is not provable
in G (or by the correct machine).
OK?
2) Assuming 1), GL can prove 1), so []1):
[]([]([]p->p) -> ([]p->p))
3) Apply Löb formula again:
[]([]([]p->p) -> ([]p->p)) -> []([]p ->p)
4) Naturally or mistakenly because I don't know if these moves are
legal through "[](p->q)->([]p->[]q)"
[](([][]p->[]p) -> ([][]p->[]p)) -> []([]p ->p) and
([][][]p-> [][]p-> [][][]p-> [][]p) -> []([]p ->p)
If that weirdness is ok then the following holds more simply because
if GL can prove [][][]p etc. -> Löb formula, then its tempting to
just:
[]p-> []([]p ->p)
5) Since []([]p ->p) is also ([][]p -> []p)
6) combining 4 & 5, GL:
[]p -> [][]p
Q.E.D., but I don't believe my memory or the appropriacy to Bruno's
exercise! Too clumsy, too many levels and vertigo, lol.
So, yes. Gotta get back to forever undecided and find the Boolos
book, I know. But it's fun as a kind of crossword puzzle activity,
even when as confused as this partially forgotten and possibly
misunderstood thing. Scratching my head for a whole hour for those
couple of lines of Kool-Aid crazy... PGC
This shows that the axiom 4 is redundant in G.
(This is known as Sambin theorem, and is proved in Smullyan's
Forever Undecided). I can give the solution, some day).
> (and []p -> []p, and p -> p) + ([](p & p) <-> []p & []q) (derivable
> in G)
>
Did you mean [](p&q) <-> []p & []q?
I'm not sure of the q and whether you can just leave out the first
bit.
Russell was right. I made a typo error, as predicted (!).
That theorem at least sounds
plausable as being about proof.
> so []p & p -> [][]p & ([]p & p)
> -> []([]p & p) & ([]p & p),
>
> thus ([]p & p) -> [][o]p (& [o]p : thus [o]p -> [o][o]p)
>
> >
> >Therefore, it cannot be that [o]p -> [o]([o]p) ???
> >
> >Something must be wrong...
Why?
> >
>
> I hope I am not too short above, (and that there is not to much
typo!)
>
> Bruno
>
And thus you've proven that for everything you know, you can know
that
you know it.
Not sure, I'd guess we're comparing G's reasoning 3rd person "I"
with reasoning of a fixed point corresponding to some sentence of G
1st person "I" in a modal/qualitative provability sense. PGC
I don't see any fixed point here.
We have both the formal (and third person) []p -> [][]p and the
first person knowledge formula [o]p -> [o][o]p, which is usually
admitted for "sufficiently introspective knower).
Bruno
This seems wrong, as the 4 colour theorem indicates. We
can prove the 4 colour theorem by means of a computer program, and it
may indeed be correct, so that we Theatetically know the 4 colour
theorem is true, but we cannot prove the proof is correct (at least
at
this stage, proving program correctness is practically impossible).
--
----------------------------------------------------------------------------
Prof Russell Standish Phone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Professor of Mathematics hpco...@hpcoders.com.au
University of New South Wales http://www.hpcoders.com.au
----------------------------------------------------------------------------
--
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-
l...@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-
l...@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/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.