h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Hi Bx4,

I'm still on old format too, so I don't know.

ttfn

Are we going to need a strimmer?

hi psi

Busy in the world so strimmer reply an uncompleted noodle as yet.. However, given Blackburn's:

//'We can see why this is so if we put it in terms of what we can call Ramsey’s ladder. This takes us from p to it is true that p, to it is really true that p, to it is really a fact that it is true that p, and if we like to it is really a fact about the independent order of things ordained by objective Platonic normative structures with which we resonate in harmony that it is true that p...Ramsey’s ladder is horizontal. The view from the top is just the same as the view from the bottom, and the view is p.'//

and your (Compatibilism 557):

//'@p means necessarily p, which is short for 'necessarily p is the case, ie p is true, recall Ramsay's Ladder//

I am unclear how you insert your modality into a Ramsey ladder and retain its bi-directional horizontality.

In a traversal from the 'top' Axiom M allows @p->p (though it would seem obvious what 'work' the modality is doing) but nothing seems to let you traverse the rungs from the 'bottom' such that p->@p unless you accept that all true propositions are necessarily true propositions.

bs

Hi Bx4,

The application of Ramsey's Ladder in my #557 applies inside the scope of the modal operator only. So if I put in some brackets to make this clearer:

@(p)== @(p is true)== @(it really is the case that p is true) etc.

So I'm not including the '@' in my application of Ramsay's Ladder in the sense of moving to terms with @ to terms without it, or vice versa.

In summary, I'm not inserting any modality into Ramsay's Ladder.

I hope that clarifies.

ttfn

hi psi

//I hope this clarifies//

I'm still a bit confused about the differences between your revised

1. @(p)== @(p is true)== @(it really is the case that p is true) etc.

and your 557 original

//'@p means necessarily p, which is short for 'necessarily p is the case, ie p is true, recall Ramsay's Ladder//

which seems to yield

2. @p==@(p is the case)==(p is true), etc.

with an apparent non-modal rung (p) is true. Is this a typo which should be read as

3. @p==@(p is the case)==@(p is true), etc.

where the order of the rungs is different from your revised 1? Is there a reason for this?

bs

Hi Bx4,

It wasn't so much a typo as an imprecision due to my typing something that attempted to convey the revised version in a way that makes sense out loud. Sorry about that it was an error. When I said:

//'@p means necessarily p, which is short for 'necessarily p is the case, ie p is true, recall Ramsay's Ladder//

I meant that 'necessarily p is the case, ie necessarily p is true'.

The original rephrase of 'p is the case' to 'p is true' was only ever intended to be inside the scope of the '@', which is why I left it off in the second clause. I can see now why that was unclear and I apologise.

The context was that you seemed to be denying that '@p' means necessarily p which by Ramsays Ladder applied inside the scope of the '@' can be translated as necessarily p is true. It seemed such a small point on which to get agreement, but alas...

ttfn

hi psi

Thanks for clarification. As to

//The context was that you seemed to be denying that '@p' means necessarily p which by Ramsays Ladder applied inside the scope of the '@' can be translated as necessarily p is true. It seemed such a small point on which to get agreement, but alas//

I'm not sure it is a small point. There are in fact a number of issues that I have with not only your modal ladder but also Blackburn's non-modal ladder.

I'll start with the latter since it allows clarification of the problem I have with your argument. Blackburn's RL using your notation has the form:

(i) p==(p is true)==(it is the case that p is true), etc.

But the truth value of p is determined by its meaning so for example if p is the analytic proposition

(a) A square has only three sides

or the synthetic proposition

(b) Water is made up of covalent molecules comprised of two molecules of hydrogen and one of sulphur

Both are false but in neither case I see anything that prevents me constructing the non-modal Ramsey ladder:

(ii) p==(p is false)==(it is the case that p is false), etc.

without being incoherent. So I am somewhat unclear why you consider that a Ramsey Ladder, given p, entails (p is true) without reference to the semantic content and hence the truth value of p.

So perhaps not such a 'small point' after all.

bs

Hi Bx4,

"I'm not sure it is a small point. There are in fact a number of issues that I have with not only your modal ladder but also Blackburn's non-modal ladder."
I wouldn't call mine a 'modal ladder' since I'm using the ladder inside the scope of the '@' only.

"(i) p==(p is true)==(it is the case that p is true), etc."
So far, so good.

"But the truth value of p is determined by its meaning so for example if p is the analytic proposition"

I think this is where you diverge from my view. I agree with your examples:

"(a) A square has only three sides

or the synthetic proposition

(b) Water is made up of covalent molecules comprised of two molecules of hydrogen and one of sulphur"

I'm fine with the above. The trouble comes here:

"Both are false but in neither case I see anything that prevents me constructing the non-modal Ramsey ladder:

(ii) p==(p is false)==(it is the case that p is false), etc."

What prevents you from doing the above is the distinction between the /meaning/ of a grammatical sentence (in logic notation or English) and its truth value. It seems to me that you are confusing these two notions above.

I'll try to explain this using one of your examples. If I say "A square has only three sides", then by RL I /mean/ "It is true that a square has only three sides". You have assumed that since it actually is false that by asserting it I must mean to assert that it is false. That's not the case at all though and we can see that the grammar has a way of distinguishing between asserting something is true and asserting it is false.

Let p = "A square has only three sides" I want to assert the truth of p so I can say:

1)"A square has only three sides"

Suppose I know p is false, then I might say:

2)"A square doesn't have only three sides"

The equivalents are:

1) p
2) ¬p

So I hope you can now see that "p==(p is false)" does not obey the rules of the grammar. In other words, "p" just doesn't mean "p is false". The fact that in your example p is actually false is irrelevant to this point, but if you can't see this, then that might be where you have gone wrong (in my view).

ttfn

hi psi

A bit busy in the world at moment. Working on a reply as time allows.

Btw, the 'improved' version of hootoo does not incorporate new italicised quotes or handle symbolic character sets any differently from old version. The improvment seems merely cosmetic.

bs

Hi Bx4,

Ah so it is entirely in keeping with the other 'upgrades'.

ttfn

...oh, and just to clarify, I see no problem with:

¬p==(p is false)==(it is the case that p is false), etc.

So Ramsay's Ladder is applicable to propositions which are false. The grammar has to be right though.

ttfn

hi psi

"Ah so it is entirely in keeping with the other 'upgrades'."

I am attempting 'steely resolve' w.r.t. the religion boards having become bored with the 'Dover Beach syndrome' but I did notice that quote boxes have reappeared though the text remains /italicised/.
Yet more progress!

hi psi

"
So Ramsay's Ladder is applicable** to propositions which are false. The grammar has to be right though."

OK, so suppose we have:

(p) 'A square has only four sides'
(q) 'A square has only three sides'

then do you agree that we can construct the following two Ramsey Ladders

(1) {p==(p is true)==(it is the case that p is true), etc.}
(2) {q==(q is false)==(it is the case that q is false), etc.}

and if so does this mean that you have dropped your earlier argument***:

"So I hope you can now see that "q==(q is false)" does not obey the rules of the grammar. In other words, "q" just doesn't mean "q is false".

** Given that Blackburn's Ramsey's Ladders are merely an extension of the conventional, though perhaps problematic view, that Ramsey was articulating a /redundancy/ theory of truth then I think we need to explore further in what sense they are 'applicable' to proposition


*** I have substituted your original 'p' by 'q' here but only to align it with (2) above.

bs

dear both,

indeed the quote boxes make for easier reading, though you are correct Bored to state that the record hasn't changed.

on that nore i am currently stuck with The Major bringing up a thread from sept 2009.........!!?? hilarious.

btw not sure if this is happeneing to you two, but most of the time i can't even access this thread and only get an error message. not that it matters since you are writing in a language wot i don't knows.....

ttfn

Hi Bx4,

"then do you agree that we can construct the following two Ramsey Ladders

(1) {p==(p is true)==(it is the case that p is true), etc.}
(2) {q==(q is false)==(it is the case that q is false), etc.}

and if so does this mean that you have dropped your earlier argument"

I think (1) is fine but (2) is not. So I'm sticking with my earlier argument; I think q==(q is false) is grammatically incorrect. The correct form is ¬q==(q is false).

If I assert p then I'm saying that 'A square has only four sides'. By RM this is equivalent to me saying 'It is true that a square has only four sides'.

If I assert q then I'm saying that 'a square has only three sides'. By RM this is equivalent to me saying 'It is true that a square has only three sides'.

So the correct ladder forms are:

(1) {p==(p is true)==(it is the case that p is true), etc.}
(2) {q==(q is true)==(it is the case that q is true), etc.}

The difference in the two cases is that in the first I am right, whereas in the second I'm wrong.

I could fix this in a number of ways. I could rewrite q as follows:
(q) 'A square does not have only three sides'

Or instead of (incorrectly) asserting q, I could assert ¬q. then the correct grammar for (2) would be:

(2) {¬q==(¬q is true)==(it is the case that ¬q is true), etc.}

On Ramsey and redundancy theories, I think there might be problems with the attempt to eliminate truth completely, but I'll have to refresh my memory as the the details. If so then it wouldn't be a surprise if Ramsey didn't in fact advocate a strict redundancy theory. I'd like to clear up this little point first though

ttfn

Hi jank,

Yes I get the error message too. I get round it by clicking on the 'My Space' link on the left.

hi jank

"the record hasn't changed."

I'm staying out off it. Not really interested in theism or anti-theism but the latter's scientific illiteracy profoundly irritating particularly their reliance on crappy vox pop tomes about 'evolutionism'

One of the 'heroes' of the tribe appears to think that 'methodological naturalism' supports atheism.

"i am currently stuck with The Major bringing up a thread from sept 2009"

Yes I noticed you were supposedly a 'liberal fundie'. Do such exist? I think he is mostly a WUM....

'error msg'

I'm not on hootoo or boards that much. but rg said that about a week ago the BBC board were regularly imploding:

http://groups.google.com/group/h2g2communityconsortium/browse_thread/thread/a1c266b2713c398f

Malware infection or a plot against the 'Save hootoo Collective'?


"you are writing in a language wot i don't knows....."

Almost certainly philosophical jabberwocky.....

Work calls

bs

hi psi


"By RM this is equivalent to me saying 'It is true that a square has only three sides"

I think this may be the nub of our disagreement.

I consider that in :

Ramsey’s ladder. This takes us from p to it is true that p, to it is really true that p, to it is really a fact that it is true that p, and if we like to it is really a fact about the independent order of things ordained by objective Platonic normative structures with which we resonate in harmony that it is true that p. For the metatheoretical minimalist, Ramsey’s ladder is horizontal. The view from the top is just the same as the view from the bottom, and the view is p.'

Blackburn is asserting that there is no 'semantic ascent' in that the phrases 'is true', it is really true that', 'it is really true that p, etc. are redundant. That is they add nothing to the meaning of p.

So Ramsey ladder simply reduces to

{p==p==p==p} which is to say {p}

Michael Dummett describes a redundancy theory thus:

'Its primary significance is as an overt rejection of the notions of truth and falsity are central to a theory of meaning: to accept the redundancy theory is to deny that a grasp of the meaning of a sentence consists in an apprehension of its truth conditions, in knowing what has to be the case for it to be true,...'
('Frege: Philosopy of Language, Duckworth, 2nd Ed.,1981 p.458)

However, while invoking Ramsey's Ladder your

"If I assert q then I'm saying that 'a square has only three sides'. By RM this is equivalent to me saying 'It is true that a square has only three sides'.'

implies a non-redundant theory where 'the meaning of a sentence consists in an apprehension of its truth conditions, in knowing what has to be the case for it to be true'.

"The correct form is ¬q==(q is false)."

Surely not since if 'is false' is redundant then you are saying `¬q==q which is incoherent.

"The difference in the two cases is that in the first I am right, whereas in the second I'm wrong.'

This appears to do more than say p is true and q is false which again implies that your theory of meaning is non-redundant in that it involves truth conditions. How is this compatible with your use of the redundant Ramsey's Ladder?

"I could rewrite q as follows:

(q) 'A square does not have only three sides'"

If you did so it would be a different proposition ¬q. rather than q.

"Or instead of (incorrectly) asserting q, I could assert ¬q."

This only works if you assume that by asserting q you are asserting that 'q is true' and not that 'q is false'. What prevents me from asserting 'A square has only three sides is false' whereby in a horizontal Ramsey's Ladder (q is false)==q since 'is false' is redundant.

It seems to me that our disagreement centres on a different reading of Blackburns Ramsey Ladder' I see it as as a set of statements which have no 'semantic ascent' because the phrases that qualify p are redundant and hence eliminable.

So I take Blackburn as saying that the /meaning/ of p does involve an appeal to truth conditions. You would seem to disagree since you hold that by asserting p you are asserting a truth condition for p. I don't see how your truth conditional interpretation of p coheres with Blackburn's redundancy interpretation of p.

(Note that I am not advocating Ramsey/Blackburn redundancy theory of meaning merely that they do and that consequentially we should deal with the Ramsey Ladder in terms of it.)

"On Ramsey and redundancy theories, I think there might be problems with the attempt to eliminate truth completely."

I don't think eliminating truth conditions from a theory of meaning precludes a theory of truth that does not appeal to the meaning of sentences understood as truth conditions.

It has been argued, by Nils-Eric Sahlin and others, that Ramsey's 'Facts and Propositions' can be read as articulating a redundancy theory of meaning as a precursor to developing a pragmatic theory of truth.

As I understand it at the time of 'Facts and Propositions' Ramsey's view, though he may have changed it later**. was that it was not propositions but beliefs s that were truth-bearers.

**Possibly in a paper 'Truth' with I have yet to track down.

written in haste so usual apologies and also for lack of strimming.

bs

morning Bored


"staying out off it"; don't blame you, i do wonder about myself at times. and admitting so year in year out may be an indication of how "bad" things are with my repetitious inclinations. but strangely it remains amusing enough on some days. not that i lack things to do, or more importantly, things that must be done....ho hum.

i fall back on my analogy (oh yes) of the pub. i used to go to my local to talk boullards with any and all. but being The Butler of my family means i just don't have time to go and talk rubbish with others. so all hail the interwebs!


"liberal fundie"; i think the Major meant another poster, but keeps confusing that person with me. that or you are quite right to point out the WUM option. it is rather bizarre to have another get quite so incensed for essentially holding a similar position.


"error msg"; yet more tech-speak-nonsense of the 'there is an error; unknown error, by which we mean we know there's an error, but we aren't sure what error, but it is most certainly a dead error' etc etc etc.

do hope the boards stay, though am spending less time on them. btw did email you to the ntl address but perhaps you're having problems with that account?


"philosophical jabberwocky"; above my pay grade unfortunately. i do have a shot at reading your exchanges with psi, but damn it, every time i think i'm getting close my brain fills up with static. but that's ok, as per Dirty Harry; "A man's gotta know his limitations."


hope the real world is treating you well, and road trips coming up? it's been ages surely.....


tot straks

hi psi

I mentioned Nils-Eric Sahlin's 'Philosophy of F.P. Ramsey' which apparently was seems to have begun the reassessment of role of redundancy theory. The chapter covering this is not available on-line nor do I have it here but I have turned up this essay by Sahlin which gives a flavour of his argument:

http://www.nilsericsahlin.se/ramsey/content.asp

In 'Ruling Passions' Blackburn (p.78) seems to agree with Sahlin's view of what Ramsey is attempting:

'F.P Ramsey whose famous paper 'Facts and Propositions' that argues it is not by staring at the word 'true' that progress is mad, but by understanding the various kinds of judgments in /behavioural/ terms.'

bs