[FOM] On >>this sentence cannot be proven true<<

[Hartley Slater]

At 12:00 PM -0400 30/7/06, Laureano Luna Caba?ero wrote:
>But assume "this very sentence cannot be proven true" or "the 
>current sentence cannot be proven true" (ad lib) succeeds in 
>achieving self-reference. Let us call it "K". Under that condition K 
>seems to express no proposition.

First some general remarks:

You are talking about an interpretation of the sentence, and so, 
first of all, you are involved not in syntactic self-reference, but 
the only possible form of self-reference, namely semantic 
self-reference.  That is why a calculus of indirect speech is 
required, to formalise remarks like 's says that p'.  It is very 
uncommon to find people considering indirect speech in connection 
with Liar, and Liar-like Paradoxes, and yet it must be brought in 
because of the impossibility of syntactic self-reference.  The 
calculus I  use originates in some of Ramsey's remarks, and was first 
given a formalisation by Goodstein in the JSL for 1958.  Amongst 
other things Goodstein demonstrated the application of his calculus 
to two Liar paradoxs, and proved its consistency.  Prior used 
Goodstein's formalisation extensively in his book 'Objects of 
Thought', but its manner of quantifying over propositions, he 
realised, was not the best.  For instance, Prior would write 
'Everything a believes is true' as '(p)(Bap -> p)', and, amongst 
other things, just how 'p' should be read in the quantifier troubled 
him (see his Ch 3).   In my Analysis 2001 paper previously mentioned, 
'Prior's Analytic Revised', I showed the Ramsey-Goodstein-Prior 
calculus could be improved by introducing variables over referring 
phrases to propositions, i.e. phrases like 'what s says', 'what a 
believes', and 'that the moon is made of green cheese' - the term 
'that' serving to nominalise the following sentence, being symbolised 
by the '*' in my previous messages.

In your case you want to say about your sentence, 'k', that it says 
that it cannot be proven that it is true, i.e. that it says that that 
k is true (*Tk) cannot be proven, i.e. Sk*-P*Tk.  But the T-scheme is 
conditional on the univocality of the sentence, as before, i.e.
(r)(Skr <-> r=*-P*Tk)  -> (Tk <-> -P*Tk),
so any contradiction derivable on the right merely leads to a denial 
of the left.  You go on:

>   1st. K cannot be proven true: if K were proven true it would be 
>simultaneously true and false.
>
>   2nd. If K expresses a proposition, then what K says has been 
>proven true in 1st, so that K must be both true and false.

For the first claim you are presumably thinking that, since we know 
that P*Tk -> Tk, but also that Tk < -> -P*Tk, hence -P*Tk.  But that 
depends on the T-scheme holding, which is not itself proven. 
Certainly the second claim would hold if it started 'if K expresses 
just the proposition that K cannot be proven true'.  For if (r)(Skr 
<-> r=*-P*Tk) then one would definitely be able to get that -P*Tk, as 
in the first, and so Tk, because the T-scheme would hold, but also 
thereby P*Tk, and so -Tk, again from the T-scheme.  But your second 
claim misses out the uniqueness clause needed to get all this.

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