Re: OUTGOEDELING A HUMAN?

LauLuna says...

As I promised, I will try to prove that, under certain conditions for
H, sentence G expresses no proposition in any linguistic code
according to which axiom 1 below holds.

I. Let 'G' denote the sentence:

'H does not believe G expresses a true proposition'

II. NOTATION

'x', 'y' are sentential variables.
'B(x)' means that H believes that x expresses a true proposition.
'K(x)' means that H knows that x expresses a true proposition.
'T(x)' means that x expresses a true proposition.
'*x' denotes the proposition which x expresses, if x expresses any.
'S(x, *y)' means that x expresses *y.

I object already. You cannot meaningfully talk about
"true proposition" without making explicit that *language*
and *interpretation*. And (according to Tarski's theorem)
no consistent interpretation can give an interpretation to
its own truth predicate. So for your notation to make sense,
you need two different interpretations, call them I_0 and I_1.
The truth predicate T(x) is truth within I_0, and T is a
predicate that resides in I_1.

In the case of beliefs, there is no need to distinguish levels,
because beliefs don't need to be consistent. In the case of truth,
you have to distinguish levels.

Now, what does it mean that H "knows" something? Presumably
it means that H believes it, and it is true. But true under
what truth predicate? We're only talking about truth in I_0
here, so I'll interpret your K(x) to mean B(x) and T(x),
where T(x) means truth in I_0.

I have *no* idea what you mean by "x expresses y". You are
quantifying over propositions? Propositions in what language?
It's not clear to me that this means anything in particular.

III. INFERENCE RULES

MP = Modus Ponens
RD = Reductio ad absurdum
&I = Introduction of the conjunction '&'
AE = elimination of the universal quantifier 'A'
K->B = for any x, from 'K(x)' infer 'B(x)'

IV. AXIOMS

From the definition of G

1. Ex S(G,*x) -> S(G, *-B(G)) Axiom 1

H's understanding of
G:

2. K(B(G)) -> K(-T(G)) Axiom 2

No, that axiom is false. The predicate T(x), if it represents
a consistent truth predicate, does not apply to sentences that
involve true. For a consistent truth predicate T(x) will be true
if x represents a true sentence in interpretatin I_0, but will
be false if x represents a sentence that is false *or* if x
represents a sentence that is not given an interpretion by I_0.
In particular T(~T(G)) will be false, regardless of whether T(G)
is true or not.

If K(x) means B(x) and T(x), then it will always be the case that

~K(~T(G))

--
Daryl McCullough
Ithaca, NY

.

Relevant Pages

  • Re: Torkel Franzen on truth
    ... But you can't have a truth predicate ... By using eiher ... contradiction we opted for the second - Tarski's truth levels. ... is just as compelling as any other axiom. ...
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  • Re: Torkel Franzen on truth
    ... language can have a truth predicate for that very language. ... is just as compelling as any other axiom. ... Daryl McCullough ...
    (sci.logic)