Re: OUTGOEDELING A HUMAN?

On Feb 20, 8:19 am, "LauLuna" <laureanol...@xxxxxxxx> wrote:
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.

No. You are defining a primitive relation as a composition of 5 more
primitive mathematical notions:

1. H (some sort of thing but not defined)
2. believes
3. x expresses
4. a true
5. proposition

You are hardcoding it to these expressions, rather than allowing
variations using the primitives, so you lack generality. How would
you represent:

x knows that y expresses z
x knows that y expresses a true proposition
x knows that y expresses a false proposition
x knows that y expresses nothing at all
H knows that x expresses nothing at all
H knows that x proves a false proposition
the proposition which x proves
the proposition which x refutes
what H knows
what x knows

'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.

Each of these is a hardcoded expression rather than a definition of
primitives that are used to define these expressions.

'S(x, *y)' means that x expresses *y.

Because of the use of *y instead of y, this says that if x is
expressible then S(y,x) means y expresses x. However, this does not
define S(y,x) if x is not expressible (a whole class of false
sentences.)

C-B

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

H's consistency with respect to
G:

3. K(-T(G)) -> -B(G) Axiom 3

4. Ax S(G,*x) & B(x) -> B(G) Axiom 4

H's reflexivity with respect to his belief regarding
G:

5. B(G) -> K(B(G)) Axiom 5

6. -B(G) -> K(-B(G)) Axiom 6

V. DEDUCTION

7. B(G)
Assumption
8. K(B(G)) MP 5, 7
9. K(-T(G)) MP 2, 8
10. -B(G) MP 3,
9
11. B(G) & -B(G) &I 7, 10
12. -B(G) RD
7-11
13. K(-B(G)) MP 6, 12
14. B(-B(G)) K->B 13
15. Ex S(G,*x)
Assumption
16. S(G,*-B(G)) MP 1, 15
17. S(G,*-B(G)) & B(-B(G)) &I 14, 16
18. S(G,*-B(G)) & B(-B(G)) -> B(G) AE 4
19. B(G) MP
17, 18
20. B(G) & -B(G) &I 12, 19
21. -Ex S(G,*x) RD 15-20

It seems this proves that if there is a linguistic code LC according
to which axiom 1 is true and if H fulfils the conditions expressed by
the axioms 2 through 6, then G expresses no proposition in LC.

Regards


.

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