On 5/31/2020 3:23 PM, Bruce Kellett wrote:
On Mon, Jun 1, 2020 at 3:12 AM 'Brent Meeker' via Everything List
<[email protected]
<mailto:[email protected]>> wrote:
On 5/30/2020 10:44 PM, Bruce Kellett wrote:
On Sun, May 31, 2020 at 2:26 AM Bruno Marchal <[email protected]
<mailto:[email protected]>> wrote:
Let us write f_n for the function from N to N computed by nth
expression.
Now, the function g defined by g(n) = f_n(n) + 1 is
computable, and is defined on all N. So it is a computable
function from N to N. It is computable because it each f_n is
computable, “+ 1” is computable, and, vy our hypothesis it
get all and only all computable functions from N to N.
But then, g has have itself an expression in that universal
language, of course. There there is a number k such that g =
f_k. OK?
But then we get that g_k, applied to k has to give f_k(k), as
g = f_k, and f_k(k) + 1, by definition of g.
That is a fairly elementary blunder. g_k applied to k, g_k(k) =
f_n(k)+1, by definition of g_k. You do not get to change the
function from f_n to f_k in the expression. It is only the
argument that changes: in other words, f_n(n) becomes f_n(k). So
you are talking nonsense.
No, I think that's OK. It's a straight substitution n->k. The
trick is that g(n) is not some well defined specific function
because n has infinite range. So none of this works in a finite
world. But it's not surprising that there is incompleteness in an
infinite theory.
Yes, I had misunderstood what g(n) was supposed to be -- it is simply
a representation of the diagonal elements of the array, plus 1. But
Bruno's attempt to use the diagonal argument here fails, because he
has to show that f_n(n)+1 is not contained in the infinite list. He
has failed to do this.
All computable functions are in the list ex hypothesi.
Brent
Bruce
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