Re: Godel's proof, truth, reality, self-awareness, and all that jazz

On Sep 25, 1:44 pm, "Jesse F. Hughes" <je...@xxxxxxxxxxxxx> wrote:
"cbr...@xxxxxxxxxxxxxxxxx" <cbr...@xxxxxxxxxxxxxxxxx> writes:
On Sep 25, 11:50 am, "Jesse F. Hughes" <je...@xxxxxxxxxxxxx> wrote:
But this doesn't seem too difficult to add (does
it?). Why not something like this?

(a) {} = {}.
(b) If A1, A2, ..., An and B1, B2, ..., Bm are sets, then
{A1, A2, ..., An} = {B1, B2, ..., Bm}
iff the following hold:

(i) m = n
(ii) for each i<=n, there is a j<=n such that Ai = Bj
(iii) for each i<=n, there is a j<=n such that Bi = Aj

Equivalently (?), we could say that two sets A and B are equal iff
f(A) = f(B), where f is his encoding mapping sets to natural numbers.

Again, it only maps sets /as strings/ to naturals; and since multiple
strings represent the same set, f(A) is a /set/ of naturals, not a
single natural.

No, I don't see that. His description of the mapping seems to presume
principles (a) and (b), I suppose, since he builds the mapping
inductively (and without concern for order of elements). He says that
he maps, e.g., {{}{{}}} to the minimal bit string representing the
elements of the set. Since {} |-> 0 and {{}} |-> 1, this set maps to
2^0 + 2^1. Similarly, the set {{{}}{}} would map to 2^1+2^0.

Anyway, what *is* lacking is a clear definition of the encoding that
would settle this question. What I wrote above is what I think he
means, but his definition is by example, which is not very useful to
settle disagreements like this. So, I will let Han defend his own
mathematics and hopefully present it more clearly.


Sigh. Another day, another mea culpa. You are correct; see my response
to Han.

The short version of his encoding, in recursive form, appears to be:

f({}) = 0
f(A) = sum(x in A) 2^f(x).

Cheers - Chas

.

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