Rote Arithmetic
Jon Awbrey
jawbrey at oakland.edu
Fri Sep 20 04:14:42 CEST 2002
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Residual Questions ...
Here is how I last defined rotes:
JA: | @ is a rote.
|
| If f is a finite function from rotes to rotes,
| say, f : index_j ~> exponent_j for j = 1 to k,
| where the "indices" and "exponents" are rotes,
| then the following tree is a rote:
|
| i_1 e_1 i_k e_k
| o----o o----o
| \ ... /
| \ | /
| \ | /
| \|/
| @
|
| where the rotes for the "indices" (source elements) i_j
| and the "exponents" (target elemnts) e_j are attached
| at the places indicated.
Strictly speaking, I probably should have stipulated that
each f is a "finite partial function" from rotes to rotes,
but since the set of rotes is infinite the partiality of f
might be taken as implicit in the fact of its being finite.
Let $N$ ("Script N") be the set of rotes.
Let (X -> Y) be the set of functions from X to Y.
Let [X -> Y] be the set of finite partial functions from X to Y.
Then we have the isomorphism $N$ ~=~ [$N$ -> $N$],
that is, every rote k : $N$ is uniquely analyzed
as a finite partial function k : $N$ -> $N$, and
every such function gets a unique gödel number
in $N$. Notice the polymorphism of types.
1 = {} = the empty function
2 = (p_1)^1 = {<1, 1>}
3 = (p_2)^1 = {<2, 1>}
4 = (p_1)^2 = {<1, 2>}
5 = (p_3)^1 = {<3, 1>}
6 = (p_1)^1 (p_2)^1 = {<1, 1>, <2, 1>}
As rooted trees, these look like this:
o--o
|
o--o o--o o--o o-o
| | | /
o--o o--o o--o o--o o-o o-o
| | | | \ /
@ @ @ @ @ @
1 2 3 4 5 6
Anyway, I think this is right, but I've learned not
to trust stuff like this until several other people
have a chance to go over it.
Jon Awbrey
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