ONT Re: Zeroth Order Ontology
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ZOO. Note 10
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Distinction and Coincidence.
I introduce here a slightly more general, slightly more formal heading
that subsumes the specific algebraic theme of generators and relations.
Let L_2 be the combinatorial species of rooted trees,
regarded modulo the arithmetic relations I_1 and I_2.
o-----------------------------------------------------------o
| |
| o o o |
| \ / | |
| @ = @ |
| |
o-----------------------------------------------------------o
| |
| ( ) ( ) = ( ) |
| |
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| Axiom I_1. Distract <---- | ----> Condense |
o-----------------------------------------------------------o
o-----------------------------------------------------------o
| |
| o |
| | |
| o |
| | |
| @ = @ |
| |
o-----------------------------------------------------------o
| |
| (( )) = |
| |
o-----------------------------------------------------------o
| Axiom I_2. Unfold <---- | ----> Refold |
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Let us run through an example of how to evaluate
a string or a tree expression modulo the initial
relations I_1 and I_2.
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| Example E_1 |
o-----------------------------------------------------------o
| |
| o o o |
| | | | |
| o o o |
| \ / | |
| o---------o |
| | |
| | |
| @ |
| |
o=============================< I_2. Refold (()) >=========o
| |
| o o |
| | | |
| o o |
| / | |
| o---------o |
| | |
| | |
| @ |
| |
o=============================< I_2. Refold (()) >=========o
| |
| o |
| | |
| o |
| | |
| o---------o |
| | |
| | |
| @ |
| |
o=============================< I_2. Refold (()) >=========o
| |
| o---------o |
| | |
| | |
| @ |
| |
o=============================< I_2. Refold (()) >=========o
| |
| @ |
| |
o=============================< QEI >=======================o
In this way, one discovers the formal equation recorded below:
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| Equation E_1 |
o-----------------------------------------------------------o
| |
| o o o |
| | | | |
| o o o |
| \ / | |
| o---------o |
| | |
| | |
| @ = @ |
| |
o-----------------------------------------------------------o
| ( (()) (()) ( (()) )) = |
o-----------------------------------------------------------o
Using the square bracket notation for a "formal equivalence class" (FEC),
one says that "( (()) (()) ( (()) ))" is in the FEC [""] = [!e!] of the
empty string, or since we are tacitly ignoring blank spaces, in the FEC
of the blank character. Consequently, ["( (()) (()) ( (()) ))"] = [""].
Jon Awbrey
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