SUO: Re: Zeroth Order Ontology
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ZOO. Note 13
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So long as we still have these maximally simple examples at the ready,
let us exploit them to introduce some of the notation and terminology
that we'll be needing later on. The syntacto-graphic entity e_3 that
we used a while ago is not really a member of the club L_2, but a bit
more like a presiding pattern whose precedents in L_2 are e_1 and e_2.
What it really is, is nothing but a name for the set {e_1, e_2} c L_2.
If you find yourself tempted to dub such a thing a "metaformula", let
me strongly counsel you to resist that occlusion of syntax, as I find
that more and more folks are getting themselves tonguetied these days
with the confusion of metatongues. Instead, I will use the neologism
"epiformula" to describe things like e_3, restricting its use to what
I make explicit here, and by this epiformula to say we escape all the
mystifying connotations that have accumulated about the prefix "meta".
o-----------------------------------------------------------o
| Epiformula e_3 |
o-----------------------------------------------------------o
| |
| x o o x x o x |
| | | | |
| x o o x x o x |
| \ / | |
| o---------o |
| | |
| | |
| @ |
| |
o-----------------------------------------------------------o
| ( ( x ( x )) (( x ) x ) ( ( x ( x x ) x ) )) |
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With this epiform, as I'll call it for the sake of saving two breaths,
I can now rename e_1 and e_2 through the use of the following devices:
1. A bare terminal node, "o", is known as a "stone".
2. A bare terminal edge, "|", is known as a "stick".
3. Let the "replacement expression" of the shape "Q[o/x]" mean
the expression that results from Q by replacing every token
of the marker "x" with a blank, that is to say, erasing "x".
4. Let the "replacement expression" of the shape "Q[|/x]" mean
the expression that results from Q by replacing every token
of the marker "x" with a stick stemming from the site of "x".
5. In the case of an expression Q that does not bear a token of the
designated marker "x", it is stipulated that Q[o/x] = Q = Q[|/x].
In the case of e_1 and e_2, we have:
e_1 = e_3 [o/x]
e_2 = e_3 [|/x]
We've already noted that [e_1] = [e_3 [o/x]] = [!e!] = [e_3 [|/x]] = [e_2].
Jon Awbrey
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