Void of Void; Absolute; Relative; Recursion

[Eodnhoj7]

Is the answer to what "or" or "not or" when LEM is applied? Why wouldn't "or" = "or" under LEM?

I think you fail to see that if LEM is to have identity, "or" is subject to identity, thus any "or" between identities allows "or" as subject to itself and in turn becomes a dichotomy where

"Or" is chosen at which case "not or" cannot occur thus resulting in an absence of contrast for "or", thus no identity for "or", while simultaneously in which case there is no "not or" and '"or" or "not or"' is negated to a recursion of '"or"or'.

Or "not or" is chosen at which point the "or" operator, being subject to identity, is negated and LEM negates itself.

Simultaneously there is either one LI or the other as
(A=A) = LI
(-A=-A) = LI

Results in "LI or LI" at the meta level and the "or" operater ceases as there is no choice in that context

While from another simultaneous angle it can be show as
"LI or -LI" at which point

LI is chosen and LI exists without LNC or LEM, as not LI, or
-LI is chosen and LNC and LEM contain no identities and have no identities thus LEM and LNC cease.

So LI = the Law of Identity, which states that any entity in logic is exactly equal to itself.
LEM = the Law of the Excluded Middle, which states that for any statement in logic, the statement is true or else the negation of the statement is true, with no middle ground.
LNC = the law of Noncontradiction, which states that a statement in logic and its contradiction cannot simultaneously be true.

The Law of Identity is generally expressed as (A=A), (B=B), (C=C), (or whatever abbreviation one chooses), etc. So are you saying that The Law of Identity = (A=A) = (B=B) = (C=C) and therefore, A=B=C? Thus, literally, ALL things are always equal to each other, whereby one can simply replace proposition (P) at will with any and all other possible propositions?

2 line segments, as a new line segment, are unequal by scale to the larger line segment, and potentially eachother, but all 3 are equal as line segments.

In the context of eachother they are unequal.
In the context of a line segment they are equal.
Equality and inequality exist simultaneously when all contexts are observed.

So

A=A
A=/=A

...simultaneously.

What is a constant fixed point is A, as A occurs in both A=A and A=/=A, or the line segment in the applied example.