Nope, I took them (without ‘missing’ or ‘ignoring’ any “parts”) as clearly illogical statements as that problem remains simply yours. Once again you evidently simply do not understand ordering.
EDIT:
The Man I am not a participator of your "
running after my own tail" closed game.
A = “(that has no predecessor) AND (that has no successor)”
B = “(that has no successor) AND (that has no predecessor)"
Every reasonable person, but you, that sees a statement like:
“((A
AND B) is different than (B
AND A)) is ordered” immediately understands that this statement is about the ability to strictly distinguish between A and B inputs under different orders, where
AND connective is used here as simultaneous existence of A;B strict inputs, which provides a strict result. The commutativity of
AND connective has no impact on the fact that the result is strict, and the result is strict because A;B inputs are strict.
Also every reasonable person, but you, that sees a statement like:
“((A
AND B) is the same as (B
AND A)) is unordered” immediately understands that this statement is about the inability to strictly distinguish between A and B inputs under different orders (because A;B values are in superposition, which prevent their strict ids), where
AND connective is used here as simultaneous existence of A;B non-strict inputs, which provides a non-strict result. The commutativity of
AND connective has no impact on the fact that the result is non-strict, and the result is non-strict because A;B inputs are non-strict.
Let's take, for example, the 2-Uncertainty
x 2-Redundancy Distinction-Tree:
Code:
(AB,AB) (AB,A) (AB,B) (AB) (A,A) (B,B) (A,B) (A) (B) ()
A * * A * * A * . A * . A * * A . . A * . A * . A . . A . .
| | | | | | | | | | | | | | | | | | | |
B *_* B *_. B *_* B *_. B ._. B *_* B ._* B ._. B *_. B ._.
(2,2) = (AB,AB)
(2,1) = (AB,A),(AB,B)
(2,0)= (AB)
(1,1) = (A,A),(B,B),(A,B)
(1,0)= (A),(B)
(0,0)= ()
Any appearance of that tree is called Distinction State (DS), where any DS is under a structure called Frame (F), for example: (AB ,B ) is a DS that is under (2,1) F. The order in each DS or F has no significance (similar to {a,b}={b,a}) but any DS is the basis of any possible order (similar to the concept of Set as being the basis of permutations).
--------------------------
Some example:
A = True
B = False
AB
AND AB has non-strict result because the input is non-strict.
AB
AND A or A
AND AB has non-strict result because the input is non-strict (the commutativity of
AND connective has no impact on the non-strict result).
AB
AND B or B
AND AB has non-strict result because the input is non-strict (the commutativity of
AND connective has no impact on the non-strict result).
A
AND B or B
AND A is strictly False (the commutativity of
AND connective has no impact on the strict result).
--------------------------
The current use of
AND connective inputs is limited only to DS (A,A),(B,B),(A,B) under F (1,1), and by this limitation we get only strict results, where the commutativity of
AND connective has no impact on the strict property of the results.
Organic Mathematics is not limited only to strict inputs, and without this limitation we get also non-strict results, where also in this case the commutativity of
AND connective has no impact on the non-strict property of the results.
In other words The Man, your "
running after my own tail" closed game, does not even scratch my arguments.
Once again you evidently simply do not understand ordering.
Once again you evidently simply do not understand symmetrical and non-strict values (which are naturally unordered) exactly because your reasoning is limited only to DS (A,A),(B,B),(A,B) under F (1,1). You simply can't get anything beyond it.
-------------------------
Furthermore, you have asked about OM results, so one of OM results is the exposure of the limitations of your reasoning to DS (A,A),(B,B),(A,B) under F (1,1). You simply can't get anything beyond it and as a result you don't understand that the commutativity of
AND connective has no impact on the result, whether it is strict or non-strict.