Hypergendered Logic Source
Hypergendered Logic — Page 116
𝑃𝐻𝑍
( ) ⇒𝐻𝑎𝑠𝑇
( ) 𝑓𝑜𝑟 𝑇∈Π 𝑍
( ).
Intensional profile identity:
𝑃𝐻𝑍1
( )≢𝑃𝐻𝑍2
( )
can remain true even when the currently observable trait sets overlap, because the source profiles and
latent/developmental organization can differ.
218. Proof that PH(NOT S) does not reduce to NOT PH(S)
A countermodel is enough.
Let a subject fail to possess the complete profile because one required trait is absent. Then
𝑆
¬𝑃𝐻𝑆
( )
is true.
But suppose the subject also fails to possess the complete non- profile. Then
𝑆
𝑃𝐻¬𝑆
(
)
is false.
Therefore
¬𝑃𝐻𝑆
( ) ⇏ 𝑃𝐻¬𝑆
(
).
Conversely, a theory can stipulate
without deriving the syntactically stronger claim that every
𝑃𝐻¬𝑆
(
)
possible way of
failing has been identified with that profile. Hence the operators must remain distinct.
𝑃𝐻𝑆
( )
219. Proof that PH(A AND B) need not equal PH(A) AND PH(B)
Choose a phenotype map with
Π 𝐴
( ) = {𝑎},
Π 𝐵
( ) = {𝑏},
and
Π 𝐴∧𝐵
(
) = {𝑎, 𝑏, 𝑒},
where is an emergent compound trait.
𝑒
A bearer can satisfy
by possessing and under their respective whole-profile identities
𝑃𝐻𝐴
( ) ∧𝑃𝐻𝐵
( )
𝑎
𝑏
while lacking . Then
is false. Thus distribution fails in at least one admissible HGL model.
𝑒
𝑃𝐻𝐴∧𝐵
(
)
220. Proof that stage plateau does not imply phenotype plateau
Construct a model with stage
SFt=SF4
for all in
years, but maturity variable
strictly increasing through puberty. Let a womanhood trait
𝑡
11, 18
[
]
𝐷𝑡
( )
be caused when
crosses a threshold while SF4 remains constant.
𝑊
𝐷𝑡
( )
Then
∆𝑆𝐹= 0
but
Source Canon: Verbatim mathematical and modal logic. Interactive navigation and operator lookups available at /#hgl-part:116.