Hypergendered Logic Source
Hypergendered Logic — Page 88
and SM^(alpha+1) = IH(SM^alpha) AND IH(NOT SF^alpha) AND PH(SM^alpha) AND NOT PH(SF^alpha) AND UD(SM^alpha) AND NOT FD(SM^alpha), for every defined ordinal alpha >= 2 This recursion has four formally distinct preservation/generation roles. Inherited own-side preservation: IHSFn or IHSMn carries forward the full own-side preceding state with its source/provenance. Inherited opposite-side non-state productivity: IH¬SMn or IH¬SFn carries forward the structured logical non-state of the newly enriched opposite-side stage. Whole own-side phenotype preservation: PHSFn or PHSMn makes the complete phenotype-profile of the preceding own-side whole an explicit positive phenotype source at the next stage. Opposite whole-profile exclusion: NOT PH(SM_n) or NOT PH(SF_n) denies complete possession of the opposite predecessor phenotype-profile. This is not positive possession of a nonstate-profile and does not erase the separately inherited structured nonstate carried by IH(NOT opposite). The combination means the ladder does not merely oscillate between fixed Boolean formulas. Each side recursively preserves its own logical/provenance history, absorbs structured information about a new opposite-side nonstate, positively possesses its own immediately preceding whole phenotype-profile, explicitly excludes the opposite predecessor whole-profile, and quality-types only the own-side predecessor through UD plus NOT-FD. 141. Why the generator does not need nested PA A tempting alternative would be , but that changes the spoken type of PA and is excluded. The 𝑃𝐴𝑃𝐴φ ( ) ( ) HGL ladder already has a recursive state-to-state operator: IH. At every stage the previous stage is placed under another IH layer. Consequently the ladder obtains unlimited finite inheritance depth while PA keeps one stable natural-language meaning. 142. Inheritance depth proof, fully spelled out Base states: dIHstageSF1=dIHstageSM1=0. Stage 2 contains IHSF1 and IH¬SM1, so dIHstageSF2=1. Likewise dIHstageSM2=1. If Stage has depth , then Stage contains applied to a depth- state, creating depth . 𝑛 𝑛−1 𝑛+ 1 𝐼𝐻 𝑛−1 ( ) 𝑛 Therefore by induction dIHstageSFn=dIHstageSMn=n−1. Examples: dIHstageSF3=2, dIHstageSF10=9, dIHstageSF100=99, dIHstageSM1000=999. This is the formal answer to the earlier saturation objection.
Source Canon: Verbatim mathematical and modal logic. Interactive navigation and operator lookups available at /#hgl-part:88.