Hypergendered Logic Source
Hypergendered Logic — Page 151
1. is the inner condition; 𝑥∧¬𝑦 2. forms the irreducible developmental mode; 𝑃𝐴·( ) 3. inner outer-state NOT forms the non-SF1 state; 4. IH inherits that non-state; 5. the outermost NOT denies that inheritance relation. No pair of adjacent operators may be silently swapped. B.13 User-style outer De Morgan ¬𝑃𝐴𝑥 ( ) ∧¬𝑃𝐴¬𝑦 ( ) ≡¬ 𝑃𝐴𝑥 ( ) ∨𝑃𝐴¬𝑦 ( ) ( ). This is valid and does not create any new PA mode. B.14 User-style OR counterpart ¬𝑃𝐴𝑥 ( ) ∨¬𝑃𝐴¬𝑦 ( ) ≡¬ 𝑃𝐴𝑥 ( ) ∧𝑃𝐴¬𝑦 ( ) ( ). B.15 IH of a De Morgan equivalent source If 𝑆≡𝐵𝑇 and HGL adopts source-normalization for Boolean equivalents, then 𝐼𝐻𝑆 ( ) and 𝐼𝐻𝑇 ( ) may be treated as equivalent descriptions of one source. This is not a distribution rule; it is substitution of an equivalent source formula. Appendix C — Forty common invalid derivations and why they fail 1. Invalid: . Why: NOT is outside PA in the premise and inside in the conclusion. ¬𝑃𝐴𝑥 ( ) ⇒𝑃𝐴¬𝑥 ( ) 2. Invalid: automatically. Why: requires an outer consistency law not supplied merely 𝑃𝐴¬𝑥 ( ) ⇒¬𝑃𝐴𝑥 ( ) by syntax. 3. Invalid: . Why: PA irreducibility. 𝑃𝐴𝐴∧𝐵 ( ) = 𝑃𝐴𝐴 ( ) ∧𝑃𝐴𝐵 ( ) 4. Invalid: . Why: PA non-homomorphism. 𝑃𝐴𝐴∨𝐵 ( ) = 𝑃𝐴𝐴 ( ) ∨𝑃𝐴𝐵 ( ) 5. Invalid: , therefore . Why: invalid causal lifting. 𝐴∧𝐵⇒𝐴 𝑃𝐴𝐴∧𝐵 ( ) ⇒𝑃𝐴𝐴 ( ) 6. Invalid: . Why: separate causes do not reconstruct a joint cause. 𝑃𝐴𝐴 ( ) ∧𝑃𝐴𝐵 ( ) ⇒𝑃𝐴𝐴∧𝐵 ( ) 7. Invalid: SF1 = PA(x) AND PA(NOT y). Why: corrected SF1 is exactly the single irreducible compound mode PA(x AND NOT y); it contains no constitutive IH/PH structure. 8. Invalid: SM1 = PA(NOT x) AND PA(y). Why: corrected SM1 is the single irreducible compound mode PA(NOT x AND y). 9. Invalid: SF2 is any old three-, four-, or six-term predecessor formula. Why: current canonical SF2 has eight conjuncts including IH, PH, UD, standalone PA routes, and NOT FD(SF1). 10. Invalid: SM2 is the old three-term PA conjunction. Why: same. 11. Invalid: . Why: projection is one-way; inheritance carries source provenance. 𝐼𝐻𝑆 ( ) = 𝑆 12. Invalid: . Why: possession does not prove inheritance. 𝑆⇒𝐼𝐻𝑆 ( ) 13. Invalid: . Why: inheritance of non-S differs from noninheritance of S. 𝐼𝐻¬𝑆 ( ) = ¬𝐼𝐻𝑆 ( ) 14. Invalid: . Why: S can be independently present. ¬𝐼𝐻𝑆 ( ) ⇒¬𝑆 15. Invalid: and , therefore . Why: ordinary consequence does not invent a new 𝐼𝐻𝑆 ( ) 𝑆⇒𝑇 𝐼𝐻𝑇 ( ) inheritance source.
Source Canon: Verbatim mathematical and modal logic. Interactive navigation and operator lookups available at /#hgl-part:151.