Three open HAL problems resolved: (1) P-hatHAL = C₀ x QBMSGR (not QBMS): factor C₀ is the origin of epsilon=alphaₑm. (2) Gap 1 closed algebraically: n=27 unique integer satisfying Postulates II+V without tachyons, forcing alphaₑm uniquely and hence Jₜotal = C₀² JGR. (3) Lᵤnif written explicitly: LHAL + LSMgᵢ, yf/C₀, lambdaH + (alphaₑm/2C₀) |Qₜheta|². Three items remain for complete proof.
Mordecai Gavila Alvarez (Fri,) studied this question.