==== H = (1,3,6,5,3,1,1) ====
PASS (A1) (random, 3 of 3 candidates accepted): C[X,Y]_3 in D, U in C[X,Y]_4: P binary and the straight ray (y, z) exists  [2.1s]
PASS (A2) C = [X^3, X^2*Y, X^2*Z, Y^3]: N_X o R_i in C, x o R_i independent mod D = C + <general G>, B_D = B_C of length 4, free-tail rank >= 3  [2.3s]
PASS (A2) C = [X^3, X^2*Y, X^2*Z, X*Z^2]: N_X o R_i in C, x o R_i independent mod D = C + <general G>, B_D = B_C of length 4, free-tail rank >= 3  [2.3s]
PASS (A3) test member: dim C(P) = 5, M = T<=2 + C(P) + S o Phi has H = (1,3,6,5,3,1,1) (random R, E)  [2.6s]
PASS (A3) test member: cubic ideal = J_{10,10}  [2.6s]
PASS (A3) test member: cubic base = five reduced points  [2.7s]
PASS (A3) test member: U contains no form with X^3Y or X^3Z (so the four other points have weight 1; excess 27 + 3 + 4 = 34)  [2.7s]
PASS (A3) (random h, 5 of 5 accepted): h o X^4 T_1 ⊆ C X^3 and h o : T_5 -> T_3/C X^3 is onto (kernel of rank 9 on T_5/X^4 T_1)  [2.8s]
  (A4) random candidates accepted by dim C: {4: 0, 5: 0}; direct members by dim C: {4: 1, 5: 1}; negative controls: 2
PASS (A4) (random + direct, both strata dim C = 4, 5, negative controls): closedness of T<=2 + D + <Phi, x o Phi, x^2 o Phi, y o Phi, H + tau> <=> S_1 o U ⊆ D and R_Z ∈ D, with H = (1,3,6,5,3,1,1) when closed  [3.5s]
PASS (A4) test member n = 3: dim C = 5, dim C_Q = 3, closed with H, cubic ideal as stated  [3.7s]
PASS    base: five reduced points (excess 27 + 3 + 4 = 34)  [3.8s]
PASS (A4) test member n = 2, P = Y^5: dim C = 5, dim C_Q = 2, closed with H, cubic ideal as stated  [4.0s]
PASS    base: the reduced points [X], [Y]; k([Y]) = 5 (Y^4 in D_4, Y^5 not in D_5): excess 27 + 3 + 2 = 32  [4.1s]
PASS (A4) test member n = 2, P = X^3Y^2: dim C = 5, dim C_Q = 2, closed with H, cubic ideal as stated  [4.3s]
PASS    base: the double point V(z, y^2) at [X]: excess 27 + 3·2 = 33  [4.3s]
==== H = (1,3,6,5,2,2,1) ====
  (B) random candidates accepted per top: {'X^6 + Y^6': 4, 'X^5*Y': 4}; negative controls (criterion false): 8
PASS (B) (random, both tops, with negative controls): with P binary, M = T<=2 + D + S o Phi has H = (1,3,6,5,2,2,1) iff q o P ∈ D and z o R ∈ D  [7.4s]
PASS (B) (random general D ⊃ C): cubic base of D = cubic base of C, of length n = dim C  [7.4s]
PASS (B) shear: under z -> z + h2(x, y) the quintic part of the dual of F changes by Z (h2 o F) (F = X^6 + Y^6)  [7.4s]
PASS (B) shear: under z -> z + h2(x, y) the quintic part of the dual of F changes by Z (h2 o F) (F = X^5*Y)  [7.4s]
PASS (C1): the substitution U_4 = V + (a/15) XY^3, R_4 = R + (a^2/120) Y^4 turns the four columns into y o V, z o V, y o R - (a/2) x o V, z o R exactly  [7.4s]
PASS (C1) (5 of 5 random cases accepted, plus one direct case; 1 closed, 5 not closed): closedness <=> the four columns lie in D, with H = (1,3,6,5,2,2,1) when closed  [7.9s]
PASS (C2) (random kappa, a; 6 of 6 accepted): (V, R) -> sum kappa_i c_i is onto W (rank 8), kernel of dimension 18  [8.0s]
PASS (C2): V -> (V_Y, V_Z) mod E0 has kernel <X^3Z, X^2Y^2> on T_4/V_4  [8.0s]
PASS (C2) (random nonzero quadric in <xz, yz, z^2, y^2>, 6 of 6 accepted): V -> h o V mod C X^2 has rank 5 on T_4/V_4  [8.1s]
PASS (C2) (random nonzero l in <y, z>, 4 of 4 accepted): V -> l o V is onto W, kernel of dimension 5  [8.1s]
PASS (C3): the member is closed, dim 20, H = (1,3,6,5,2,2,1)  [8.3s]
PASS (C3): the four columns have rank 3 modulo E0 and lie in D (rank-3 locus of Y)  [8.3s]
PASS (C3): f_1, f_2, f_3 annihilate M  [8.3s]
PASS (C3): local length 35: dim S/(C + m^N) = dim S/(C + m^(N+1)) = 35 at N = 10, and standard basis (ds) = 35  [48.8s]
PASS (C3): residual scheme of length 15 = 35 - 20  [52.8s]
PASS (C3): length 10 on x = 0, length 15 on z = 0 (local, ds); 2·10 + 15 = 35  [52.8s]
total time 52.8 s
ALL CHECKS PASSED
