Part W: base-free pencil W_2 = <y^2, z^2> (Q_D = Y Z): weighted (4,4,5) linkage
  ok   seed 1: H = [1, 3, 5, 5, 5, 2], gamma_f = 0, CI(f + 3 b^4, K_{y^2}, g) local length 40 <= 40
  ok   seed 2: H = [1, 3, 5, 5, 5, 2], gamma_f = 0, CI(f + 3 b^4, K_{y^2}, g) local length 40 <= 40
  ok   seed 3: H = [1, 3, 5, 5, 5, 2], gamma_f = 0, CI(f + 3 b^4, K_{y^2}, g) local length 40 <= 40
  ok   seed 4: H = [1, 3, 5, 5, 5, 2], gamma_f = 0, CI(f + 3 b^4, K_{y^2}, g) local length 40 <= 40
  ok   Part W: 4 accepted samples
  ok   Part W anchor certified by stabilization: 40 (N = 10) <= 40
Part R: residual stratum W_2 = z<y,z> (Q_D = Y^2), D_4 = binary quartics, D_5 = <Y^5, G>, theta_m(Y) = 0
  ok   G = Y^3*Z^2 + Z^5: H = [1, 3, 5, 5, 5, 2], f_3 in x^2 S_1 after the shear: -2/45*x^2*y - 7/100*x^2*z
  ok     theta_{yz}(Y) = 0, theta_{z^2}(Y) = 0 (residual: both 0); J_4 = 0: no element with binary quartic part y^4 + ...
  ok     Z4 exists: y^4
  ok     x Z4, z Z4 in I; y Z4 not in I
  ok     dim I/m^6 = 35, dim N/m^6 = 32 (= dim - 3: f, K, yK span a complement)
  ok     (R1) tau = 1: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = 2: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = -3: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R2) tau = 0: CI(f + 3 b^4, K_{yz} + 5 K_{z^2}, g) local length 42 (= 42: the bad point [Y])
  ok     (R3) tau = 1, t = 3: CI(f + t b^4, K + tau Z4 + 5 K_{z^2}, g) local length 40 <= 40
  ok     (R3) tau = 2, t = -1: CI(f + t b^4, K + tau Z4 + 7 K_{z^2}, g) local length 40 <= 40
  ok   G = Y^2*Z^3 - 2*Y*Z^4 + Z^5: H = [1, 3, 5, 5, 5, 2], f_3 in x^2 S_1 after the shear: -4/115*x^2*y - 121/345*x^2*z
  ok     theta_{yz}(Y) = 0, theta_{z^2}(Y) = 0 (residual: both 0); J_4 = 0: no element with binary quartic part y^4 + ...
  ok     Z4 exists: y^4
  ok     x Z4, z Z4 in I; y Z4 not in I
  ok     dim I/m^6 = 35, dim N/m^6 = 32 (= dim - 3: f, K, yK span a complement)
  ok     (R1) tau = 1: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = 2: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = -3: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R2) tau = 0: CI(f + 3 b^4, K_{yz} + 5 K_{z^2}, g) local length 42 (= 42: the bad point [Y])
  ok     (R3) tau = 1, t = 3: CI(f + t b^4, K + tau Z4 + 5 K_{z^2}, g) local length 40 <= 40
  ok     (R3) tau = 2, t = -1: CI(f + t b^4, K + tau Z4 + 7 K_{z^2}, g) local length 40 <= 40
  ok   G = Y^4*Z + Z^5: H = [1, 3, 5, 5, 5, 2], f_3 in x^2 S_1 after the shear: -1/20*x^2*y + 8/55*x^2*z
  ok     theta_{yz}(Y) = 0, theta_{z^2}(Y) = 0 (residual: both 0); J_4 = 0: no element with binary quartic part y^4 + ...
  ok     Z4 exists: y^4 - 1/5*z^4
  ok     x Z4, z Z4 in I; y Z4 not in I
  ok     dim I/m^6 = 35, dim N/m^6 = 32 (= dim - 3: f, K, yK span a complement)
  ok     (R1) tau = 1: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = 2: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = -3: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R2) tau = 0: CI(f + 3 b^4, K_{yz} + 5 K_{z^2}, g) local length 42 (= 42: the bad point [Y])
  ok     (R3) tau = 1, t = 3: CI(f + t b^4, K + tau Z4 + 5 K_{z^2}, g) local length 40 <= 40
  ok     (R3) tau = 2, t = -1: CI(f + t b^4, K + tau Z4 + 7 K_{z^2}, g) local length 40 <= 40
  ok   G = Y^3*Z^2 + 5*Y*Z^4 - Z^5: H = [1, 3, 5, 5, 5, 2], f_3 in x^2 S_1 after the shear: 1/60*x^2*y - 1/860*x^2*z
  ok     theta_{yz}(Y) = 0, theta_{z^2}(Y) = 0 (residual: both 0); J_4 = 0: no element with binary quartic part y^4 + ...
  ok     Z4 exists: y^4
  ok     x Z4, z Z4 in I; y Z4 not in I
  ok     dim I/m^6 = 35, dim N/m^6 = 32 (= dim - 3: f, K, yK span a complement)
  ok     (R1) tau = 1: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = 2: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = -3: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R2) tau = 0: CI(f + 3 b^4, K_{yz} + 5 K_{z^2}, g) local length 42 (= 42: the bad point [Y])
  ok     (R3) tau = 1, t = 3: CI(f + t b^4, K + tau Z4 + 5 K_{z^2}, g) local length 40 <= 40
  ok     (R3) tau = 2, t = -1: CI(f + t b^4, K + tau Z4 + 7 K_{z^2}, g) local length 40 <= 40
  ok   G = Y^4*Z + 3*Y^3*Z^2 + Y^2*Z^3 + Z^5: H = [1, 3, 5, 5, 5, 2], f_3 in x^2 S_1 after the shear: -2/15*x^2*y + 38/165*x^2*z
  ok     theta_{yz}(Y) = 0, theta_{z^2}(Y) = 0 (residual: both 0); J_4 = 0: no element with binary quartic part y^4 + ...
  ok     Z4 exists: y^4 - 1/5*z^4
  ok     x Z4, z Z4 in I; y Z4 not in I
  ok     dim I/m^6 = 35, dim N/m^6 = 32 (= dim - 3: f, K, yK span a complement)
  ok     (R1) tau = 1: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = 2: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R1) tau = -3: slid ideal has colength 21 and Hilbert function [1, 3, 5, 5, 5, 2]
  ok     (R2) tau = 0: CI(f + 3 b^4, K_{yz} + 5 K_{z^2}, g) local length 42 (= 42: the bad point [Y])
  ok     (R3) tau = 1, t = 3: CI(f + t b^4, K + tau Z4 + 5 K_{z^2}, g) local length 40 <= 40
  ok     (R3) tau = 2, t = -1: CI(f + t b^4, K + tau Z4 + 7 K_{z^2}, g) local length 40 <= 40
  ok   Part R: 5 accepted samples
  ok   Part R anchor certified by stabilization: 40 (N = 10) <= 40
Part U: universal identities
  ok   U1: the functional <., z o G> on binary quartics is proportional to <., Y^4> iff a2 = a3 = a4 = a5 = 0: [24*a1, 12*a2, 12*a3, 24*a4, 120*a5]
  ok   U2: y z^2 kills Y^5 and Y^4 Z
ALL CHECKS PASSED
