Let x0 of type ι → (ι → ι) → (ι → ι) → ι → ο be given.
Let x1 of type ι be given.
Let x2 of type ι → ι be given.
Let x3 of type ι → ι be given.
Let x4 of type ι be given.
Assume H0: ∀ x5 : ι → ι . (∀ x6 . x6 ∈ x1 ⟶ x2 x6 = x5 x6) ⟶ ∀ x6 : ι → ι . (∀ x7 . x7 ∈ x1 ⟶ x3 x7 = x6 x7) ⟶ x0 x1 x5 x6 x4 = x0 x1 x2 x3 x4.
Apply pack_u_u_e_0_eq2 with
x1,
x2,
x3,
x4,
λ x5 x6 . x0 x5 (ap (ap (pack_u_u_e x1 x2 x3 x4) 1)) (ap (ap (pack_u_u_e x1 x2 x3 x4) 2)) (ap (pack_u_u_e x1 x2 x3 x4) 3) = x0 x1 x2 x3 x4.
Apply pack_u_u_e_3_eq2 with
x1,
x2,
x3,
x4,
λ x5 x6 . x0 x1 (ap (ap (pack_u_u_e x1 x2 x3 x4) 1)) (ap (ap (pack_u_u_e x1 x2 x3 x4) 2)) x5 = x0 x1 x2 x3 x4.
Apply H0 with
ap (ap (pack_u_u_e x1 x2 x3 x4) 1),
ap (ap (pack_u_u_e x1 x2 x3 x4) 2) leaving 2 subgoals.
The subproof is completed by applying pack_u_u_e_1_eq2 with x1, x2, x3, x4.
The subproof is completed by applying pack_u_u_e_2_eq2 with x1, x2, x3, x4.