Let x0 of type ι → (ι → ι → ι) → (ι → ι → ι) → (ι → ι) → ι be given.
Let x1 of type ι be given.
Let x2 of type ι → ι → ι be given.
Let x3 of type ι → ι → ι be given.
Let x4 of type ι → ι be given.
Assume H0: ∀ x5 : ι → ι → ι . (∀ x6 . x6 ∈ x1 ⟶ ∀ x7 . x7 ∈ x1 ⟶ x2 x6 x7 = x5 x6 x7) ⟶ ∀ x6 : ι → ι → ι . (∀ x7 . x7 ∈ x1 ⟶ ∀ x8 . x8 ∈ x1 ⟶ x3 x7 x8 = x6 x7 x8) ⟶ ∀ x7 : ι → ι . (∀ x8 . x8 ∈ x1 ⟶ x4 x8 = x7 x8) ⟶ x0 x1 x5 x6 x7 = x0 x1 x2 x3 x4.
Apply pack_b_b_u_0_eq2 with
x1,
x2,
x3,
x4,
λ x5 x6 . x0 x5 (decode_b (ap (pack_b_b_u x1 x2 x3 x4) 1)) (decode_b (ap (pack_b_b_u x1 x2 x3 x4) 2)) (ap (ap (pack_b_b_u x1 x2 x3 x4) 3)) = x0 x1 x2 x3 x4.
Apply H0 with
decode_b (ap (pack_b_b_u x1 x2 x3 x4) 1),
decode_b (ap (pack_b_b_u x1 x2 x3 x4) 2),
ap (ap (pack_b_b_u x1 x2 x3 x4) 3) leaving 3 subgoals.
The subproof is completed by applying pack_b_b_u_1_eq2 with x1, x2, x3, x4.
The subproof is completed by applying pack_b_b_u_2_eq2 with x1, x2, x3, x4.
The subproof is completed by applying pack_b_b_u_3_eq2 with x1, x2, x3, x4.