Let x0 of type ι → (ι → ι → ι) → (ι → ι) → (ι → ι → ο) → 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 be given.
Let x4 of type ι → ι → ο be given.
Assume H0:
∀ x5 : ι → ι → ι . (∀ x6 . x6 ∈ x1 ⟶ ∀ x7 . x7 ∈ x1 ⟶ x2 x6 x7 = x5 x6 x7) ⟶ ∀ x6 : ι → ι . (∀ x7 . x7 ∈ x1 ⟶ x3 x7 = x6 x7) ⟶ ∀ x7 : ι → ι → ο . (∀ x8 . x8 ∈ x1 ⟶ ∀ x9 . x9 ∈ x1 ⟶ iff (x4 x8 x9) (x7 x8 x9)) ⟶ x0 x1 x5 x6 x7 = x0 x1 x2 x3 x4.
Apply pack_b_u_r_0_eq2 with
x1,
x2,
x3,
x4,
λ x5 x6 . x0 x5 (decode_b (ap (pack_b_u_r x1 x2 x3 x4) 1)) (ap (ap (pack_b_u_r x1 x2 x3 x4) 2)) (decode_r (ap (pack_b_u_r x1 x2 x3 x4) 3)) = x0 x1 x2 x3 x4.
Apply H0 with
decode_b (ap (pack_b_u_r x1 x2 x3 x4) 1),
ap (ap (pack_b_u_r x1 x2 x3 x4) 2),
decode_r (ap (pack_b_u_r x1 x2 x3 x4) 3) leaving 3 subgoals.
The subproof is completed by applying pack_b_u_r_1_eq2 with x1, x2, x3, x4.
The subproof is completed by applying pack_b_u_r_2_eq2 with x1, x2, x3, x4.
Let x5 of type ι be given.
Assume H1: x5 ∈ x1.
Let x6 of type ι be given.
Assume H2: x6 ∈ x1.
Apply pack_b_u_r_3_eq2 with
x1,
x2,
x3,
x4,
x5,
x6,
λ x7 x8 : ο . iff (x4 x5 x6) x7 leaving 3 subgoals.
The subproof is completed by applying H1.
The subproof is completed by applying H2.
The subproof is completed by applying iff_refl with x4 x5 x6.