Let x0 of type ι be given.
Let x1 of type ι → ι → ι be given.
Let x2 of type ι → ι → ι be given.
Let x3 of type ι → ι be given.
Apply H0 with
λ x4 . x4 = pack_b_b_u x0 x1 x2 x3 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x3 x5 ∈ x0 leaving 2 subgoals.
Let x4 of type ι be given.
Let x5 of type ι → ι → ι be given.
Assume H1: ∀ x6 . x6 ∈ x4 ⟶ ∀ x7 . x7 ∈ x4 ⟶ x5 x6 x7 ∈ x4.
Let x6 of type ι → ι → ι be given.
Assume H2: ∀ x7 . x7 ∈ x4 ⟶ ∀ x8 . x8 ∈ x4 ⟶ x6 x7 x8 ∈ x4.
Let x7 of type ι → ι be given.
Assume H3: ∀ x8 . x8 ∈ x4 ⟶ x7 x8 ∈ x4.
Apply pack_b_b_u_inj with
x4,
x0,
x5,
x1,
x6,
x2,
x7,
x3,
∀ x8 . x8 ∈ x0 ⟶ x3 x8 ∈ x0 leaving 2 subgoals.
The subproof is completed by applying H4.
Assume H5:
and (and (x4 = x0) (∀ x8 . x8 ∈ x4 ⟶ ∀ x9 . x9 ∈ x4 ⟶ x5 x8 x9 = x1 x8 x9)) (∀ x8 . x8 ∈ x4 ⟶ ∀ x9 . x9 ∈ x4 ⟶ x6 x8 x9 = x2 x8 x9).
Apply H5 with
(∀ x8 . x8 ∈ x4 ⟶ x7 x8 = x3 x8) ⟶ ∀ x8 . x8 ∈ x0 ⟶ x3 x8 ∈ x0.
Assume H6:
and (x4 = x0) (∀ x8 . x8 ∈ x4 ⟶ ∀ x9 . x9 ∈ x4 ⟶ x5 x8 x9 = x1 x8 x9).
Apply H6 with
(∀ x8 . x8 ∈ x4 ⟶ ∀ x9 . x9 ∈ x4 ⟶ x6 x8 x9 = x2 x8 x9) ⟶ (∀ x8 . x8 ∈ x4 ⟶ x7 x8 = x3 x8) ⟶ ∀ x8 . x8 ∈ x0 ⟶ x3 x8 ∈ x0.
Assume H7: x4 = x0.
Assume H8: ∀ x8 . x8 ∈ x4 ⟶ ∀ x9 . x9 ∈ x4 ⟶ x5 x8 x9 = x1 x8 x9.
Assume H9: ∀ x8 . x8 ∈ x4 ⟶ ∀ x9 . x9 ∈ x4 ⟶ x6 x8 x9 = x2 x8 x9.
Assume H10: ∀ x8 . x8 ∈ x4 ⟶ x7 x8 = x3 x8.
Apply H7 with
λ x8 x9 . ∀ x10 . x10 ∈ x8 ⟶ x3 x10 ∈ x8.
Let x8 of type ι be given.
Assume H11: x8 ∈ x4.
Apply H10 with
x8,
λ x9 x10 . x9 ∈ x4 leaving 2 subgoals.
The subproof is completed by applying H11.
Apply H3 with
x8.
The subproof is completed by applying H11.
Let x4 of type ι → ι → ο be given.
The subproof is completed by applying H1.