Let x0 of type ι be given.
Let x1 of type ι → ι be given.
Let x2 of type ι → ι be given.
Let x3 of type ι → ο be given.
Let x4 of type ι → ο be given.
Apply H0 with
λ x5 . x5 = pack_u_u_p_p x0 x1 x2 x3 x4 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x2 x6 ∈ x0 leaving 2 subgoals.
Let x5 of type ι be given.
Let x6 of type ι → ι be given.
Assume H1: ∀ x7 . x7 ∈ x5 ⟶ x6 x7 ∈ x5.
Let x7 of type ι → ι be given.
Assume H2: ∀ x8 . x8 ∈ x5 ⟶ x7 x8 ∈ x5.
Let x8 of type ι → ο be given.
Let x9 of type ι → ο be given.
Apply pack_u_u_p_p_inj with
x5,
x0,
x6,
x1,
x7,
x2,
x8,
x3,
x9,
x4,
∀ x10 . x10 ∈ x0 ⟶ x2 x10 ∈ x0 leaving 2 subgoals.
The subproof is completed by applying H3.
Assume H4:
and (and (and (x5 = x0) (∀ x10 . x10 ∈ x5 ⟶ x6 x10 = x1 x10)) (∀ x10 . x10 ∈ x5 ⟶ x7 x10 = x2 x10)) (∀ x10 . x10 ∈ x5 ⟶ x8 x10 = x3 x10).
Apply H4 with
(∀ x10 . x10 ∈ x5 ⟶ x9 x10 = x4 x10) ⟶ ∀ x10 . x10 ∈ x0 ⟶ x2 x10 ∈ x0.
Assume H5:
and (and (x5 = x0) (∀ x10 . x10 ∈ x5 ⟶ x6 x10 = x1 x10)) (∀ x10 . x10 ∈ x5 ⟶ x7 x10 = x2 x10).
Apply H5 with
(∀ x10 . x10 ∈ x5 ⟶ x8 x10 = x3 x10) ⟶ (∀ x10 . x10 ∈ x5 ⟶ x9 x10 = x4 x10) ⟶ ∀ x10 . x10 ∈ x0 ⟶ x2 x10 ∈ x0.
Assume H6:
and (x5 = x0) (∀ x10 . x10 ∈ x5 ⟶ x6 x10 = x1 x10).
Apply H6 with
(∀ x10 . x10 ∈ x5 ⟶ x7 x10 = x2 x10) ⟶ (∀ x10 . x10 ∈ x5 ⟶ x8 x10 = x3 x10) ⟶ (∀ x10 . x10 ∈ x5 ⟶ x9 x10 = x4 x10) ⟶ ∀ x10 . x10 ∈ x0 ⟶ x2 x10 ∈ x0.
Assume H7: x5 = x0.
Assume H8: ∀ x10 . x10 ∈ x5 ⟶ x6 x10 = x1 x10.
Assume H9: ∀ x10 . x10 ∈ x5 ⟶ x7 x10 = x2 x10.
Assume H10: ∀ x10 . x10 ∈ x5 ⟶ x8 x10 = x3 x10.
Assume H11: ∀ x10 . x10 ∈ x5 ⟶ x9 x10 = x4 x10.
Apply H7 with
λ x10 x11 . ∀ x12 . x12 ∈ x10 ⟶ x2 x12 ∈ x10.
Let x10 of type ι be given.
Assume H12: x10 ∈ x5.
Apply H9 with
x10,
... leaving 2 subgoals.