Let x0 of type ι be given.
Let x1 of type ι → ι → ι be given.
Let x2 of type ι → ι be given.
Let x3 of type ι → ι → ο be given.
Let x4 of type ι be given.
Apply H0 with
λ x5 . x5 = pack_b_u_r_e 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 → ι be given.
Assume H1: ∀ x7 . x7 ∈ x5 ⟶ ∀ x8 . x8 ∈ x5 ⟶ x6 x7 x8 ∈ x5.
Let x7 of type ι → ι be given.
Assume H2: ∀ x8 . x8 ∈ x5 ⟶ x7 x8 ∈ x5.
Let x8 of type ι → ι → ο be given.
Let x9 of type ι be given.
Assume H3: x9 ∈ x5.
Apply pack_b_u_r_e_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 H4.
Assume H5:
and (and (and (x5 = x0) (∀ x10 . x10 ∈ x5 ⟶ ∀ x11 . x11 ∈ x5 ⟶ x6 x10 x11 = x1 x10 x11)) (∀ x10 . x10 ∈ x5 ⟶ x7 x10 = x2 x10)) (∀ x10 . x10 ∈ x5 ⟶ ∀ x11 . x11 ∈ x5 ⟶ x8 x10 x11 = x3 x10 x11).
Apply H5 with
x9 = x4 ⟶ ∀ x10 . x10 ∈ x0 ⟶ x2 x10 ∈ x0.
Assume H6:
and (and (x5 = x0) (∀ x10 . x10 ∈ x5 ⟶ ∀ x11 . x11 ∈ x5 ⟶ x6 x10 x11 = x1 x10 x11)) (∀ x10 . x10 ∈ x5 ⟶ x7 x10 = x2 x10).
Apply H6 with
(∀ x10 . x10 ∈ x5 ⟶ ∀ x11 . x11 ∈ x5 ⟶ x8 x10 x11 = x3 x10 x11) ⟶ x9 = x4 ⟶ ∀ x10 . x10 ∈ x0 ⟶ x2 x10 ∈ x0.
Assume H7:
and (x5 = x0) (∀ x10 . x10 ∈ x5 ⟶ ∀ x11 . x11 ∈ x5 ⟶ x6 x10 x11 = x1 x10 x11).
Apply H7 with
(∀ x10 . x10 ∈ x5 ⟶ x7 x10 = x2 x10) ⟶ (∀ x10 . x10 ∈ x5 ⟶ ∀ x11 . x11 ∈ x5 ⟶ x8 x10 x11 = x3 x10 x11) ⟶ x9 = x4 ⟶ ∀ x10 . x10 ∈ x0 ⟶ x2 x10 ∈ x0.
Assume H8: x5 = x0.
Assume H9: ∀ x10 . x10 ∈ x5 ⟶ ∀ x11 . x11 ∈ x5 ⟶ x6 x10 x11 = x1 x10 x11.
Assume H10: ∀ x10 . x10 ∈ x5 ⟶ x7 x10 = x2 x10.