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