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 → ο be given.
Let x4 of type ι be given.
Let x5 of type ι be given.
Apply H0 with
λ x6 . x6 = pack_b_b_r_e_e x0 x1 x2 x3 x4 x5 ⟶ ∀ x7 . x7 ∈ x0 ⟶ ∀ x8 . x8 ∈ x0 ⟶ x1 x7 x8 ∈ x0 leaving 2 subgoals.
Let x6 of type ι be given.
Let x7 of type ι → ι → ι be given.
Assume H1: ∀ x8 . x8 ∈ x6 ⟶ ∀ x9 . x9 ∈ x6 ⟶ x7 x8 x9 ∈ x6.
Let x8 of type ι → ι → ι be given.
Assume H2: ∀ x9 . x9 ∈ x6 ⟶ ∀ x10 . x10 ∈ x6 ⟶ x8 x9 x10 ∈ x6.
Let x9 of type ι → ι → ο be given.
Let x10 of type ι be given.
Assume H3: x10 ∈ x6.
Let x11 of type ι be given.
Assume H4: x11 ∈ x6.
Apply pack_b_b_r_e_e_inj with
x6,
x0,
x7,
x1,
x8,
x2,
x9,
x3,
x10,
x4,
x11,
x5,
∀ x12 . x12 ∈ x0 ⟶ ∀ x13 . x13 ∈ x0 ⟶ x1 x12 x13 ∈ x0 leaving 2 subgoals.
The subproof is completed by applying H5.
Assume H6:
and (and (and (and (x6 = x0) (∀ x12 . x12 ∈ x6 ⟶ ∀ x13 . x13 ∈ x6 ⟶ x7 x12 x13 = x1 x12 x13)) (∀ x12 . x12 ∈ x6 ⟶ ∀ x13 . x13 ∈ x6 ⟶ x8 x12 x13 = x2 x12 x13)) (∀ x12 . x12 ∈ x6 ⟶ ∀ x13 . x13 ∈ x6 ⟶ x9 x12 x13 = x3 x12 x13)) (x10 = x4).
Apply H6 with
x11 = x5 ⟶ ∀ x12 . x12 ∈ x0 ⟶ ∀ x13 . x13 ∈ x0 ⟶ x1 x12 x13 ∈ x0.
Assume H7:
and (and (and (x6 = x0) (∀ x12 . x12 ∈ x6 ⟶ ∀ x13 . x13 ∈ x6 ⟶ x7 x12 x13 = x1 x12 x13)) (∀ x12 . x12 ∈ x6 ⟶ ∀ x13 . x13 ∈ x6 ⟶ x8 x12 x13 = x2 x12 x13)) (∀ x12 . x12 ∈ x6 ⟶ ∀ x13 . x13 ∈ x6 ⟶ x9 x12 x13 = x3 x12 x13).
Apply H7 with
x10 = x4 ⟶ x11 = x5 ⟶ ∀ x12 . x12 ∈ x0 ⟶ ∀ x13 . x13 ∈ x0 ⟶ x1 x12 x13 ∈ x0.
Assume H8:
and (and (x6 = x0) (∀ x12 . x12 ∈ x6 ⟶ ∀ x13 . x13 ∈ x6 ⟶ x7 x12 x13 = x1 x12 x13)) (∀ x12 . ... ⟶ ∀ x13 . ... ⟶ x8 x12 x13 = x2 x12 ...).