Let x0 of type ι → ο be given.
Claim L1:
∀ x1 x2 x3 . x0 x1 ⟶ x0 x2 ⟶ Hom_b_b_e x1 x2 x3 ⟶ x3 ∈ setexp (ap x2 0) (ap x1 0)
Let x1 of type ι be given.
Let x2 of type ι be given.
Let x3 of type ι be given.
Assume H1: x0 x1.
Assume H2: x0 x2.
Apply H0 with
x1,
λ x4 . Hom_b_b_e x4 x2 x3 ⟶ x3 ∈ setexp (ap x2 0) (ap x4 0) leaving 2 subgoals.
The subproof is completed by applying H1.
Let x4 of type ι be given.
Let x5 of type ι → ι → ι be given.
Assume H3: ∀ x6 . x6 ∈ x4 ⟶ ∀ x7 . x7 ∈ x4 ⟶ x5 x6 x7 ∈ x4.
Let x6 of type ι → ι → ι be given.
Assume H4: ∀ x7 . x7 ∈ x4 ⟶ ∀ x8 . x8 ∈ x4 ⟶ x6 x7 x8 ∈ x4.
Let x7 of type ι be given.
Assume H5: x7 ∈ x4.
Apply H0 with
x2,
λ x8 . Hom_b_b_e (pack_b_b_e x4 x5 x6 x7) x8 x3 ⟶ x3 ∈ setexp (ap x8 0) (ap (pack_b_b_e x4 x5 x6 x7) 0) leaving 2 subgoals.
The subproof is completed by applying H2.
Let x8 of type ι be given.
Let x9 of type ι → ι → ι be given.
Assume H6: ∀ x10 . x10 ∈ x8 ⟶ ∀ x11 . x11 ∈ x8 ⟶ x9 x10 x11 ∈ x8.
Let x10 of type ι → ι → ι be given.
Assume H7: ∀ x11 . x11 ∈ x8 ⟶ ∀ x12 . x12 ∈ x8 ⟶ x10 x11 x12 ∈ x8.
Let x11 of type ι be given.
Assume H8: x11 ∈ x8.
Apply unknownprop_10267d1d502dfe147b6d457a56b873d600165a3fde1bf35785171c85aa221639 with
x4,
x8,
x5,
x6,
x9,
x10,
x7,
x11,
x3,
λ x12 x13 : ο . x13 ⟶ x3 ∈ setexp (ap (pack_b_b_e x8 x9 x10 x11) 0) (ap (pack_b_b_e x4 x5 x6 x7) 0).
Assume H9:
and (and (and (x3 ∈ setexp x8 x4) (∀ x12 . x12 ∈ x4 ⟶ ∀ x13 . x13 ∈ x4 ⟶ ap x3 (x5 x12 x13) = x9 (ap x3 x12) (ap x3 x13))) (∀ x12 . x12 ∈ x4 ⟶ ∀ x13 . x13 ∈ x4 ⟶ ap x3 (x6 x12 x13) = x10 (ap x3 x12) (ap x3 x13))) (ap x3 x7 = x11).
Apply pack_b_b_e_0_eq2 with
x8,
x9,
x10,
x11,
λ x12 x13 . x3 ∈ setexp x12 (ap (pack_b_b_e x4 x5 x6 x7) 0).
Apply pack_b_b_e_0_eq2 with
x4,
x5,
x6,
x7,
λ x12 x13 . x3 ∈ setexp x8 x12.
Apply H9 with
x3 ∈ setexp x8 x4.
Assume H10:
and (and (x3 ∈ setexp x8 x4) (∀ x12 . ... ⟶ ∀ x13 . ... ⟶ ap x3 (x5 x12 x13) = x9 (ap x3 x12) (ap x3 ...))) ....
Apply unknownprop_cb7abf829499aec888363ff9292dd7680786c42dc92f10fdd88dc16ada048723 with
x0,
λ x1 . ap x1 0,
Hom_b_b_e.
The subproof is completed by applying L1.