Let x0 of type ο be given.
Apply H0 with
pack_r 0 (λ x1 x2 . False).
Let x1 of type ο be given.
Apply H1 with
λ x2 . lam 0 (λ x3 . 0).
Apply andI with
struct_r (pack_r 0 (λ x2 x3 . False)),
∀ x2 . struct_r x2 ⟶ and (BinRelnHom (pack_r 0 (λ x3 x4 . False)) x2 ((λ x3 . lam 0 (λ x4 . 0)) x2)) (∀ x3 . BinRelnHom (pack_r 0 (λ x4 x5 . False)) x2 x3 ⟶ x3 = (λ x4 . lam 0 (λ x5 . 0)) x2) leaving 2 subgoals.
The subproof is completed by applying pack_struct_r_I with
0,
λ x2 x3 . False.
Let x2 of type ι be given.
Apply H2 with
λ x3 . and (BinRelnHom (pack_r 0 (λ x4 x5 . False)) x3 ((λ x4 . lam 0 (λ x5 . 0)) x3)) (∀ x4 . BinRelnHom (pack_r 0 (λ x5 x6 . False)) x3 x4 ⟶ x4 = (λ x5 . lam 0 (λ x6 . 0)) x3).
Let x3 of type ι be given.
Let x4 of type ι → ι → ο be given.
Apply andI with
BinRelnHom (pack_r 0 (λ x5 x6 . False)) (pack_r x3 x4) ((λ x5 . lam 0 (λ x6 . 0)) (pack_r x3 x4)),
∀ x5 . BinRelnHom (pack_r 0 (λ x6 x7 . False)) (pack_r x3 x4) x5 ⟶ x5 = (λ x6 . lam 0 (λ x7 . 0)) (pack_r x3 x4) leaving 2 subgoals.
Apply unknownprop_4e486761c3790f4990f398ce8c16ea7ac5915924a294f8e5b06e45030e68e983 with
0,
x3,
λ x5 x6 . False,
x4,
lam 0 (λ x5 . 0),
λ x5 x6 : ο . x6.
Apply andI with
lam 0 (λ x5 . 0) ∈ setexp x3 0,
∀ x5 . x5 ∈ 0 ⟶ ∀ x6 . x6 ∈ 0 ⟶ (λ x7 x8 . False) x5 x6 ⟶ x4 (ap (lam 0 (λ x7 . 0)) x5) (ap (lam 0 (λ x7 . 0)) x6) leaving 2 subgoals.
Apply lam_Pi with
0,
λ x5 . x3,
λ x5 . 0.
Let x5 of type ι be given.
Assume H3: x5 ∈ 0.
Apply FalseE with
(λ x6 . 0) x5 ∈ (λ x6 . x3) x5.
Apply EmptyE with
x5.
The subproof is completed by applying H3.
Let x5 of type ι be given.
Assume H3: x5 ∈ 0.
Apply FalseE with
∀ x6 . x6 ∈ 0 ⟶ (λ x7 x8 . False) x5 x6 ⟶ x4 (ap (lam 0 (λ x7 . 0)) x5) (ap (lam 0 (λ x7 . 0)) x6).
Apply EmptyE with
x5.
The subproof is completed by applying H3.
Let x5 of type ι be given.
Apply unknownprop_4e486761c3790f4990f398ce8c16ea7ac5915924a294f8e5b06e45030e68e983 with
0,
x3,
λ x6 x7 . False,
x4,
x5,
λ x6 x7 : ο . x7 ⟶ x5 = (λ x8 . lam 0 (λ x9 . 0)) (pack_r x3 x4).
Assume H3:
and (x5 ∈ setexp x3 0) (∀ x6 . ... ⟶ ∀ x7 . ... ⟶ ... ⟶ x4 (ap x5 x6) (ap ... ...)).