Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Apply unknownprop_2dc9e26456bb41cf29ae85d63c3e8887dbdb5a0601149d8bd58cb1df95ffb8a5 with
x0,
λ x3 . BinRelnHom x3 x1 x2 = (x2 = 0) leaving 2 subgoals.
The subproof is completed by applying H0.
Let x3 of type ι be given.
Let x4 of type ι → ι → ο be given.
Assume H2: x3 = 0.
Apply unknownprop_2dc9e26456bb41cf29ae85d63c3e8887dbdb5a0601149d8bd58cb1df95ffb8a5 with
x1,
λ x5 . BinRelnHom (pack_r x3 x4) x5 x2 = (x2 = 0) leaving 2 subgoals.
The subproof is completed by applying H1.
Let x5 of type ι be given.
Let x6 of type ι → ι → ο be given.
Assume H3: x5 = 0.
Apply unknownprop_4e486761c3790f4990f398ce8c16ea7ac5915924a294f8e5b06e45030e68e983 with
x3,
x5,
x4,
x6,
x2,
λ x7 x8 : ο . x8 = (x2 = 0).
Apply H2 with
λ x7 x8 . and (x2 ∈ setexp x5 x8) (∀ x9 . x9 ∈ x8 ⟶ ∀ x10 . x10 ∈ x8 ⟶ x4 x9 x10 ⟶ x6 (ap x2 x9) (ap x2 x10)) = (x2 = 0).
Apply H3 with
λ x7 x8 . and (x2 ∈ setexp x8 0) (∀ x9 . x9 ∈ 0 ⟶ ∀ x10 . x10 ∈ 0 ⟶ x4 x9 x10 ⟶ x6 (ap x2 x9) (ap x2 x10)) = (x2 = 0).
Apply prop_ext_2 with
and (x2 ∈ setexp 0 0) (∀ x7 . x7 ∈ 0 ⟶ ∀ x8 . x8 ∈ 0 ⟶ x4 x7 x8 ⟶ x6 (ap x2 x7) (ap x2 x8)),
x2 = 0 leaving 2 subgoals.
Assume H4:
and (x2 ∈ setexp 0 0) (∀ x7 . x7 ∈ 0 ⟶ ∀ x8 . x8 ∈ 0 ⟶ x4 x7 x8 ⟶ x6 (ap x2 x7) (ap x2 x8)).
Apply H4 with
x2 = 0.
Assume H6:
∀ x7 . x7 ∈ 0 ⟶ ∀ x8 . x8 ∈ 0 ⟶ x4 x7 x8 ⟶ x6 (ap x2 x7) (ap x2 x8).
Apply Pi_eta with
0,
λ x7 . 0,
x2,
λ x7 x8 . x7 = 0 leaving 2 subgoals.
The subproof is completed by applying H5.
Apply Empty_eq with
lam 0 (λ x7 . ap x2 x7).
Let x7 of type ι be given.
Assume H7:
x7 ∈ lam 0 (λ x8 . ap x2 x8).
Apply lamE with
0,
λ x8 . ap x2 x8,
x7,
False leaving 2 subgoals.
The subproof is completed by applying H7.
Let x8 of type ι be given.
Assume H8:
(λ x9 . and (x9 ∈ 0) (∃ x10 . and (x10 ∈ ap x2 x9) (x7 = setsum x9 x10))) x8.
Apply H8 with
False.
Assume H9: x8 ∈ 0.
Apply FalseE with
(∃ x9 . and (x9 ∈ ap x2 x8) (x7 = setsum x8 x9)) ⟶ False.
Apply EmptyE with
x8.
The subproof is completed by applying H9.
Assume H4: x2 = 0.
Apply andI with
x2 ∈ setexp 0 0,
∀ x7 . ... ⟶ ∀ x8 . ... ⟶ x4 ... ... ⟶ x6 (ap x2 x7) (ap x2 x8) leaving 2 subgoals.