Apply H0 with
False.
Let x0 of type ι be given.
Apply H1 with
False.
Let x1 of type ι → ι be given.
Apply H2 with
False.
Apply unknownprop_c86200a2eefb0ff844f50b29d5cbeaa2ee14856a2db63542bcbf63218f0d0f1e with
x0,
λ x2 . (∀ x3 . PtdPred x3 ⟶ and (UnaryPredHom x2 x3 (x1 x3)) (∀ x4 . UnaryPredHom x2 x3 x4 ⟶ x4 = x1 x3)) ⟶ False leaving 2 subgoals.
The subproof is completed by applying H3.
Let x2 of type ι be given.
Let x3 of type ι → ο be given.
Let x4 of type ι be given.
Assume H4: x4 ∈ x2.
Assume H5: x3 x4.
Apply H6 with
pack_p 2 (λ x5 . True),
False leaving 2 subgoals.
The subproof is completed by applying L7.
Apply neq_0_1.
set y5 to be 0
set y6 to be 1
Claim L12: ∀ x7 : ι → ο . x7 y6 ⟶ x7 y5
Let x7 of type ι → ο be given.
Assume H12: x7 1.
set y8 to be ...
Apply beta with
x4,
λ x9 . 0,
y6,
λ x9 x10 . y8 x10 x9 leaving 2 subgoals.
The subproof is completed by applying H4.
set y9 to be ...
set y10 to be ...
Claim L13: ∀ x11 : ι → ο . x11 y10 ⟶ x11 y9
Let x11 of type ι → ο be given.
set y12 to be ...
Apply H9 with
lam x7 (λ x13 . 0),
λ x13 x14 . y12 (ap ... ...) ... leaving 2 subgoals.
set y11 to be λ x11 . y10
Apply L13 with
λ x12 . y11 x12 y10 ⟶ y11 y10 x12 leaving 2 subgoals.
Assume H14: y11 y10 y10.
The subproof is completed by applying H14.
set y13 to be
ap (lam y9 (λ x13 . 1)) y11
Claim L14: ∀ x14 : ι → ο . x14 y13 ⟶ x14 y12
Let x14 of type ι → ο be given.
Assume H14:
x14 (ap (lam y10 (λ x15 . 1)) y12).
set y15 to be λ x15 . x14
set y16 to be
λ x16 x17 . y15 (ap x16 y12) (ap x17 y12)
Apply L10 with
lam y10 (λ x17 . 1),
λ x17 x18 . y16 x18 x17 leaving 2 subgoals.
The subproof is completed by applying H12.
The subproof is completed by applying H14.
set y14 to be λ x14 . y13
Apply L14 with
λ x15 . y14 x15 y13 ⟶ y14 y13 x15 leaving 2 subgoals.
Assume H15: y14 y13 y13.
The subproof is completed by applying H15.
Apply beta with
y11,
λ x15 . 1,
y13,
λ x15 . y14 leaving 2 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying L14.
Let x7 of type ι → ι → ο be given.
Apply L12 with
λ x8 . x7 x8 y6 ⟶ x7 y6 x8.
Assume H13: x7 y6 y6.
The subproof is completed by applying H13.