Let x0 of type ι be given.
Let x1 of type ι be given.
Assume H0: x1 ∈ x0.
Let x2 of type ι be given.
Assume H1: x2 ∈ x0.
Assume H2: x1 = x2 ⟶ ∀ x3 : ο . x3.
Assume H3:
∀ x3 . x3 ∈ x0 ⟶ or (x3 = x1) (x3 = x2).
Let x3 of type ο be given.
Assume H4:
∀ x4 : ι → ι . bij 2 x0 x4 ⟶ x3.
Apply H4 with
λ x4 . ap (lam 2 (λ x5 . If_i (x5 = 0) x1 x2)) x4.
Apply bijI with
2,
x0,
λ x4 . ap (lam 2 (λ x5 . If_i (x5 = 0) x1 x2)) x4 leaving 3 subgoals.
Let x4 of type ι be given.
Assume H5: x4 ∈ 2.
Apply cases_2 with
x4,
λ x5 . (λ x6 . ap (lam 2 (λ x7 . If_i (x7 = 0) x1 x2)) x6) x5 ∈ x0 leaving 3 subgoals.
The subproof is completed by applying H5.
Apply tuple_2_0_eq with
x1,
x2,
λ x5 x6 . x6 ∈ x0.
The subproof is completed by applying H0.
Apply tuple_2_1_eq with
x1,
x2,
λ x5 x6 . x6 ∈ x0.
The subproof is completed by applying H1.
Let x4 of type ι be given.
Assume H5: x4 ∈ 2.
Let x5 of type ι be given.
Assume H6: x5 ∈ 2.
Apply cases_2 with
x4,
λ x6 . (λ x7 . ap (lam 2 (λ x8 . If_i (x8 = 0) x1 x2)) x7) x6 = (λ x7 . ap (lam 2 (λ x8 . If_i (x8 = 0) x1 x2)) x7) x5 ⟶ x6 = x5 leaving 3 subgoals.
The subproof is completed by applying H5.
Apply cases_2 with
x5,
λ x6 . (λ x7 . ap (lam 2 (λ x8 . If_i (x8 = 0) x1 x2)) x7) 0 = (λ x7 . ap (lam 2 (λ x8 . If_i (x8 = 0) x1 x2)) x7) x6 ⟶ 0 = x6 leaving 3 subgoals.
The subproof is completed by applying H6.
Assume H7:
(λ x6 . ap (lam 2 (λ x7 . If_i (x7 = 0) x1 x2)) x6) 0 = (λ x6 . ap (lam 2 (λ x7 . If_i (x7 = 0) x1 x2)) x6) 0.
Let x6 of type ι → ι → ο be given.
Assume H8: x6 0 0.
The subproof is completed by applying H8.
Apply tuple_2_0_eq with
x1,
x2,
λ x6 x7 . x7 = ap (lam 2 (λ x8 . If_i (x8 = 0) x1 x2)) 1 ⟶ 0 = 1.
Apply tuple_2_1_eq with
x1,
x2,
λ x6 x7 . x1 = x7 ⟶ 0 = 1.
Assume H7: x1 = x2.
Apply FalseE with
0 = 1.
Apply H2.
The subproof is completed by applying H7.
Apply cases_2 with
x5,
λ x6 . (λ x7 . ap (lam 2 (λ x8 . If_i (x8 = 0) x1 x2)) x7) 1 = (λ x7 . ap (lam 2 (λ x8 . If_i (x8 = 0) x1 x2)) x7) x6 ⟶ 1 = x6 leaving 3 subgoals.
The subproof is completed by applying H6.
Apply tuple_2_0_eq with
x1,
x2,
λ x6 x7 . ap (lam 2 ...) 1 = ... ⟶ 1 = 0.