Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Assume H0: x0 ∈ 3.
Assume H1: x1 ∈ 3.
Assume H2: x2 ∈ 3.
Assume H3: x0 = x1 ⟶ ∀ x3 : ο . x3.
Assume H4: x0 = x2 ⟶ ∀ x3 : ο . x3.
Assume H5: x1 = x2 ⟶ ∀ x3 : ο . x3.
Apply PigeonHole_nat_bij with
3,
λ x3 . ap (lam 3 (λ x4 . If_i (x4 = 0) x0 (If_i (x4 = 1) x1 x2))) x3 leaving 3 subgoals.
Apply nat_ordsucc with
2.
The subproof is completed by applying nat_2.
Let x3 of type ι be given.
Assume H6: x3 ∈ 3.
Apply cases_3 with
x3,
λ x4 . ap (lam 3 (λ x5 . If_i (x5 = 0) x0 (If_i (x5 = 1) x1 x2))) x4 ∈ 3 leaving 4 subgoals.
The subproof is completed by applying H6.
Apply tuple_3_0_eq with
x0,
x1,
x2,
λ x4 x5 . x5 ∈ 3.
The subproof is completed by applying H0.
Apply tuple_3_1_eq with
x0,
x1,
x2,
λ x4 x5 . x5 ∈ 3.
The subproof is completed by applying H1.
Apply tuple_3_2_eq with
x0,
x1,
x2,
λ x4 x5 . x5 ∈ 3.
The subproof is completed by applying H2.
Let x3 of type ι be given.
Assume H6: x3 ∈ 3.
Let x4 of type ι be given.
Assume H7: x4 ∈ 3.
Apply cases_3 with
x3,
λ x5 . ap (lam 3 (λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 x2))) x5 = ap (lam 3 (λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 x2))) x4 ⟶ x5 = x4 leaving 4 subgoals.
The subproof is completed by applying H6.
Apply cases_3 with
x4,
λ x5 . ap (lam 3 (λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 x2))) 0 = ap (lam 3 (λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 x2))) x5 ⟶ 0 = x5 leaving 4 subgoals.
The subproof is completed by applying H7.
Assume H8:
ap (lam 3 (λ x5 . If_i (x5 = 0) x0 (If_i (x5 = 1) x1 x2))) 0 = ap (lam 3 (λ x5 . If_i (x5 = 0) x0 (If_i (x5 = 1) x1 x2))) 0.
Let x5 of type ι → ι → ο be given.
Assume H9: x5 0 0.
The subproof is completed by applying H9.
Apply tuple_3_0_eq with
x0,
x1,
x2,
λ x5 x6 . (λ x7 . x6 = ap (lam 3 (λ x8 . If_i (x8 = 0) x0 (If_i (x8 = 1) x1 x2))) x7 ⟶ 0 = x7) 1.
Apply tuple_3_1_eq with
x0,
x1,
x2,
λ x5 x6 . x0 = x6 ⟶ 0 = 1.
Assume H8: x0 = x1.
Apply H3 with
0 = 1.
The subproof is completed by applying H8.
Apply tuple_3_0_eq with
x0,
x1,
x2,
λ x5 x6 . (λ x7 . x6 = ap (lam 3 (λ x8 . If_i (x8 = 0) x0 (If_i ... ... ...))) ... ⟶ 0 = x7) 2.