Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Let x3 of type ι be given.
Assume H0: x0 ∈ 4.
Assume H1: x1 ∈ 4.
Assume H2: x2 ∈ 4.
Assume H3: x3 ∈ 4.
Assume H4: x0 = x1 ⟶ ∀ x4 : ο . x4.
Assume H5: x0 = x2 ⟶ ∀ x4 : ο . x4.
Assume H6: x0 = x3 ⟶ ∀ x4 : ο . x4.
Assume H7: x1 = x2 ⟶ ∀ x4 : ο . x4.
Assume H8: x1 = x3 ⟶ ∀ x4 : ο . x4.
Assume H9: x2 = x3 ⟶ ∀ x4 : ο . x4.
Apply PigeonHole_nat_bij with
4,
λ x4 . ap (lam 4 (λ x5 . If_i (x5 = 0) x0 (If_i (x5 = 1) x1 (If_i (x5 = 2) x2 x3)))) x4 leaving 3 subgoals.
Apply nat_ordsucc with
3.
Apply nat_ordsucc with
2.
The subproof is completed by applying nat_2.
Let x4 of type ι be given.
Assume H10: x4 ∈ 4.
Apply cases_4 with
x4,
λ x5 . ap (lam 4 (λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 (If_i (x6 = 2) x2 x3)))) x5 ∈ 4 leaving 5 subgoals.
The subproof is completed by applying H10.
Apply tuple_4_0_eq with
x0,
x1,
x2,
x3,
λ x5 x6 . x6 ∈ 4.
The subproof is completed by applying H0.
Apply tuple_4_1_eq with
x0,
x1,
x2,
x3,
λ x5 x6 . x6 ∈ 4.
The subproof is completed by applying H1.
Apply tuple_4_2_eq with
x0,
x1,
x2,
x3,
λ x5 x6 . x6 ∈ 4.
The subproof is completed by applying H2.
Apply tuple_4_3_eq with
x0,
x1,
x2,
x3,
λ x5 x6 . x6 ∈ 4.
The subproof is completed by applying H3.
Let x4 of type ι be given.
Assume H10: x4 ∈ 4.
Let x5 of type ι be given.
Assume H11: x5 ∈ 4.
Apply cases_4 with
x4,
λ x6 . ap (lam 4 (λ x7 . If_i (x7 = 0) x0 (If_i (x7 = 1) x1 (If_i (x7 = 2) x2 x3)))) x6 = ap (lam 4 (λ x7 . If_i (x7 = 0) x0 (If_i (x7 = 1) x1 (If_i (x7 = 2) x2 x3)))) x5 ⟶ x6 = x5 leaving 5 subgoals.
The subproof is completed by applying H10.
Apply cases_4 with
x5,
λ x6 . ap (lam 4 (λ x7 . If_i (x7 = 0) x0 (If_i (x7 = 1) x1 (If_i (x7 = 2) x2 x3)))) 0 = ap (lam 4 (λ x7 . If_i (x7 = 0) x0 (If_i (x7 = 1) x1 (If_i (x7 = 2) x2 x3)))) x6 ⟶ 0 = x6 leaving 5 subgoals.
The subproof is completed by applying H11.
Assume H12:
ap (lam 4 (λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 (If_i (x6 = 2) x2 x3)))) 0 = ap (lam 4 (λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 (If_i (x6 = 2) ... ...)))) 0.