Let x0 of type ι be given.
Let x1 of type ο be given.
Assume H1: ∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ ∀ x4 . x4 ∈ x0 ⟶ (x2 = x3 ⟶ ∀ x5 : ο . x5) ⟶ (x2 = x4 ⟶ ∀ x5 : ο . x5) ⟶ (x3 = x4 ⟶ ∀ x5 : ο . x5) ⟶ x1.
Apply H0 with
x1.
Let x2 of type ι → ι be given.
Apply H2 with
x1.
Assume H3: ∀ x3 . x3 ∈ 3 ⟶ x2 x3 ∈ x0.
Assume H4: ∀ x3 . x3 ∈ 3 ⟶ ∀ x4 . x4 ∈ 3 ⟶ x2 x3 = x2 x4 ⟶ x3 = x4.
Apply H1 with
x2 0,
x2 1,
x2 2 leaving 6 subgoals.
Apply H3 with
0.
The subproof is completed by applying In_0_3.
Apply H3 with
1.
The subproof is completed by applying In_1_3.
Apply H3 with
2.
The subproof is completed by applying In_2_3.
Assume H5: x2 0 = x2 1.
Apply neq_0_1.
Apply H4 with
0,
1 leaving 3 subgoals.
The subproof is completed by applying In_0_3.
The subproof is completed by applying In_1_3.
The subproof is completed by applying H5.
Assume H5: x2 0 = x2 2.
Apply neq_0_2.
Apply H4 with
0,
2 leaving 3 subgoals.
The subproof is completed by applying In_0_3.
The subproof is completed by applying In_2_3.
The subproof is completed by applying H5.
Assume H5: x2 1 = x2 2.
Apply neq_1_2.
Apply H4 with
1,
2 leaving 3 subgoals.
The subproof is completed by applying In_1_3.
The subproof is completed by applying In_2_3.
The subproof is completed by applying H5.