Let x0 of type ι be given.
Let x1 of type ι → ι be given.
Assume H1: ∀ x2 . x2 ∈ x0 ⟶ x1 x2 ∈ x0.
Assume H2: ∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ x1 x2 = x1 x3 ⟶ x2 = x3.
Apply and3I with
∀ x2 . x2 ∈ x0 ⟶ x1 x2 ∈ x0,
∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ x1 x2 = x1 x3 ⟶ x2 = x3,
∀ x2 . x2 ∈ x0 ⟶ ∃ x3 . and (x3 ∈ x0) (x1 x3 = x2) leaving 3 subgoals.
The subproof is completed by applying H1.
The subproof is completed by applying H2.
Let x2 of type ι be given.
Assume H3: x2 ∈ x0.
Apply dneg with
∃ x3 . and (x3 ∈ x0) (x1 x3 = x2).
Assume H4:
not (∃ x3 . and (x3 ∈ x0) (x1 x3 = x2)).
Apply PigeonHole_nat with
x0,
λ x3 . If_i (x3 = x0) x2 (x1 x3) leaving 3 subgoals.
The subproof is completed by applying H0.
Let x3 of type ι be given.
Apply xm with
x3 = x0,
If_i (x3 = x0) x2 (x1 x3) ∈ x0 leaving 2 subgoals.
Assume H6: x3 = x0.
Apply If_i_1 with
x3 = x0,
x2,
x1 x3,
λ x4 x5 . x5 ∈ x0 leaving 2 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H3.
Assume H6: x3 = x0 ⟶ ∀ x4 : ο . x4.
Apply If_i_0 with
x3 = x0,
x2,
x1 x3,
λ x4 x5 . x5 ∈ x0 leaving 2 subgoals.
The subproof is completed by applying H6.
Apply H1 with
x3.
Apply ordsuccE with
x0,
x3,
x3 ∈ x0 leaving 3 subgoals.
The subproof is completed by applying H5.
Assume H7: x3 ∈ x0.
The subproof is completed by applying H7.
Assume H7: x3 = x0.
Apply H6 with
x3 ∈ x0.
The subproof is completed by applying H7.
Let x3 of type ι be given.
Let x4 of type ι be given.
Apply xm with
x3 = x0,
If_i (x3 = x0) x2 (x1 x3) = If_i (x4 = x0) x2 (x1 x4) ⟶ x3 = x4 leaving 2 subgoals.
Assume H9: x3 = x0.
Apply xm with
x4 = x0,
If_i (x3 = x0) x2 (x1 x3) = If_i (x4 = x0) x2 (x1 x4) ⟶ x3 = x4 leaving 2 subgoals.
Assume H10: x4 = x0.
Assume H11:
If_i (x3 = x0) x2 (x1 x3) = If_i (x4 = x0) x2 (x1 x4).
Apply H10 with
λ x5 x6 . x3 = x6.
The subproof is completed by applying H9.
Assume H10: x4 = x0 ⟶ ∀ x5 : ο . x5.
Apply If_i_1 with
x3 = x0,
x2,
x1 x3,
λ x5 x6 . x6 = If_i (x4 = x0) x2 (x1 x4) ⟶ x3 = x4 leaving 2 subgoals.
The subproof is completed by applying H9.
Apply If_i_0 with
x4 = x0,
x2,
x1 x4,
λ x5 x6 . x2 = x6 ⟶ x3 = x4 leaving 2 subgoals.
The subproof is completed by applying H10.