Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι → ι be given.
Let x3 of type ι → ι be given.
Assume H0: ∀ x4 . x4 ∈ x0 ⟶ x2 x4 = x3 x4.
Apply H1 with
bij x0 x1 x3.
Assume H2:
and (∀ x4 . x4 ∈ x0 ⟶ x2 x4 ∈ x1) (∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x2 x4 = x2 x5 ⟶ x4 = x5).
Apply H2 with
(∀ x4 . x4 ∈ x1 ⟶ ∃ x5 . and (x5 ∈ x0) (x2 x5 = x4)) ⟶ bij x0 x1 x3.
Assume H3: ∀ x4 . x4 ∈ x0 ⟶ x2 x4 ∈ x1.
Assume H4: ∀ x4 . x4 ∈ x0 ⟶ ∀ x5 . x5 ∈ x0 ⟶ x2 x4 = x2 x5 ⟶ x4 = x5.
Assume H5:
∀ x4 . x4 ∈ x1 ⟶ ∃ x5 . and (x5 ∈ x0) (x2 x5 = x4).
Apply bijI with
x0,
x1,
x3 leaving 3 subgoals.
Let x4 of type ι be given.
Assume H6: x4 ∈ x0.
Apply H0 with
x4,
λ x5 x6 . x5 ∈ x1 leaving 2 subgoals.
The subproof is completed by applying H6.
Apply H3 with
x4.
The subproof is completed by applying H6.
Let x4 of type ι be given.
Assume H6: x4 ∈ x0.
Let x5 of type ι be given.
Assume H7: x5 ∈ x0.
Apply H0 with
x4,
λ x6 x7 . x6 = x3 x5 ⟶ x4 = x5 leaving 2 subgoals.
The subproof is completed by applying H6.
Apply H0 with
x5,
λ x6 x7 . x2 x4 = x6 ⟶ x4 = x5 leaving 2 subgoals.
The subproof is completed by applying H7.
Apply H4 with
x4,
x5 leaving 2 subgoals.
The subproof is completed by applying H6.
The subproof is completed by applying H7.
Let x4 of type ι be given.
Assume H6: x4 ∈ x1.
Apply H5 with
x4,
∃ x5 . and (x5 ∈ x0) (x3 x5 = x4) leaving 2 subgoals.
The subproof is completed by applying H6.
Let x5 of type ι be given.
Assume H7:
(λ x6 . and (x6 ∈ x0) (x2 x6 = x4)) x5.
Apply H7 with
∃ x6 . and (x6 ∈ x0) (x3 x6 = x4).
Assume H8: x5 ∈ x0.
Assume H9: x2 x5 = x4.
Let x6 of type ο be given.
Assume H10:
∀ x7 . and (x7 ∈ x0) (x3 x7 = x4) ⟶ x6.
Apply H10 with
x5.
Apply andI with
x5 ∈ x0,
x3 x5 = x4 leaving 2 subgoals.
The subproof is completed by applying H8.
Apply H0 with
x5,
λ x7 x8 . x7 = x4 leaving 2 subgoals.
The subproof is completed by applying H8.
The subproof is completed by applying H9.