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