Apply nat_complete_ind with
λ x0 . ∀ x1 . equip x0 x1 ⟶ ∀ x2 : ι → ι . (∀ x3 . x3 ∈ x1 ⟶ x2 x3 ∈ x1) ⟶ (∀ x3 . x3 ∈ x1 ⟶ x2 (x2 x3) = x3) ⟶ (∃ x3 . and (x3 ∈ x1) (and (x2 x3 = x3) (∀ x4 . x4 ∈ x1 ⟶ x2 x4 = x4 ⟶ x3 = x4))) ⟶ odd_nat x0.
Let x0 of type ι be given.
Assume H1:
∀ x1 . x1 ∈ x0 ⟶ ∀ x2 . equip x1 x2 ⟶ ∀ x3 : ι → ι . (∀ x4 . x4 ∈ x2 ⟶ x3 x4 ∈ x2) ⟶ (∀ x4 . x4 ∈ x2 ⟶ x3 (x3 x4) = x4) ⟶ (∃ x4 . and (x4 ∈ x2) (and (x3 x4 = x4) (∀ x5 . x5 ∈ x2 ⟶ x3 x5 = x5 ⟶ x4 = x5))) ⟶ odd_nat x1.
Let x1 of type ι be given.
Let x2 of type ι → ι be given.
Assume H3: ∀ x3 . x3 ∈ x1 ⟶ x2 x3 ∈ x1.
Assume H4: ∀ x3 . x3 ∈ x1 ⟶ x2 (x2 x3) = x3.
Assume H5:
∃ x3 . and (x3 ∈ x1) (and (x2 x3 = x3) (∀ x4 . x4 ∈ x1 ⟶ x2 x4 = x4 ⟶ x3 = x4)).
Apply H5 with
odd_nat x0.
Let x3 of type ι be given.
Assume H6:
(λ x4 . and (x4 ∈ x1) (and (x2 x4 = x4) (∀ x5 . x5 ∈ x1 ⟶ x2 x5 = x5 ⟶ x4 = x5))) x3.
Apply H6 with
odd_nat x0.
Assume H7: x3 ∈ x1.
Assume H8:
and (x2 x3 = x3) (∀ x4 . x4 ∈ x1 ⟶ x2 x4 = x4 ⟶ x3 = x4).
Apply H8 with
odd_nat x0.
Assume H9: x2 x3 = x3.
Assume H10: ∀ x4 . x4 ∈ x1 ⟶ x2 x4 = x4 ⟶ x3 = x4.
Apply H2 with
odd_nat x0.
Let x4 of type ι → ι be given.
Assume H12:
bij x0 x1 x4.
Apply H12 with
odd_nat x0.
Assume H13:
and (∀ x5 . x5 ∈ x0 ⟶ x4 x5 ∈ x1) (∀ x5 . x5 ∈ x0 ⟶ ∀ x6 . x6 ∈ x0 ⟶ x4 x5 = x4 x6 ⟶ x5 = x6).
Apply H13 with
(∀ x5 . x5 ∈ x1 ⟶ ∃ x6 . and (x6 ∈ x0) (x4 x6 = x5)) ⟶ odd_nat x0.
Assume H14: ∀ x5 . x5 ∈ x0 ⟶ x4 x5 ∈ x1.
Assume H15: ∀ x5 . ... ⟶ ∀ x6 . x6 ∈ x0 ⟶ x4 x5 = x4 x6 ⟶ x5 = x6.