Let x0 of type ι → ι be given.
Apply nat_ind with
λ x1 . (∀ x2 . x2 ∈ x1 ⟶ nat_p (x0 x2)) ⟶ Pi_nat x0 x1 = 0 ⟶ ∃ x2 . and (x2 ∈ x1) (x0 x2 = 0) leaving 2 subgoals.
Assume H0:
∀ x1 . x1 ∈ 0 ⟶ nat_p (x0 x1).
Apply FalseE with
∃ x1 . and (x1 ∈ 0) (x0 x1 = 0).
Apply neq_1_0.
Let x1 of type ι → ι → ο be given.
set y2 to be λ x2 . x1 x2 0 ⟶ x1 0 x2
Claim L2: y2 0
Assume H2: y2 0 0.
The subproof is completed by applying H2.
Apply Pi_nat_0 with
x0,
λ x3 x4 . (λ x5 . y2) x4 x3.
Apply H1 with
λ x3 . y2.
The subproof is completed by applying L2.
Let x1 of type ι be given.
Assume H1:
(∀ x2 . x2 ∈ x1 ⟶ nat_p (x0 x2)) ⟶ Pi_nat x0 x1 = 0 ⟶ ∃ x2 . and (x2 ∈ x1) (x0 x2 = 0).
Apply Pi_nat_S with
x0,
x1,
λ x2 x3 . x3 = 0 ⟶ ∃ x4 . and (x4 ∈ ordsucc x1) (x0 x4 = 0) leaving 2 subgoals.
The subproof is completed by applying H0.
Claim L4:
∀ x2 . x2 ∈ x1 ⟶ nat_p (x0 x2)Let x2 of type ι be given.
Assume H4: x2 ∈ x1.
Apply H2 with
x2.
Apply ordsuccI1 with
x1,
x2.
The subproof is completed by applying H4.
Apply mul_nat_0_inv with
Pi_nat x0 x1,
x0 x1,
∃ x2 . and (x2 ∈ ordsucc x1) (x0 x2 = 0) leaving 5 subgoals.
Apply nat_p_omega with
Pi_nat x0 x1.
Apply Pi_nat_p with
x0,
x1 leaving 2 subgoals.
The subproof is completed by applying H0.
The subproof is completed by applying L4.
Apply nat_p_omega with
x0 x1.
Apply H2 with
x1.
The subproof is completed by applying ordsuccI2 with x1.
The subproof is completed by applying H3.
Apply H1 with
∃ x2 . and (x2 ∈ ordsucc x1) (x0 x2 = 0) leaving 3 subgoals.
The subproof is completed by applying L4.
The subproof is completed by applying H5.
Let x2 of type ι be given.
Assume H6:
(λ x3 . and (x3 ∈ x1) (x0 x3 = 0)) x2.
Apply H6 with
∃ x3 . and (x3 ∈ ordsucc x1) (x0 x3 = 0).
Assume H7: x2 ∈ x1.
Assume H8: x0 x2 = 0.
Let x3 of type ο be given.
Assume H9:
∀ x4 . and (x4 ∈ ordsucc x1) (x0 x4 = 0) ⟶ x3.
Apply H9 with
x2.
Apply andI with
x2 ∈ ordsucc x1,
x0 x2 = 0 leaving 2 subgoals.
Apply ordsuccI1 with
x1,
x2.
The subproof is completed by applying H7.
The subproof is completed by applying H8.
Assume H5: x0 x1 = 0.
Let x2 of type ο be given.
Assume H6:
∀ x3 . and (x3 ∈ ordsucc x1) (x0 x3 = 0) ⟶ x2.
Apply H6 with
x1.
Apply andI with
x1 ∈ ordsucc x1,
x0 x1 = 0 leaving 2 subgoals.
The subproof is completed by applying ordsuccI2 with x1.
The subproof is completed by applying H5.