Apply H0 with
False.
Let x0 of type ι be given.
Apply H1 with
False.
Assume H2:
x0 ∈ omega.
Apply equip_sym with
primes,
x0,
False leaving 2 subgoals.
The subproof is completed by applying H3.
Let x1 of type ι → ι be given.
Apply H4 with
False.
Assume H5:
and (∀ x2 . x2 ∈ x0 ⟶ x1 x2 ∈ primes) (∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ x1 x2 = x1 x3 ⟶ x2 = x3).
Apply H5 with
(∀ x2 . x2 ∈ primes ⟶ ∃ x3 . and (x3 ∈ x0) (x1 x3 = x2)) ⟶ False.
Assume H6:
∀ x2 . x2 ∈ x0 ⟶ x1 x2 ∈ primes.
Assume H7: ∀ x2 . x2 ∈ x0 ⟶ ∀ x3 . x3 ∈ x0 ⟶ x1 x2 = x1 x3 ⟶ x2 = x3.
Assume H8:
∀ x2 . x2 ∈ primes ⟶ ∃ x3 . and (x3 ∈ x0) (x1 x3 = x2).
Apply L11 with
ordsucc (Pi_nat x1 x0) leaving 2 subgoals.
Apply and3I with
ordsucc (Pi_nat x1 x0) ∈ omega,
1 ∈ ordsucc (Pi_nat x1 x0),
∀ x2 . x2 ∈ omega ⟶ divides_nat x2 (ordsucc (Pi_nat x1 x0)) ⟶ or (x2 = 1) (x2 = ordsucc (Pi_nat x1 x0)) leaving 3 subgoals.
Apply nat_p_omega with
ordsucc (Pi_nat x1 x0).
Apply nat_ordsucc with
Pi_nat x1 x0.
The subproof is completed by applying L10.
The subproof is completed by applying L12.
Let x2 of type ι be given.
Assume H13:
x2 ∈ omega.
Apply orIL with
x2 = 1,
x2 = ordsucc (Pi_nat x1 x0).
Apply ordinal_In_Or_Subq with
1,
x2,
x2 = 1 leaving 4 subgoals.
Apply nat_p_ordinal with
1.
The subproof is completed by applying nat_1.
Apply nat_p_ordinal with
x2.
Apply omega_nat_p with
x2.
The subproof is completed by applying H13.
Assume H15: 1 ∈ x2.
Apply prime_nat_divisor_ex with
x2,
x2 = 1 leaving 3 subgoals.
Apply omega_nat_p with
x2.
The subproof is completed by applying H13.
The subproof is completed by applying H15.
Let x3 of type ι be given.
Apply H16 with
x2 = 1.
Apply L11 with
x3,
x2 = 1 leaving 2 subgoals.
The subproof is completed by applying H17.
Apply unknownprop_f799b99d854d7bca6941dc7751c0c00a5bf29ac2d7e070aa318a7a7ed9ce8fa0 with
x3,
x2,
ordsucc (Pi_nat x1 x0) leaving 2 subgoals.
The subproof is completed by applying H18.
The subproof is completed by applying H14.
Assume H15: x2 ⊆ 1.
Apply nat_le1_cases with
x2,
x2 = 1 leaving 4 subgoals.
Apply omega_nat_p with
x2.
The subproof is completed by applying H13.
The subproof is completed by applying H15.
Assume H16: x2 = 0.
Apply FalseE with
x2 = 1.
Apply H14 with
False.
Apply H18 with
False.
Let x3 of type ι be given.