English
Same statement as above: for nonzero a,b, (a·b).factorization = a.factorization + b.factorization.
Русский
То же самое: для ненулевых a,b, факторизация ab равна сумме факторизаций.
LaTeX
$$$ (a \\cdot b).factorization = a.factorization + b.factorization \\quad (a \\neq 0,\\; b \\neq 0)$$$
Lean4
theorem factorization_le_iff_dvd {d n : ℕ} (hd : d ≠ 0) (hn : n ≠ 0) : d.factorization ≤ n.factorization ↔ d ∣ n :=
by
constructor
· intro hdn
set K := n.factorization - d.factorization with hK
use K.prod (· ^ ·)
rw [← factorization_prod_pow_eq_self hn, ← factorization_prod_pow_eq_self hd, ←
Finsupp.prod_add_index' pow_zero pow_add, hK, add_tsub_cancel_of_le hdn]
· rintro ⟨c, rfl⟩
rw [factorization_mul hd (right_ne_zero_of_mul hn)]
simp