English
Let M be a commutative monoid. For n ∈ ℕ, f : Fin (n + 1) → M and a,b : Fin n, the product over i ∈ uIcc (a.castSucc) (b.castSucc) equals the product over i ∈ uIcc a b of f i.castSucc.
Русский
Пусть M — коммутативный моноид. Пусть f : Fin(n+1) → M и a,b : Fin n, тогда произведение по i ∈ uIcc(a.castSucc)(b.castSucc) равно произведению по i ∈ uIcc a b от f i.castSucc.
LaTeX
$$$$ \\displaystyle \\prod_{i \\in \\mathrm{uIcc}(a^{\\mathrm{castSucc}}, b^{\\mathrm{castSucc}})} f(i) = \\prod_{i \\in \\mathrm{uIcc} a b} f(i^{\\mathrm{castSucc}}) $$$$
Lean4
@[to_additive]
theorem prod_uIcc_castSucc (f : Fin (n + 1) → M) (a b : Fin n) :
∏ i ∈ uIcc a.castSucc b.castSucc, f i = ∏ i ∈ uIcc a b, f i.castSucc := by simp [← map_castSuccEmb_uIcc]
