English
For id-family, a ∉ span_K s is equivalent to independence on insert a s together with a ∉ s.
Русский
Для последовательности id не принадлежит span_K s тогда независимость на insert a s вместе с a ∉ s.
LaTeX
$$$\text{LinearIndepOn}_K(\text{id},s) \Rightarrow \Bigl( a \notin \operatorname{span}_K(s) \Rightarrow \text{LinearIndepOn}_K(\text{id},\operatorname{insert} a\,s) \land a \notin s \Bigr)$$$
Lean4
theorem linearIndepOn_id_insert (hxs : x ∉ s) :
LinearIndepOn K id (insert x s) ↔ LinearIndepOn K id s ∧ x ∉ Submodule.span K s :=
(linearIndepOn_insert (f := id) hxs).trans <| by simp