English
For a subsingleton R, IsTranscendenceBasis R x iff Nonempty ι.
Русский
Пусть R сводимо к едва ли не единственному элементу; тогда IsTranscendenceBasis R x эквивалентно не пустоте множества индексов ι.
LaTeX
$$$\\mathrm{IsTB}_R(x) \\;\\iff\\; ι \\neq \\varnothing$$$
Lean4
theorem isTranscendenceBasis_iff_of_subsingleton [Subsingleton R] (x : ι → A) : IsTranscendenceBasis R x ↔ Nonempty ι :=
by
have := Module.subsingleton R A
refine
⟨fun h ↦ ?_, fun h ↦
⟨.of_subsingleton, fun s hs hx ↦ hx.antisymm fun a _ ↦ ⟨Classical.arbitrary _, Subsingleton.elim ..⟩⟩⟩
by_contra! hι
have := h.2 {0} .of_subsingleton
simp [range_eq_empty, eq_comm (a := ∅)] at this