Proof using.
intros. find_apply_lem_hyp argmax_elim. intuition.
Qed.

Lemma argmax_one_different :
forall A (A_eq_dec : forall x y : A, {x = y} + {x <> y}) f g (l : list A) a,
(forall x, In x l -> a <> x -> f x = g x) ->
(forall x, In x l -> f x <= g x) ->
(argmax g l = argmax f l \/
argmax g l = Some a).