Section condensing.
(** - Easy to see but not easy to prove: if $x=\sum_{i=1}^n a_ib_i$ then in fact $x=\sum_{i=1}^{n'} a_ib'_i$ where all $b'_i$ are distinct vectors.*)

Lemma lin_comb_with_distinct_vectors :
forall (F : field) (V : vectorspace F) (x : V) (B : part_set V),
is_lin_comb x B ->
exists n : Nat,
(exists a : seq n F,
(exists v : seq n B,
x =' sum (mult_by_scalars a (Map_embed v)) in _ /\ distinct v)).