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(* 02110-1301 USA                                                     *)
(** %\subsection*{ support :  counting\_elements.v }%*)
Set Implicit Arguments.
Unset Strict Implicit.
From Algebra Require Export Diff.
From Algebra Require Export Singleton.
Require Export has_n_elements.
From Algebra Require Export Union.
Require Export const.
(** - Once we have has_n_elements, it would seem to be easy to define has_at_least_n_elements and has_at_most_n_elements. But one has to take a little care. We could not just say, e.g. [  Definition has_at_least_n_elements := ] [    [n:nat;A:Setoid](EXT m:nat | (le n m) /\ (has_n_elements m A)).  ] since the set may be infinite, and then there is no such m to be found.  - But we may adapt the definition of has_n_elements slightly. It said that we can make a sequence [v] of length [n] with all elements distinct---[(distinct v)]; covering the whole set---[(seq_set v)='(full v)]. Now we just drop this last requirement so we only need to point out a sequence of $n$ different elements:  %\label{hasatleastnelements}% *)
Definition has_at_least_n_elements (n : Nat) (A : Setoid) :=
exists v : seq n A, distinct v.
(** - Note that, contrarily, we may not just drop the other requirement to obtain  [Definition has_at_most_n_elements_wrong := [n:nat;A:Setoid](EXT v:(seq n A) | (full A)='(seq_set v)).]  since this doesn't work out well for empty sets: it has at most n elements for any n, but with this definition we'd need to exhibit an n-length sequence, which we can't do for $n>0$.*)
Definition has_at_most_n_elements (n : Nat) (A : Setoid) :=
exists m : nat, m <= n /\ has_n_elements m A.

Lemma has_at_most_n_elements_comp :
forall (A : Setoid) (n m : Nat) (B C : part_set A),
has_at_most_n_elements n B ->
B =' C in _ -> n =' m in _ -> has_at_most_n_elements m C.