(**)
intros k. apply aaron_stump_cse545 with(Q:=fun z=>Q z->exists m,Q m/\forall n
,Q n->m<=n)(n:=k). 2:auto. intros n H H1. elim(classic(forall i,i<n->~Q i)).
intro H2. exists n. split. auto. intros j H3. apply NNPP. red;intro H4. apply(
H2 j). intuition. auto.
(**)
intro H2. assert(H3:exists i:nat,i<n/\Q i). apply NNPP. red;intro H3. apply H2.
intros i H4. red;intro H5. apply H3. exists i. auto. elim H3. intros ii H4.
elim(H ii). intros m H5. exists m. auto. tauto. tauto. Qed.

(* Specifications: *)
(****************************************************************************** You can actually prove this without using constructive_definite_description, you can use Yevgeniy Marakov's constructive_indefinite_description_nat from Coq.Logic.ConstructiveEpsilon, which is purely constructive -- probably don't even need classic really to do this proof. ******************************************************************************)
Lemma cntabl_P0(U:Type)(C:Ensemble(Ensemble U))
:cntabl _ C->
exists g:forall c,C c->nat,
(forall c c' c'',g c c'=g c c'')
/\forall c cc c' cc',g c c'=g cc cc'->c=cc.