Proof using.
move => m.
elim => //=.
case => /= n m' l IH n' m0 H_in.
rewrite /pt_map_name_msg /=.
case H_m: (pt_map_msg m') => [m1|].
have H_in_f := H_in (n, m').
rewrite /= in H_in_f.
move => H_in_map.
case: H_in_map => H_in_map.
inversion H_in_map.
rewrite H_in_f in H_m; last by left.
by rewrite -H1.
move: H_in_map.
apply: (IH _ m0) => //.
move => nm H_in'.
apply: H_in.
by right.
apply: (IH _ m0) => //.
move => nm H_in'.
apply: H_in => //.
by right.
Qed.

Lemma pt_map_in_in :
forall m m0 n l,
(forall nm, In nm l -> snd nm = m) ->
~ In (n, m) l ->
~ In (tot_map_name n, m0) (filterMap pt_map_name_msg l).