Proof using.
induction tr; intuition; unfold trace_mutual_exclusion in *; simpl in *;
repeat break_match; subst; intuition.
Qed.

Lemma last_holder'_no_out_inv :
forall tr h c n,
last_holder' h (tr ++ [(c, inr [])]) = Some n ->
last_holder' h tr = Some n.