Proof using.
induction tr; intros; simpl in *; repeat break_match; intuition (auto; try congruence).
Qed.

(* Specifications: *)
Lemma last_holder'_unlock_none :
forall tr h c,
last_holder' h tr = Some c ->
last_holder' h (tr ++ [(Client c, inl Unlock)]) = None.
Lemma last_holder'_unlock_none :
forall tr h c,
last_holder' h tr = Some c ->
last_holder' h (tr ++ [(Client c, inl Unlock)]) = None.