Proof using.
induction tr; intros; simpl in *; repeat break_match; subst;
intuition; try congruence; eauto.
Qed.

Lemma last_holder'_neq_unlock_extend :
forall tr h n c,
last_holder' h tr = Some n ->
n <> c ->
last_holder' h (tr ++ [(Client c, inl Unlock)]) = Some n.