Proof using.
induction ll; intros; simpl in *; intuition.
repeat break_match; simpl in *; intuition eauto.
Qed.

Lemma in_contradicting_leader_logs_on_leader_not_in_log :
forall t' l ll t e,
In (t', l) (contradicting_leader_logs_on_leader ll t e) ->
In e l -> False.