Proof using.
induction l1; intros; simpl in *; auto.
break_if; simpl in *; eauto.
Qed.

Lemma log_to_ks'_a_equiv :
forall l ks ks',
a_equiv clientId_eq_dec ks ks' ->
a_equiv clientId_eq_dec (log_to_ks' l ks) (log_to_ks' l ks').