Deixei A={L∣Lis one-counter and L¯ is also one-counter}A={L∣Lis one-counter and L¯ is also one-counter}A= \{L \mid L \;\text{is one-counter and \(\bar{L}\) is also one-counter} \} Claramente, Deterministic one-counter⊆ADeterministic one-counter⊆A\text{Deterministic one-counter} \subseteq A É o...