\[lemma:pumping\].

We first consider the case that (3) holds. In order to satisfy $ x_1 x_2^i x_3 x_4^i x_5 y \in {\mathtt{END}}$ for $ i\geq 0 $, $ x_2 x_4 $ must not have the symbol 1 (otherwise, $ x_1 x_2^0 x_3 x_4^0 x_5 y \not\in {\mathtt{END}}$). Thus, $ x_2 $ and $ x_4 $ are of the form $ 2^t $ or $ 0^t $ for some constant $ t $. If $ x_2 = 2^{t_1} $ and $ x_4 = 2^{t_2} $, $ x_1 x_2^i x_3 x_4^i x_5 y \not\in {\mathtt{END}}$ for $ i\neq 1$. Similarly, $ x_2 = 0^{t_1} $ and $ x_4 = 0^{t_2} $ cannot occur. Thus, the only possible choice is $ x_2 = 2^{t_1} $ and $ x_4 = 0^{t_2} $ for some $ t_1 $ and $ t_2 $. In order to satisfy $x_1 x_2^i x_3 x_4^i x_5 y \in {\mathtt{END}}$ for $ i\geq 0 $, $ t_1 = t_2$ must hold. However, This causes $ x_1 x_2^i x_3 x_4^i x_5 z \not\in {\mathtt{END}}$ for $ i\neq 1 $. Thus, (3) does not hold.

Next, we consider the case that (4) holds. Since $ |x_2 x_3 | \leq C $, $ x_2 