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How to prove using pumping lemma {0^n OR 1^2n OR 2^3n | n >= 0} is not context free

This isnt the same language as {0^n1^2n2^3n | n >= 0} as this language the numbers need to be in order.

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It is not specified but I'll assume the alphabet to be $\Sigma = \{0,1,2\}$ and also that your strange notation means the language

\begin{align*} L&= \{x \in \Sigma{}^* : x=0^n \lor x=1^{2n} \lor x=2^{3n}, n \in \mathbb{N}\} \\ &=\{0^n : n \in \mathbb{N}\}\cup\{1^{2n} : n \in \mathbb{N}\}\cup\{2^{3n} : n \in \mathbb{N}\} \end{align*}

Well, the answer is that actually it is context-free since each of those three languages is trivially CF and the union of CFLs is also CF.

But it is also Regular! By definition a language is said to be regular if it is accepted by some $DFA$ or, equivalently, if it is expressible by a regular expression.

In our case, $L = 0^*+(11)^*+(222)^*$.

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