Is the language regular A2 = {w1w2w3 | w1, w2, w3 ϵ {0, 1}* }? How to prove?

So I think the above language is regular. I tried using pumping lemma but pumping up or down, changes the value of w1 but has no relation with w2 or w3. The resulting string after pumping will also be in the language. I think the language w1w2w3 is just similar to {0,1}* . I can create a NFA for {0,1}* , so that is why the language is regular. w1w2w3 is just like a concatenation of w1,w2,w3 but all of them are {0,1}* so the NFA for the language {0,1}* is also the NFA that accepts the language A2. Is my thinking so far correct? Will the NFA accept the language?

2. The sentence "$$\{0,1\}^*$$ can represent the NFA for the language" makes absolutely no sense.
3. Your language is equal to $$\{0,1\}^*$$ (you can choose any $$w_1$$, and $$w_2 = w_3 = \varepsilon$$).
• For your first comment: there were no NFA in your initial post. In your edit, this is in fact a DFA, and it does not have an initial state, so it recognizes the empty language. For your second comment: you cannot do the same thing with $\{w_1w_2w_2\mid w_1, w_2\in \{0,1\}^*\}$, because if you choose $w_1$ to be any non-empty word, you cannot choose it to be the empty word at the same time. Oct 20, 2023 at 5:00