Asked By : Andrew Reynolds
Answered By : Roi Divon
- $|vwx| leq n$
- $|vx| geq 1$
- for all $i geq 0$: $uv^iwx^iy in L$
Because $L$ is over an unary alphabet, we can change the order of the sub-words and the word ($z$) will not change, meaning we can also write $z=wvxuy$. So. for all $i geq 0$, $uv^iwx^iy = wv^ix^iuy = w(vx)^iuy$. Let’s write a little different: $u’ = w, v’ = vx, w’ = uy$. and we have that $u'(v’)^iw’ in L$. It’s easy to see that $|u’v’| leq n$ and $|v’| geq 1$, So we can conclude that for the same $n$, the conditions of pumping lemma for regular languages holds. It might be worth mentioning that every CFL of unary alphabet is also regular (We know that even though a language satisfies the pumping lemma for regular/CF languages, it does’t mean that the language is regular/CF). It can be shown using Parikh’s theorem, and showing that for every semi-linear set $S subseteq mathbb{N} $ there is a regular language $L$ such that $Psi (L)=S$ (or $p(L)$ using wikipedia’s notations)
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