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--- lfa:2024:lab08 [2024/11/25 03:05]
cata_chiru created
+++ lfa:2024:lab08 [2024/11/28 01:33] (current)
stefan.sterea Corrected pumping lemma formulation
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 ** Pumping Lemma **
-Let L be an infinite regular language. Then, for $\forall w \in L$, $\exists n \in \mathbf{N}$, $ |w| \ge n $, $ w = xyz $, $ |xy| \le n $ and $ y \neq \varepsilon $, such that $ \forall k \ge 0, w_{k} = xy^{k}z \in L$.
+Let L be an infinite regular language. Then, $\exists n \in \mathbb{N}$: $ \forall w\ \ \text{s.t.}\ \ |w| \ge n $, $ w = xyz $, $ |xy| \le n $ and $ y \neq \varepsilon $, such that $ \forall k \ge 0, w_{k} = xy^{k}z \in L$.
 </note>
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 ** Complement of Pumping Lemma **
-Let L be an infinite language. If $\forall n \in \mathbf{N}$, $\exists w_{n} \in L $ with $ |w| \ge n $ such that regardless of how $ w_{n} $ is split into $ w_{n} = xyz $ with $ |xy| \le n $ and $ y \neq \varepsilon $, $\exists k \ge 0 $ such that $ w_{n} = xy^{k}z \notin L $, then L in **not** a regular language.
+Let L be an infinite language. If $\forall n \in \mathbb{N}$, $\exists w_{n} \in L $ with $ |w| \ge n $ such that regardless of how $ w_{n} $ is split into ($ \forall x, y, z \in \Sigma^* $) $ w_{n} = xyz $ with $ |xy| \le n $ and $ y \neq \varepsilon $,  $\exists k \ge 0 $ such that $ w_{n} = xy^{k}z \notin L $, then L in **not** a regular language.
 </note>