# Show that the Language L is not regular (pumping lemma)

$$L = \{cda^nb^n\mid n\in \Bbb N\} \cup \{a,b,d\}^*$$

Assuming $$L$$ is regular then there exist a pumping length $$n$$ for $$L$$. Lets use w = $$cda^nb^n$$.

Thus $$w \in L$$ and $$|w| = 2n+2$$ $$\implies$$ $$|w| \geq n$$.

$$w$$ can be splitted into three pieces $$w = xyz$$ with the following conditions:

• $$|xy| \leq n$$
• $$|y| \geq 1$$

Let be $$x = cda^j$$ and $$y = a^k$$ with $$k+j\leq n$$ and $$k\geq 1$$.

Now choose $$i = 2$$ so that $$xy^iz = xyyz = cda^ja^ka^ka^{n-j-k}b^n = cda^{n+k}b^n$$

$$w$$ has now more $$a$$‘s than $$b$$‘s (considering that $$k$$ is at least 1) $$\implies w \notin L \implies$$ $$L$$ is not regular.

Is that enough for the proof?