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?