Why does removing all copies of a letter preserve regularity? Let $P(a,L)$ remove every $a$ in $L$, for example
$$ P(a,\{ab,aab,aaab,bba\}) = \{b,bb\}. $$

How to show that if $L$ is a regular language then $P(a, L)$ is also a regular language?



My attempt:

If $L$ is regular then $P(a, L)$ is regular. Namely, let $D = (Q,Σ,δ,q_0,F)$ be a DFA for $L$. 

Assume every state in $L$ is reachable from $q_0$. Then we can turn $D$ into an NFA $N$ for $P(a, L)$ by adding a new state $q_{−1}$ which becomes the NFA’s start state. For each state $q ∈ L$, we add an epsilon transition from $q_{−1}$ to $q$.

Therefore, define $N$ to be 
$( L ∪ \{q_{−1}\}, Σ , δ′, q_{−1},F)$



Is this the correct way to prove it?
Given an NFA or a DFA (like in your proof, if you want), we replace each arc with letter $a$ by $\varepsilon$. We obtain an NFA for $P(a,L)$. So, $P(a,L)$ must be regular.

In details, each trace in the newly constructed NFA can be transformed back to $a$-full instance of $L$ .