$L_4 = \{ w | w \text{ does not contain symbol a immediately followed by symbol b} \} $
where: $ \Sigma = \{ a, b, c \} $
So far I believe I have a defined regular grammar for this language, however I am having trouble coming up with an equivalent regular expression.
$<Language\_4> ::= a <AOrC> $
$<Language\_4> ::= b <Language\_4> $
$<Language\_4> ::= c <Language\_4> $
$<Language\_4> ::= \epsilon $
$<AOrC> ::= a <Language\_4> $
$<AOrC> ::= c <Language\_4> $
$<AOrC> ::= \epsilon $
Via a standard regex grammar this could be done via a lookahead, however I don't believe that the lookahead operation is a valid regular language operation. I'm having trouble proving that if it is true though.
This is for a homework assignment, so I'm not expecting a full answer just some pointers in the right direction.
Edit
I went through this set of answers: How to convert finite automata to regular expressions?
I then used Arden's Lemma as it seemed the most straight-forward.
First I defined:
$Q_0 = aQ_1 \cup bQ_0 \cup cQ_0 \cup \epsilon $
$Q_1 = cQ_0 \cup aQ_1 \cup \epsilon $
$ Q_1 = cQ_0 \cup aQ_1 \cup \epsilon $
Apply Arden's Lemma:
$ U = a $, $ L = Q_1 $, $ V = cQ_0 $
Then I get:
$Q_1 = a * c Q_0 \cup \epsilon $
By then substituting $Q_1$ into $Q_0$ I get:
$Q_0 = a[a*cQ_0 \cup \epsilon] \cup bQ_0 \cup cQ_0 \cup \epsilon $
$Q_0 = \underbrace{(aa * ac \cup b \cup c)}_U \underbrace{Q_0}_L \cup \underbrace{a \cup \epsilon }_V $
$Q_0 = (aa * ac \cup b \cup c)* a$
Is this a correct application of the lemma?
EDIT 2
I've tested the regex more and it doesn't match the desired language. I think there may be an error in my application of the lemma. I'm going to try the application again in case I missed something.
