Prove that this language isn't regular using closure properties of regular languages:
$L=\{a^{j+1}b^kc^{j-k}:j,k,j-k\ge 0\}$
$L$ is over $\sum=\{a,b,c\}$.
My Solution:
We define the homomorphism $h:\sum \to \{a,b\}^*$ this way:
$h(a)=a$
$h(b)=b$
$h(c)=b$
And now we get that $h(L)=\{a^{j+1}b^j:j \ge 0\}$.
Note: I know that this language is irregular and it's very easy to prove using the pumping lemma and here the question ends.
But I've been trying to do something as a challenge to reach the known irregular language $\{a^jb^j:j \ge 0\}$.
But I got stuck trying to get rid of the extra $a$, I've thought of concatenating the regular language $\{b\}$, like this $h(L)\cdot \{b\}$ which still gives me the same language but $j\ge 1$.
I haven't been able to think of a way of doing it using homomorphism, is there a way to delete exactly $1$ $a$ from that language?
Would appreciate any feedback, thanks in advance!
Yes, you can use the fact that each quotient of a regular language is regular. Suppose that $K = \{a^{j+1}b^j \mid j \geqslant 0 \}$ is regular. Then its left quotient by $a$, which is $$a^{-1}K = \{u \mid au \in K\} = \{a^jb^j \mid j \geqslant 0 \}$$ would be regular, a contradiction.