Using the pumping lemma for $L=\{a^{p}b^{q} ∣ 0 ≤ p ≤ q\}$ I need to prove that $L$ is irregular. I already have proven the irregularity for $L=\{a^{p}b^{q} ∣ 0 ≤ p < q\}$.
I have a gut feeling that these proofs will be identical. Is this correct?
A short summary of what I think the proof would look like:
1) define a word $w = a^{m}b^{m}$ which should be in $L$ and $|w| \geq m$
2) pumping lemma makes $w = xyz, y \neq \epsilon, |xy|≤ m$
3) define $x$, $y$ and $z$ (important is that $y$ would only consist of $a$'s)
4) after pumping realise that the amount of $a$'s is not valid with the amount of $b$'s in the language
5) Q.E.D. (contradiction)
Let $A = \{a, b\}$, $L_1 =\{a^{p}b^{q} \mid 0 \leqslant p \leqslant q\}$ and $L_2 =\{a^{p}b^{q} \mid 0 \leqslant p < q\}$. Then $$ a^{-1}L_1 = \{u \in A^* \mid au \in L_1\} = L_2 $$ It is known that any quotient of a regular language is again regular. Therefore, if $L_1$ were regular, then $L_2$ would also be regular. Since you proved that $L_2$ is not regular, you are done.