the pumping lemma is not sufficient

Here is the single most important honesty caveat about the pumping lemma. It is a NECESSARY condition for regularity — every regular language satisfies it — but it is NOT a SUFFICIENT one. In plain terms: 'every regular language pumps' is true, but 'every language that pumps is regular' is FALSE. There exist non-regular languages that nonetheless satisfy the entire pumping condition, so passing the pumping test certifies nothing positive.

Why the gap? The pumping lemma captures only ONE consequence of finite memory — the existence of a repeatable loop on long strings. A devious language can hand you a trivially pumpable structure (so the loop condition is always satisfiable) while still encoding a constraint that finite memory cannot enforce. The standard counterexample is L = { a^i b^j c^k : if i = 1 then j = k } over {a,b,c}, together with all strings where i is not 1. Every long string can be pumped (pump in the a-region when i is not 1; when i = 1 the j = k constraint is not threatened by the available split) — yet L is not regular, as Myhill-Nerode reveals.

The practical upshot is a strict rule of use: only the FAILURE of the pumping lemma proves anything, and what it proves is non-regularity. If you try the pumping lemma and the language stubbornly keeps pumping, you have learned nothing about whether it is regular — you must reach for Myhill-Nerode, which is both necessary AND sufficient and therefore never gives a false 'pass'. The same one-directional warning applies to the pumping lemma for context-free languages.