Euclid's lemma $$\forall p[Prime(p) \rightarrow \forall (a, b) [p|ab \rightarrow p|a \lor p|b]] $$ Converse $$\forall p[\forall (a, b) [p|ab \rightarrow p|a \lor p|b] \rightarrow Prime(p)]$$ The proof here makes sense to me. I can't grasp how do concrete examples like p = 9, a = 18, b = 9 fit here?
The assertion here is this: given a natural number $p$, if for every pair $(a,b)$ of natural numbers, when $p\mid ab$, then $p\mid a$ or $p\mid b$, then $p$ is prime. Note the presence of the word “every”. So, indeed, if $p=b=9$ and $a=18$, you do have $p\mid ab\implies p\mid a\vee p\mid b$. And indeed $9$ is not a prime number. However, $(18,9)$ is just one pair of natural numbers. And $9$ is not prime because, for instance, $9\mid3\times6$, but $9\nmid3$ and $9\nmid6$.