TITLE: Contrapositive: $\forall\; n > 1, n:$ composite $\implies\exists\; p$ (prime) s.t. $p \leq \sqrt n$ and $p\mid n$
QUESTION [5 upvotes]: Usually, I find it a cakewalk to write the contrapositive, but the following statement is quite complex for the task:

For all integers $n > 1$, if $n$ is not prime, then there exists a prime number $p$ such that  $p \leq \sqrt n$ and $n$ is divisible by $p$.

Is it "There exists no prime number $p$ such that $p \leq \sqrt n$ and $p \, | \,  n$  given that $(n > 1)$ for a prime integer $n$"? The weird thing here is that the first universal statement (for all integers $n$) was not converted into a conditional, which makes me uncomfortable.
I'd appreciate some guidance.

REPLY [2 votes]: Let's write it as FO formula: $$(n>1)\wedge(\mbox{ $n$ is not prime})\Longrightarrow \exists p((p \mbox{ is prime}) \wedge (p\leq n)\wedge (p|n)) $$ So the contapositive statement will be $$\sim( \exists p((p \mbox{ is prime}) \wedge (p\leq n)\wedge (p|n)))\Longrightarrow \sim ((n>1)\wedge(\mbox{ $n$ is not prime}))$$ which is same as $$ \forall p((p \mbox{ is not prime}) \vee (p> n)\vee (p\not|n))\Longrightarrow  ((n\leq1)\vee(\mbox{ $n$ is  prime}))$$