Contrapositive: $\forall\; n > 1, n:$ composite $\implies\exists\; p$ (prime) s.t. $p \leq \sqrt n$ and $p\mid n$ 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.
 A: You have the statement:

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 $p$.

This is a statement of the form $$\text{Let }\;n\in \mathbb Z, n > 1: \quad \forall n\left [\lnot P(n) \implies \exists p (Q(n, p) \land R(n, p))\right]$$
It's contrapositive is:

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

Which is a statement of the form $$\text{Let}\;n\in \mathbb Z, n > 1:\quad  \forall n [\lnot \exists p(Q(n,p) \land R(n,p)) \implies P(n))$$
A: The contrapositive is: Let $n > 1$ be an integer. If there does not exist a prime $p$ such that $p \leq \sqrt{n}$ and $n$ is divisible by $p$, then $n$ is prime. 
Or in other words: let $n > 1$. If for all primes $p$, either $p > \sqrt{n}$ or $n$ is not divisible by $p$, then $n$ is prime.
We are starting out with an arbitrary integer $n > 1$. The contrapositive to "if $n$ is not prime, then there exists a prime number $p$ such that $p≤ \sqrt{n}$ and $n$ is divisible by $p$" is  "if for all primes $p$, either $p > \sqrt{n}$ or $n$ is not divisible by $p$, then $n$ is prime." The contrapositive to $A \implies B$ is "if not $B$, then not $A$." But your original statement was of the form "For all $n > 1$, $A$ implies $B$." So the "contrapositive" would be "For all $n > 1$, not $B$ implies not $A$."
A: 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 $$ \forall p((p \mbox{ is not prime}) \vee (p> n)\vee (p\not|n))\Longrightarrow  ((n\leq1)\vee(\mbox{ $n$ is  prime}))$$
