A conjecture about the prime counting function Using this lemma it can be proved that 
$\Delta(m,n)=\pi(m\cdot n)-\pi(m)\cdot\pi(n)+1$ (where $\pi$ is the prime counting function) 
is a function 
$\Delta:\mathbb N\times\mathbb N\to\mathbb N$. 
Reformulated conjecture:

Given $m\in\mathbb N,m>2$, then $n\in \mathbb N$ is an odd prime if 
  $\Delta(m,n-1)>\Delta(m,n)<\Delta(m,n+1)$ and if $n$ is an odd prime then $\Delta(m,n-1)\ge\Delta(m,n)\le\Delta(m,n+1)$.

The conjecture is tested for $m,n<3000$.

Counterexample: 
$5879$ is a prime, but $\Delta(5878,5879)=1525672>\Delta(5878,5878)=1523414$
 A: Suppose $\Delta(m,n-1)>\Delta(m,n)$, then 
$$\pi(mn-m)-\pi(m)\pi(n-1)+1>\pi(mn)-\pi(m)\pi(n)+1\\\pi(m)\pi(n)-\pi(m)\pi(n-1)>\pi(mn)-\pi(mn-m)\geq 0\\\pi(n)>\pi(n-1)$$ so $n$ must be prime, since $\pi(x)$ increases on $n$.
If $n$ is prime, then $\pi(n)=\pi(n-1)+1$, so we can transform $\Delta(m,n-1)\geq\Delta(m,n)$ as follows:
$$\pi(mn-m)-\pi(m)\pi(n-1)+1\geq\pi(mn)-\pi(m)\pi(n)+1\\
\pi(m)\pi(n)-\pi(m)\pi(n-1)\geq\pi(mn)-\pi(mn-m)\\
\pi(m)\geq\pi(mn)-\pi(mn-m)$$
This is equivalent to specific cases of second Hardy-Littlewood conjecture. Obviously we don't get full generality, but nevertheless I seriously doubt anything is known on this. Even though the mentioned conjecture is believed to be false, this special case might just as well be true.
Lastly, if $n$ is odd prime, then $n+1$ isn't, so $\pi(n+1)=\pi(n)$, so transforming $\Delta(m,n+1)\geq\Delta(m,n)$ we easily get
$$\pi(mn+m)\geq\pi(mn)$$
which is clearly true.
To sum up:

First part of your conjecture is true. Indeed, already $\Delta(m,n-1)>\Delta(m,n)$ implies $n$ is prime.
Primality of $n$ implies $\Delta(m,n+1)\geq\Delta(m,n)$, but whether it implies $\Delta(m,n-1)\geq\Delta(m,n)$ is most likely an open problem.

