I'm trying to prove that for all $a,b,c,d\in\mathbb{N}$ we have that \begin{align*}\tag{*} (a<b)\wedge (c<d)\Longrightarrow a \cdot c < b\cdot d. \end{align*}
I know that for all $a,b,c,d\in\mathbb{N}$ we have that \begin{align*}\tag{**} (a\leq b)\wedge (c\leq d)\Longrightarrow a\cdot c \leq b\cdot d \end{align*} and, for $c\neq 0$, we also have \begin{align*}\tag{***} a<b \Longleftrightarrow a\cdot c < b\cdot c. \end{align*}
Using $(***)$ we can show that $(*)$ holds for $b\neq 0$ and $c\neq 0$. But how can we prove that $(*)$ holds for all $a,b,c,d\in\mathbb{N}$? Using $(**)$ we can show that $a\cdot c \leq b\cdot d$ for all $a,b,c,d\in\mathbb{N}$. But in order for $(*)$ to hold, we then have to show that $a\cdot c \neq b\cdot d$. Anyone got an idea?
$$ bd - ac = bd - ad + ad - ac = (b-a)d + a (d-c) $$
from your conditions, $b-a, d, a, d-c$ are all positive, the two products are positive, and the sum of those is positive. So $bd - ac >0$