Consider the primal linear program
\begin{align} \operatorname{min} c^T &x \\ A&x\leq b \\ &x \geq 0 \end{align}
In the dual problem, we associate each constraint $i$ with dual variable $y_i \leq 0$ and each variable $j$ with constraint $a_j^T y \leq c_j$. Therefore, the dual is
\begin{align} \operatorname{max} b^T &y \\ A^T&y\leq c \\ &y \leq 0 \end{align}
Now if the primal is a maximization problem the memorization rule is a little different.
\begin{align} \operatorname{max} c^T &x\\ A&x\leq b \\ &x \geq 0 \end{align}
has the dual
\begin{align} \operatorname{min} b^T &y\\ A^T&y\geq c \\ &y \geq 0 \end{align}
I didn't want to memorize the rule for the primal max case, so I thought I could just transform the maximization problem to minimization problem and follow the steps from there.
\begin{align} \operatorname{max} c^T &x \to - \operatorname{min} -c^T x\\ A&x\leq b \\ &x \geq 0 \end{align}
Which has the dual
\begin{align} - \operatorname{max} b^T &y \\ A^T&y\leq -c \\ &y \leq 0 \end{align}
Now if I change the objective and multiply each row with $-1$, I get \begin{align} \operatorname{min} -b^T &y \\ -A^T&y\geq c \\ -&y \geq 0 \end{align}
I cannot reason that this is equivalent to $\operatorname{min}\{ b^T y \,|\, A^T y \geq c, y \geq 0\}$. Did I miss something?
So you are asking why \begin{align} \operatorname{min} -b^T &y \\ -A^T&y\geq c \\ -&y \geq 0 \end{align}
is equivalent to \begin{align} \operatorname{min} b^T &y\\ A^T&y\geq c \\ &y \geq 0 \;? \end{align}
This can be seen by substituting $y$ with $-y$.
To make it more explicit:
As an intermediate step we can write down \begin{align} \operatorname{min} b^T &(-y) \\ A^T&(-y)\geq c \\ &(-y) \geq 0. \end{align}
This is equivalent to the first system by linearity, and then if you replace $y$ with $(-y)$, you can see that it is equivalent to the second system.