Suppose I have the following cdf for random variable $e$: $F_e(t)=Pr(e<t)$. The support of $e$ is the real line.
I could always do the following identity transformation by adding a parameter $r$ on both sides of the inequality and rearrange: $F_e(t)=Pr(e<t)=Pr(r<t+r-e)=Pr(r<t+m)=Pr(-m<t-r)=F_{-m}(t-r)$, where $m=r-e$.
Now suppose I do partial derivative of $F_{-m}(t-r)$ with respect to $r$, if I do it using $F_{-m}(t-r)$, it gives me $\frac{\partial F_{-m}(t-r)}{\partial r}=f_{-m}(t-r)\times(-1)$, where $f_{-m}(\cdot)$ is the pdf of $-m$. However, if I do partial derivative with respect to $F_e(t)$, since it does not depend on $r$,$\frac{\partial F_e(t)}{\partial r}=0$, which is obviously a contradiction because $F_e(t)=F_{-m}(t-r)$ and thus they should have the same partial derivatives.
What's wrong with my derivation and reasoning, and how to correctly calculate $\frac{\partial F_{-m}(t-r)}{\partial r}$ (and show that it indeed equals zero)?