Does the logistic function uniquely satisfy these three conditions? Given
$$r(t)=\frac{f(t)}{1-F(t)} \tag{Eq. 1}$$
where
$$f(t)=\frac{dF}{dt} \tag{Eq. 2}$$
and the conditions:
$$\lim_{t\rightarrow \infty} r(t)=1 \tag{Eq. 3}$$
$$\lim_{t\rightarrow \infty} F(t)=1 \tag{Eq. 4}$$
$$\lim_{t\rightarrow \infty} f(t)=1-F(t) \tag{Eq. 5}$$
I can think of just one function $F$ satisfying these three conditions--the logistic function:
$$F(t)=\frac{1}{1+e^{-t}}  \tag{Eq. 6}$$
(which can also be expressed $F(t)=r(t)$)
Is this is the only function satisfying these conditions?  If so, is there a way to prove it?

$$
r(t) = \frac{F'(t)}{1-F(t)}
$$
$$
r(t)\,dt = \frac{dF}{1-F}
$$
$$
\int r(t)\,dt = -\log(1-F(t)) + C
$$
$$
1-\exp\left(-\int r(t)\,dt+C\right)=F(t).
$$
So you can put as many things in the role of $F$ as will fit in the role of $r$, i.e. things satisfying your $\mathrm{Eq.}\ 3$.
