Define $g_0(x)$ and $g_1(x)$ so that $$g_0(x) = f(x)\cdot [1-F(x + \delta)] \\ g_1(x) = f(x-c)\cdot[1-F(x-\delta)],$$ where $f(x) = F'(x)$. I know that $F(x)$ is a cdf. Can one can uniquely solve for $F(x)$, $c$, and $\delta$ given knowledge of $g_0(x)$ and $g_1(x)$, at least under certain conditions? Any pointers on how to approach this problem would be useful.
Context: It is known that if $\{X_i\}_{i=1}^n$ are independent random variables, then the distribution of each $X_i$ can be uniquely determined from the joint distribution of the minimum value of $\{X_i\}$ and the identity of the minimum [1] (the problem of the "identified minimum"). The above question is a variant of this problem. Suppose that $\{Y_i\}_{i=1}^n$ are iid, $X_i = c_i + Y_i$ (for real numbers $c_i$), and we observe $(X_i, i)$ if $u_i - X_i \ge u_j - X_j$ for all $j \neq i$. The above equations come from the case where $n = 2$, $c_1 = u_1 = 0$, $\delta \equiv u_2 - c_2$, and $F(\cdot)$ is the cdf of $Y_i$.
[1] https://www.tandfonline.com/doi/abs/10.1080/00401706.1970.10488742