Do $X,Z$ and $Y,Z$ have the same density if $X, Y \sim \text{Unif}(0,1)$, $X \perp \!\!\! \perp Y$, and $Z = X + Y$ Let


*

*$X, Y \sim \text{Unif}(0,1)$

*$X \perp \!\!\! \perp Y$, and

*$Z = X + Y$
Is it true that $f_{X\mid Z=z}(s)$ and $f_{Y\mid Z=z}(s)$ for all $s$ (i.e. they have the same density)?
Disclosure: this is part of a homework question, but the TA said we can just claim this is true "by symmetry" without any additional work. There is more to the problem. I wanted to know if there is a more rigorous proof that this is true.
Intuitively, this seems true since $Z = X + Y$. (It doesn't seem true in general, for example of $Z = 2X + Y$). My approach to show this is to show
$$
\mathbb{P}\{X \leq s, Z \leq t\} = \mathbb{P}\{Y \leq s, Z \leq t\} \quad \forall s, t
$$
and so
\begin{align}
\mathbb{P}\{X \leq s, Z \leq t\} &= \mathbb{P}\{Y \leq s, Z \leq t\}
\\
\mathbb{P}\{X \leq s, X+Y \leq t\} &= \mathbb{P}\{Y \leq s, X+Y \leq t\}
\\
\mathbb{P}\{X \leq s, Y \leq t-X\} &= \mathbb{P}\{Y \leq s, X \leq t-Y\}
\end{align}
Now we've put the joint CDF of $X$ and $Y$. But at this point I am stuck.

For $0\leq t-s \leq 1$, we have
\begin{align}
\mathbb{P}\left\{X\leq s,Y\leq t-X\right\}&=\mathbb{P}\left\{X\leq s,X\leq t-Y\right\} \\
&= \int_{0}^{t-s} \mathbb{P}\left\{X\leq s\right\}dy + \int_{t-s}^{1} \mathbb{P}\left\{X\leq t-y\right\}dy \\
&= \int_{0}^{t-s} s dy + \int_{t-s}^{1}  (t-y)dy \\
&= s(t-s) +  t(1-t+s)-(1-(t-s)^2)/2 = st-s^2/2-(1-t)^2/2. 
\end{align}
So, we have,
\begin{equation}
\mathbb{P}\left\{X\leq s,Y\leq t-X\right\}=\begin{cases}
 st-s^2/2-(1-t)^2/2,& \text{ if } 0\leq t-s\leq 1\\
0, & \text{ otherwise}. 
\end{cases}
\end{equation}
