I would like to compare two proper times of two inertial observers in Special Relativity.
Let's consider three inertial observers A, O and B freely moving on a line in a flat space-time, with B moving with a constant velocity $v$ with respect to A and O moving with a constant velocity $w$ with respect to A.
In the rest frame of O, A will move with the velocity $-w$ (always), and, by choosing $w$ carefully, B can move with the same velocity but in the opposite direction (+$w$). I just created an in-between frame in which the situation is symmetrical. Applying the basic time dilation formula in the rest frame of O for A and B respectively, I find $d\tau_{A} = d\tau_{B}$.
Mathematical proof
I want the velocity of O seen from B to be $-w$. I therefore need to solve the following system based on the velocity addition formula: $$ -w = \frac{w-v}{1-\frac{wv}{c^{2}}}$$
This leads to
$$w = \frac{c^{2}}{v}\left(1-\sqrt{1-\left(\frac{v}{c}\right)^{2}}\right)$$
In the rest frame of O associated with the coordinate time $t_{o}$, we have:
$$d\tau_{A} = \frac{dt_{o}}{\gamma(-w)}$$ $$d\tau_{B} = \frac{dt_{o}}{\gamma(w)}$$
With $\tau_{A}$ and $\tau_{B}$ the proper times of A and B, and $\gamma(w) = 1/\sqrt{1-(w/c)^{2}}$
Leading to $d\tau_{A} = d\tau_{B}$. Ok.
Now let's go back in the rest frame of A. We associate a coordinate time $t$ to it. We therefore have:
$$d\tau_{B} = \frac{dt}{\gamma(v)}$$
Similarly, in the rest frame of B associated with a coordinate time $t'$ :
$$d\tau_{A} = \frac{dt'}{\gamma(-v)}$$
Which leads to $dt=dt'$. Or we know that:
$$dt' = \gamma(v)(dt-\frac{v}{c^{2}}dx)$$
It seems like there is an inconsistency. Please let me know what is wrong, thank you!