Let $S$ be an Inertial Frame. Consider two timelike separated events in S: $\mathcal{A}$ and $\mathcal{B}$. Consider two massive moving particles in $S$: $\mathcal{P1}$ and $\mathcal{P2}$ wich both have trajectories beggining at $\mathcal{A}$ and ending in $\mathcal{B}$. $\mathcal{P1}$ have a non-accelerated motion, with a _constant_ 3-velocity relative to $S$. $\mathcal{P2}$ have an accelerated trajectory with instantaneous 3-velocity relative to $S$. Inertial Observers can disagree if particle $\mathcal{P1}$ is moving or not, but all inertial observers will agree that particle $\mathcal{P1}$ is in movement.

Consider that the clocks moving with the particles and the clock o $S$ starts at 0, and the particles start to move in theis respective trajectories. Since $\mathcal{P1}$ measures a proper time $\tau $ and $\mathcal{P1}$ measure another proper time $\tau '$ between $\mathcal{A}$ and $\mathcal{B}$, I'm trying to prove that all observers agree that $\tau > \tau '$. I'm using the metric signature $(+,-,-,-)$. Also I'll assume one-dimensional spacial motion without loss of generality and using $c=1$.

_ATTEMPT OF A PROOF:_

Let $K$ be the rest-frame of $\mathcal{P1}$ an $K'$ be some instantaneous rest-frame of $\mathcal{P1}$

Consider two events in $K'$ infinitesimally separeted such that the spacetime interval in $K'$ is given by $ds^2 = d\tau ' ^2$. These infinitesimally separated events in $K$ have spacetime interval given by $ds^2 = d\tau' ^2 = d\tau^2 - dx^2$, where $dx$ is the spacial displacement of those events measured by $K$. Then we have that

\begin{align}

d\tau ' ^2 &= d\tau ^2 - dx^2 \\ &= d\tau^2\left[ 1 - \left( \frac{dx}{d\tau}\right)^2\right] \\ &= d\tau ^2 (1 - v^2)

\end{align}

\begin{equation}

\Longrightarrow d\tau' = d\tau \sqrt{1 - v^2} \Longrightarrow d\tau' < d\tau

\end{equation}

where $v$ is the isntantaneous velocity of $\mathcal{P2$}$ measured by $\mathcal{P1}$.

Since $\tau = \int_{\tau _1}^{\tau _2} d\tau$ and $\tau ' = \int_{\tau _1}^{\tau _2} \sqrt{1 - v^2} d\tau$ and $\sqrt{1-v^2} \le 1$ , we conclude that $\tau > \tau '$.

_WHERE THE PROBLEM LIES:_

I can use the same argument as above, by considering there are two infinitesimally separated events in the rest frame $K$ of $\mathcal{P1}$ such that the spacetime interval measured is $ds^2 = d\tau^2$ this interval measured by $K'$, the instantaneous rest-frame of $\mathcal{P2}$ gives the opposite relation

\begin{equation}

d\tau^2 = d\tau' ^2 - dx' ^2

\end{equation}

giving me the opposite result. But this is inconsistent since all particles agree that $\mathcal{P2}$ is moving because he's accelerated. Where's my mistake. How to correct this inconsistence?