Uniqueness of the curve given the curvature and the torsion i was reading the book of Docarmo of differential geometry and I Have a question at the end of the proof that given the curvature and the torsion of a curve, the curve it´s unique , I only omitted the part where it shows that a rigid motion does not alter the curvature and torsion
My question is in the red rectangle, why this equality it´s true? i did not understand it, sorry for my questions...

The author proved that the derivative of $|t-\bar{t}|^2+|n-\bar{n}|^2+|b-\bar{b}|^2$ is $0$. When the derivative of a function is $0$, it means the function is constant. But since the expression is identifiably $0$ at the initial point $s_0$, it must be identically $0$ for all $s$. The only way that's possible is if its constituent parts $|t-\bar{t}|^2,|n-\bar{n}|^2,|b-\bar{b}|^2$ are all $0$. The only way $|t-\bar{t}|^2$ is always $0$ is if $t=\bar{t}$ always holds. But $t=d\alpha/ds$ and $\bar{t}=d\bar{\alpha}/ds$, so we have the equality in red.
