The main point to get from Special Relativity's so-called Twin Paradox, is that imaginary elapsed coordinate time, Δt, is different than physically relevant elapsed proper time, Δτ.
Coordinate time is dependent on an imaginary choice of inertial coordinates, so varies from frame to frame by the Lorentz transform.
$$\Delta \vec{x}'_{AB} = \Delta \vec{x}_{AB} + \frac{1}{\vec{u}^2} \left( \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} - 1 \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \vec{u} + \frac{\vec{u}}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}}\Delta t_{AB} \\
\Delta t'_{AB} = \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}}\Delta t_{AB} + \frac{1}{c^2 \sqrt{1 - \frac{\vec{u}^2}{c^2}}} \vec{u} \cdot \Delta \vec{x}_{AB} $$
Proper time, while path-dependent, is independent of the choice of inertial coordinates use to describe a time-like trajectory through space-time:
$$\Delta \tau_{AB} = \int_{A}^{B} d\tau = \int_{A}^{B} \sqrt{1 - \frac{\left( \vec{v}(t) \right)^2}{c^2}} dt = \int_{A}^{B} \sqrt{1 - \frac{\left( \vec{v}'( t' ) \right)^2}{c^2}} dt'$$
This follows because the relativistic interval likewise is independent of the choice of inertial coordinates use to describe a time-like separation in space-time:
$$
c^2 \left( \Delta t'_{AB} \right)^2 - \left( \Delta \vec{x}'_{AB} \right)^2
\\ = c^2 \left( \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}}\Delta t_{AB} + \frac{1}{c^2 \sqrt{1 - \frac{\vec{u}^2}{c^2}}} \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2 - \left( \Delta \vec{x}_{AB} + \frac{1}{\vec{u}^2} \left( \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} - 1 \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \vec{u} + \frac{\vec{u}}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} \Delta t_{AB} \right)^2
\\ = \frac{c^2}{1 - \frac{\vec{u}^2}{c^2}} \left(\Delta t_{AB} \right)^2
+ \frac{2}{1 - \frac{\vec{u}^2}{c^2}} \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \left(\Delta t_{AB} \right)
+ \frac{1}{c^2 } \frac{1}{1 - \frac{\vec{u}^2}{c^2}} \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2
\\ \quad \quad
- \left( \Delta \vec{x}_{AB} \right)^2
- \frac{1}{\vec{u}^2} \left( \frac{1}{1 - \frac{\vec{u}^2}{c^2}} - \frac{2}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} + 1 \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2
- \frac{\vec{u}^2}{1 - \frac{\vec{u}^2}{c^2}} \left( \Delta t_{AB} \right)^2
\\ \quad \quad
- \frac{2}{\vec{u}^2} \left( \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} - 1 \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2
- \frac{2}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \Delta t_{AB}
-2 \left( \frac{1}{1 - \frac{\vec{u}^2}{c^2}} - \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \Delta t_{AB}
\\ = \frac{c^2 - \vec{u}^2}{1 - \frac{\vec{u}^2}{c^2}} \left(\Delta t_{AB} \right)^2 - \left( \Delta \vec{x}_{AB} \right)^2
+ \left( \frac{1}{c^2 - \vec{u}^2} + \frac{1}{\vec{u}^2} - \frac{c^2}{\left( \vec{u}^2 \right) \left( c^2 - \vec{u}^2 \right) } \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2
\\ = c^2\left(\Delta t_{AB} \right)^2 - \left( \Delta \vec{x}_{AB} \right)^2
$$
Coordinate time is dependent on an imaginary choice of inertial coordinates, so varies from frame to frame by the Lorentz transform.
$$\Delta \vec{x}'_{AB} = \Delta \vec{x}_{AB} + \frac{1}{\vec{u}^2} \left( \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} - 1 \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \vec{u} + \frac{\vec{u}}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}}\Delta t_{AB} \\
\Delta t'_{AB} = \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}}\Delta t_{AB} + \frac{1}{c^2 \sqrt{1 - \frac{\vec{u}^2}{c^2}}} \vec{u} \cdot \Delta \vec{x}_{AB} $$
Proper time, while path-dependent, is independent of the choice of inertial coordinates use to describe a time-like trajectory through space-time:
$$\Delta \tau_{AB} = \int_{A}^{B} d\tau = \int_{A}^{B} \sqrt{1 - \frac{\left( \vec{v}(t) \right)^2}{c^2}} dt = \int_{A}^{B} \sqrt{1 - \frac{\left( \vec{v}'( t' ) \right)^2}{c^2}} dt'$$
This follows because the relativistic interval likewise is independent of the choice of inertial coordinates use to describe a time-like separation in space-time:
$$
c^2 \left( \Delta t'_{AB} \right)^2 - \left( \Delta \vec{x}'_{AB} \right)^2
\\ = c^2 \left( \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}}\Delta t_{AB} + \frac{1}{c^2 \sqrt{1 - \frac{\vec{u}^2}{c^2}}} \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2 - \left( \Delta \vec{x}_{AB} + \frac{1}{\vec{u}^2} \left( \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} - 1 \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \vec{u} + \frac{\vec{u}}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} \Delta t_{AB} \right)^2
\\ = \frac{c^2}{1 - \frac{\vec{u}^2}{c^2}} \left(\Delta t_{AB} \right)^2
+ \frac{2}{1 - \frac{\vec{u}^2}{c^2}} \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \left(\Delta t_{AB} \right)
+ \frac{1}{c^2 } \frac{1}{1 - \frac{\vec{u}^2}{c^2}} \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2
\\ \quad \quad
- \left( \Delta \vec{x}_{AB} \right)^2
- \frac{1}{\vec{u}^2} \left( \frac{1}{1 - \frac{\vec{u}^2}{c^2}} - \frac{2}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} + 1 \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2
- \frac{\vec{u}^2}{1 - \frac{\vec{u}^2}{c^2}} \left( \Delta t_{AB} \right)^2
\\ \quad \quad
- \frac{2}{\vec{u}^2} \left( \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} - 1 \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2
- \frac{2}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \Delta t_{AB}
-2 \left( \frac{1}{1 - \frac{\vec{u}^2}{c^2}} - \frac{1}{\sqrt{1 - \frac{\vec{u}^2}{c^2}}} \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right) \Delta t_{AB}
\\ = \frac{c^2 - \vec{u}^2}{1 - \frac{\vec{u}^2}{c^2}} \left(\Delta t_{AB} \right)^2 - \left( \Delta \vec{x}_{AB} \right)^2
+ \left( \frac{1}{c^2 - \vec{u}^2} + \frac{1}{\vec{u}^2} - \frac{c^2}{\left( \vec{u}^2 \right) \left( c^2 - \vec{u}^2 \right) } \right) \left( \vec{u} \cdot \Delta \vec{x}_{AB} \right)^2
\\ = c^2\left(\Delta t_{AB} \right)^2 - \left( \Delta \vec{x}_{AB} \right)^2
$$
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