Einstein asserts in his 1905 paper that if two separated clocks at rest in the same frame are synchronized, and one is put in motion at a constant velocity v, it will be out of sync upon arrival at the second clock. No turn around is involved - so since the amount of time difference depends upon the velocity and the length of travel - we can presume that during the travel time the clock that is put in motion runs slower. Now we take two clocks A and B at the origin of the xy axis and a third clock C located at some distance L along the positive x axis. A, B and C are all initially sychronized while at rest in the same frame - then A and B are accelerated to a uniform velocity v in the direction of C (along the x axis). When A and B reach their cruise velocity (no longer accelerating), it is presumed that their reference frame is equally as good as that of C which has not moved. As A and B pass C, B is accelerated in the -X direction until B reaches a velocity v relative to A. Therefore, from C's inertial perspective, clock A will be running slower as measured by clock C. From A's perspective, B will be running slower as measured by the rate of clock A. So we have A running slower than C and B running slow than A. But since B is no longer in motion with respect to C, B and C should be running at the same rate.