I am sorry I misunderstood you. It looks like you are saying that the source and the receiver are both moving at 10 km/sec.origin..

ok we agree LS=c..

So, when travelling LS for 1 second it travels 300.000km correct ?

So, instead it has to travel 300.010 for we have 10 km/s as a starting speed..

And it will take therefore 300.010/300.000 correct ? (1 s + 1(30.000) second right ?

Now on the way back when emitted from that exact 300.010 km, and we are still travelling at that same speed

the point of origin will have moved 10 km closer when the light arrives again right ?

It'l be at 20km..300.010-20 km right ?

So asially when it travels back it reaches the destination sooner..299.990 km travel..

Since LS is constant it'll therefore be less then 1s, 299.990/300.000 <1 second yes ?

this MUST be true when LS is constant , and we agreed on that already..

Now, the distance travelled to and fro will be 300.010+299.990 km altogether yes ?

The start and return point are still 300.000 km apart from each other..from beginning till end..

So yes, IF LS =c and we agreed on that the entire journey will take 2 second right ?

Now, is there a difference in timing when you compare AB and BA against each other ?

300.010 to, 299.990 fro..

That is a very simple question. It would take exactly 1 second from the source to the receiver and it would take exactly 1 second to return. Both the source and the receiver are in the same inertial frame so they both would agree on the time and distance the light traveled.

Consider this. Assume you wake up on what appears to be an airplane but your not sure how or why you are there. Is there any test you can do (other than looking out the window) that will prove that you are either flying at 500 mph or that you are on the ground and you are being tricked into thinking you are flying? In other words are you moving at 0 mph or 500 mph.