What will the observer observe? Consider the case where observer $A$ is at rest, and observer $ B$ is moving with speed $\frac{c}{2}$ (where $c$ is the speed of light) propagating a wave with wave speed $c$. So my question is what will be the wave speed observed by observer $A$(which is at rest)? 
I'm really curious to know the answer.
 A: Observer A will see the wave propagating at speed $c$ (the same speed which Observer B sees). This is essentially one of the basic postulates of special relativity: the speed of light (in vacuum) has the same value as measured in all inertial frames.
Even if Observer B moved at speed $0.999c$ relative to Observer A, they would still both see the wave propagate at speed $c$ relative to themselves.
The relevant formula in special relativity should be the velocity-addition formula if you want to look at how you might compute this, but really the velocity-addition formula is constructed so as to reproduce the postulate mentioned above, so arguing from first principles perhaps makes more sense here.
Note that this is very different from Galilean velocity-addition, which would give the intuitive result that observer A sees the wave propagating at speed $\frac{3}{2} c$.
Also I assume here that all the motion is just in one dimension.
A: Einstein's two axioms for special relativity : 

1. PRINCIPLE OF RELATIVITY: The laws of physics are identical in all inertial frames, or, equivalently, the outcome of any physical experiment is the
  same when performed with identical initial conditions relative to any inertial frame. 



2. LAW OF LIGHT PROPAGATION: Light signals in vacuum are propagated rectilinearly, with the same
  speed c, at all times, in all directions, in all inertial frames. 

