I Phase Speed of Wave in non-relativistic Doppler Shift Derivation

[unified]

TL;DR

The phase speed of a wave in the derivation of the non-relativistic Doppler shift does not change between reference frames. Shouldn't the Galilean transformation apply?

Consider the situation where an observer at rest on the ground measures the frequency of a siren which is moving away from the observer at speed $v_{Ex}$. Let $v_w$ be the speed of the sound wave. Let $\lambda_0$, $f_0$, $\lambda_D$, and $f_D$ be the wavelengths and frequencies measured by the emitter and ground observer. Let T be the wave's period measured by the ground observer. Following the standard non-relativistic doppler shift derivation, $f_D$ = $\frac{v_w}{\lambda_D}$ = $\frac{v_w}{\lambda_0 + v_{Ex}T}$ = $\frac{v_w}{\frac{v_w}{f_0} + \frac{v_{Ex}}{f_0}}$ = $\frac{f_0}{1 + \frac{v_{Ex}}{v_w}}$.

My question, is why is $\lambda_0$ = $\frac{v_w}{f_0}$? If the wave speed on the ground is $v_w$, shouldn't the wave speed as measured by the emitter be calculated using the Galilean transformation? Instead it is the same value as measured by the ground observer.

 

[unified]

To answer my own question, we are comparing the frequency measured by the ground observer -- who is at rest relative to the medium air -- with that measured by an observer moving with the siren and at rest relative to the air. Since they are both at rest relative to the air, they will measure the sound to have the same speed $v_w$.