special relativity - Why is rapidity additive?

With the rapidity $\phi$ defined so that $\frac{v}{c}=\tanh{\phi}$, say you have 3 parallel moving reference frames $S$, $S'$ and $S''$ with a constant but different velocity/rapidity.

If the rapidity of $S'$ in relation to $S$ is $\phi_1$,

and the rapidity of $S''$ in relation to $S'$ is $\phi_2$,
then the rapidity of $S''$ in relation to $S$ is $\phi_1+\phi_2$

I don't really see why this is the case, it's probably something really simple I'm missing because there's no further explanation in my syllabus.

Answer

Alright here's a very crude heuristic(definitely not a proof) on why(without using velocity composition) rapidity is additive.

Impose these two constraints:

1) the constancy of speed of light in all frames. This means that the rapidity of light $\tanh\phi_c=1$ is the same in all frames of reference.

Consider this case:

You observe a frame of reference $S'$ moving with $\tanh \phi_1=v$ with respect to $S$ followed by a light beam with $\tanh\phi_c=1$. we know the rapidity of light in the $S'$ frame is the same as in $S$. Given this fact we'd like to find out what is the relation $f(\phi_1,\phi_c)$ between the old and new rapidities in the new frame $S'$ such that:

$$\tanh( f(\phi_1,\phi_c))=1$$

2) The second constraint is that the rapidity of an object at rest is zero. So if there's an object moving with velocity $v$ in $S$, it's at rest in $S'$. That is

$$\tanh( f(\phi_1,\phi_1))=0$$

Then it's reasonable to conjecture that $f(\phi_1,\phi_2)=\phi_2 -\phi_1$

As you can check yourself using the hyperbolic identity

$$\tanh (\phi_1 +\phi_2)=\dfrac{\tanh \phi_1+ \tanh \phi_2}{1+\tanh \phi_1 \tanh \phi_2}$$

Update: I was asked in the comments by user Prahar that I jumped into the conclusion too quickly. why for instance the functions

$$(\phi_2 -\phi_1)^2$$ Or say $$(\phi_2 -\phi_1)^n$$ Or say $$\ln(\dfrac{\phi_2}{\phi_1})$$ are invalid?Although they satisfy the first two constraints?

The reason has to do with another constraint:

3)The principle of relativity dictates the following: If you're $S$ and $S'$ is moving with $\tanh \phi_1 =v$ according to $S$, then according to $S'$, $S$ is moving with $-v$ so that

$$\tanh (f(\phi_1,0))=-v$$ This is only satisfied for $$\phi_2 -\phi_1$$

Because for

$$(\phi_2 -\phi_1)^2$$ or any other $n$, We have $$\tanh ( (f(\phi_1,0))= \tanh (0 -\phi_1)^2= \tanh \phi_1^2$$ which is evidently not equal to $-v$ because $\tanh \phi_1^2$ is positive(whereas it should have been negative) and also because it's never equal to $v$, since $\tanh$ is a one to one function.