See the bottom section for a discussion of the posted answer.
My question is at the end of this section, above the horizontal rule.
Warning: my use of proper time $\Delta\tau$ and proper distance $\Delta s$
are a bit tricky in the following. I am resolving arbitrary intervals
in the $\bar{\mathcal{S}}$ system into sums of space-like and time-like
components, along the $\bar{\mathcal{S}}$ basis vectors.
The goal is to establish a formula transforming components $\left\{ \Delta\bar{t},\Delta\bar{x}\right\} $
of space-time intervals given in the platform $\bar{\mathcal{S}}$
inertial coordinate system to components $\left\{ \Delta t,\Delta x\right\} $
of the onboard $\mathcal{S}$ inertial coordinate system when $\mathcal{S}$
moves with speed $v$ in the positive $x$ direction relative to $\bar{\mathcal{S}}$.
Based on the assumption that space-time is homogeneous we know our
transformation must be linear, which leads to the formal matrix equation
$$
\begin{bmatrix}\mathrm{c}\Delta t\\
\Delta x
\end{bmatrix}=\begin{bmatrix}e^{t}{}_{\bar{t}} & e^{t}{}_{\bar{x}}\\
e^{x}{}_{\bar{t}} & e^{x}{}_{\bar{x}}
\end{bmatrix}\begin{bmatrix}\mathrm{c}\Delta\bar{t}\\
\Delta\bar{x}
\end{bmatrix}.
$$
The plan is to first treat the case in which $\Delta\bar{x}=0$ in
order to establish the first column of our transformation matrix.
Then we treat the case of $\Delta\bar{t}=0$ to get the second column.
By purely formal considerations we establish for the case of $\Delta\bar{x}=0,\Delta\bar{t}=\Delta\tau$
the following
$$
\begin{bmatrix}\mathrm{c}\Delta t\\
\Delta x
\end{bmatrix}=\begin{bmatrix}e^{t}{}_{\bar{t}} & e^{t}{}_{\bar{x}}\\
e^{x}{}_{\bar{t}} & e^{x}{}_{\bar{x}}
\end{bmatrix}\begin{bmatrix}\mathrm{c}\Delta\bar{t}\\
0
\end{bmatrix}=\mathrm{c}\Delta\bar{t}\begin{bmatrix}e^{t}{}_{\bar{t}}\\
e^{x}{}_{\bar{t}}
\end{bmatrix}=\mathrm{c}\Delta\tau\begin{bmatrix}e^{t}{}_{\bar{t}}\\
e^{x}{}_{\bar{t}}
\end{bmatrix}
$$
$$
\mathrm{c}\Delta t=e^{t}{}_{\bar{t}}\mathrm{c}\Delta\bar{t}=e^{t}{}_{\bar{t}}\mathrm{c}\Delta\tau
$$
$$
\implies e^{t}{}_{\bar{t}}=\frac{\Delta t}{\Delta\bar{t}}=\frac{\Delta t}{\Delta\tau}
$$
$$
\Delta x=e^{x}{}_{\bar{t}}\mathrm{c}\Delta\bar{t}=-v\Delta\bar{t}
$$
$$
\implies e^{x}{}_{\bar{t}}=\frac{\Delta x}{\mathrm{c}\Delta\bar{t}}=-\frac{v}{\mathrm{c}}\frac{\Delta t}{\Delta\tau}=-\frac{v}{\mathrm{c}}e^{t}{}_{\bar{t}}.
$$
And for the case of $\Delta\bar{t}=0,\Delta\bar{x}=\Delta s>0$ we
have
$$
\begin{bmatrix}\mathrm{c}\Delta t\\
\Delta x
\end{bmatrix}=\begin{bmatrix}e^{t}{}_{\bar{t}} & e^{t}{}_{\bar{x}}\\
e^{x}{}_{\bar{t}} & e^{x}{}_{\bar{x}}
\end{bmatrix}\begin{bmatrix}0\\
\Delta\bar{x}
\end{bmatrix}=\Delta\bar{x}\begin{bmatrix}e^{t}{}_{\bar{x}}\\
e^{x}{}_{\bar{x}}
\end{bmatrix}=\Delta s\begin{bmatrix}e^{t}{}_{\bar{x}}\\
e^{x}{}_{\bar{x}}
\end{bmatrix}
$$
$$
\mathrm{c}\Delta t=e^{t}{}_{\bar{x}}\Delta\bar{x}=e^{t}{}_{\bar{t}}\mathrm{c}\Delta s
$$
$$
\implies e^{t}{}_{\bar{x}}=\frac{\mathrm{c}\Delta t}{\Delta\bar{x}}=\frac{\mathrm{c}\Delta t}{\Delta s}
$$
$$
\Delta x=e^{x}{}_{\bar{x}}\Delta\bar{x}=e^{x}{}_{\bar{x}}\Delta s
$$
$$
\implies e^{x}{}_{\bar{x}}=\frac{\Delta x}{\Delta\bar{x}}=\frac{\Delta x}{\Delta s}.
$$
Combining results and introducing the abbreviations $\gamma=\Delta t/\Delta\tau$
and $\beta=v/\mathrm{c}$ gives
$$
\begin{bmatrix}e^{t}{}_{\bar{t}} & e^{t}{}_{\bar{x}}\\
e^{x}{}_{\bar{t}} & e^{x}{}_{\bar{x}}
\end{bmatrix}=\begin{bmatrix}\frac{\Delta t}{\Delta\bar{t}} & \frac{\mathrm{c}\Delta t}{\Delta\bar{x}}\\
\frac{\Delta x}{\mathrm{c}\Delta\bar{t}} & \frac{\Delta x}{\Delta\bar{x}}
\end{bmatrix}=\begin{bmatrix}\gamma & \frac{\mathrm{c}\Delta t}{\Delta s}\\
-\beta\gamma & \frac{\Delta x}{\Delta s}
\end{bmatrix}.
$$
Using a scenario in the case of $\Delta\bar{x}=0,$ we find an expression
for $\gamma$ in terms of the boost speed $v$. For simplicity we
shall assume the coordinates of our initial event are all zero, and
the $z$ direction is not involved. We also use the principle of relativity
to equate transverse components. Thus
$$
0=t_{o}=\bar{t}_{o}=x_{o}=\bar{x}_{o}=y_{o}=\bar{y}_{o}=z
$$
$$
y=\bar{y}
$$
$$
\Delta\bar{t}=\bar{t}-\bar{t}_{o}=\bar{t}
$$
$$
\Delta t=t-t_{o}=t,\text{etc.}
$$
Consider the initial event $\mathcal{E}_{o}$ to be the emission of
a flash of light, and the terminating event $\mathcal{E}_{\tau}=\left\{ \bar{t}=\tau,\bar{x}=0,\bar{y}=y=\mathrm{c}\tau\right\} $
to be its detection by a device located directly above the source
in the platform system. In the onboard system, the detector is seen
to move by $x=-vt;$ so the light travels along the hypotenuse of
length $h=\mathrm{c}t.$ From this we get
$$
h=\mathrm{c}t=\sqrt{x^{2}+y^{2}}=\sqrt{\left(-vt\right)^{2}+\left(\mathrm{c}\tau\right)^{2}}
$$
$$
t^{2}\left(1-\beta^{2}\right)=\tau^{2}
$$
$$
\frac{t}{\tau}=\frac{1}{\sqrt{1-\beta^{2}}}=\gamma
$$
The next step is where I get stuck. The objective is to establish
expressions for the second column independently of my derivation of
those for the first column. I need a scenario in which two events
are simultaneous in $\bar{\mathcal{S}},$ separated by $\Delta\bar{x}=\Delta s.$
From that I want to find the values for the second column.
Here's my current effort:
Let event $\mathcal{E}_{o}=\left\{ \bar{t}_{o}=0,\bar{x}_{o}=0\right\} $
be the emission of a flash from the center of the platform such that
it arrives at the ends of the platform (events $\mathcal{E}_{r}$
at the rear and $\mathcal{E}_{f}$ at the front) as the center of
the train coincides with the emitter $\mathcal{E}_{a}$. Since the
center of the train is to the left of the source event, the onboard
time $\Delta t_{f}$ that the flash takes to reach the front of the
platform will be less than the time $\Delta t_{r}$ that the flash
takes to reach the rear of the platform. From this we conclude that
$t_{f}<t_{a}<t_{r}.$ Arguing on the basis of homogeneity and symmetry,
we also conclude that $t_{r}-t_{a}=t_{a}-t_{f}.$
In order to keep track of things, lets put marks on the train recording
the $\mathcal{S}$ 3-space locations of $\mathcal{E}_{o},\mathcal{E}_{r},\mathcal{E}_{f}$.
On the platform $\bar{x}$ coordinate values are measured from the
center of the platform. Likewise, the onboard $x$ values are measured
from the center of the train. We can set the onboard clock to zero
at the time of $\mathcal{E}_{o}=\left\{ t_{o}=0,x_{o}=-vt_{a}\right\} .$
At time $\bar{t}_{a}$ the platform observer system measures the $\bar{x}$
separations of the marks on the train. These ratios of those distances
will be the same as the corresponding ratios of the onboard proper
distances between the marks.
In the drawing, the onboard system is red. The platform is blue, and
light is yellow.
Can this be done without resorting to the use of a transverse displacement?
My Diagram of the Scenario in the Answer
I labeled things differently, but an interested reader should be able to sort things out. My diagram only treats the first part of the answer. The takeaway is that the two heavy-lined triangles are "spacetime similar".
This is a table of equations relating features of the diagram. Fixed points in the "stationary" frame $\mathscr{S}$ are represented by the worldlines $\mathcal{P,Q,R}.$ Fixed points in the "moving" $\bar{\mathscr{S}}$ frame are $\mathcal{\bar{P},\bar{Q},\bar{R}}.$ Superscripts indicate components, and their decoration indicates the inertial system the component is resolved in. So (with $y,z$ dimensions suppressed), for example, $\bar{\mathcal{P}}^{x}\left[t_{4}\right]$ is the 3-space $\mathscr{S}$ position of the point (worldline) $\bar{\mathcal{P}}$ resting in $\bar{\mathscr{S}}$ at the $\mathscr{S}$ time $t_4$ of event $\mathcal{E}_4.$ Observe that $\bar{\mathcal{P}}\left[t_{4}\right]$ is the unlabeled event at the intersection of the $\bar{\mathcal{P}}$ worldline and the line connecting events $\mathcal{E}_4$ and $\mathcal{E}_5,$ just above the heavy red line. Events $\mathcal{L}_{e}$ and $\mathcal{L}_{d}$ are light emission and light detection, respectively.
$\dagger_{1}$ Omitting the subscript on the time parameter says that the measurement is constant over time.
This is a table summarizing key relationships, and relating different forms of notation. The right two boxes relate my symbols to those used in the posted answer.
And, finally, a rewrite of the first steps of calculation in the posted answer. This should take you far enough to appreciated the "spacetime similar triangles" argument.
Best Answer
0. Summary and Outline of the following derivations
Yes1.
$\def\ReceivedAndReflected{{\hphantom{.}\text r \hphantom{.}}\llap{\bigcirc}}$ $\def\SimultaneousTo{{\hphantom{.}\text s \hphantom{.}}\llap{\bigcirc}}$ $\def\CoincidentWith{{\hphantom{.}\text c \hphantom{.}}\llap{\bigcirc}}$ The proof presented below (completed in section 3) uses two lemmas:
$$\left( \frac{AB}{PQ} \right) = \sqrt{1 - \beta^2},$$
where $c \, \beta$ denotes the constant and mutually equal speed of all members of the one inertial system wrt. members of the other. Further:
$$ \left( \frac{\tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith J \, ]}}{\tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo J \CoincidentWith A \, ]}} \right) $$
between $A$'s duration from its indications of having met and passed $P$ until its indication of having met and passed $J$ and $P$'s duration from its indication of having met and passed $A$ until its indications simultaneous to $J$'s of having met and passed $A$. As to be expected (for a derivation of what's a.k.a. "time dilation, in SR"), the ratio between these durations is obtained as
$$ \left( \frac{\tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith J \, ]}}{\tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo J \CoincidentWith A \, ]}} \right) = \sqrt{1 - \beta^2}. $$
These derivations are accompanied by some suggestive illustrations ("one-dimensional snapshots") as memory aides, but presently without much additional verbiage (in order for the length of my answer to be not too excessive).
In section 3 follows a proof of invariance of spacelike interval $s^2[ \, \varepsilon_{AP}, \varepsilon_{BQ} \, ]$ as representative of general cases. Section 4 concludes with a corresponding proof of invariance of timelike interval $s^2[ \, \varepsilon_{AP}, \varepsilon_{AK} \, ]$.
The derivations and proofs are coordinate-free, and notation has been chosen accordingly. The present derivation steps may not yet be optimal ...
1: Well -- Yes, provided that inertial systems can be considered given; i.e. by W. Rindler's characterization: "An inertial frame is simply an infinite set of point particles sitting still in space relative to each other."; such that a (non-zero) distance can be attributed to any pair of such point particles who are members of the same inertial system.
However: a (comprehensible, satisfying) demonstration of exactly which point particles are "sitting still relative to each other", if at all, is likely not available by considering only 1+1 dimensions, i.e. only one spatial dimension along with one temporal dimension.
1. Comparison of distances between inertial systems
$$\matrix{ :::::::::::::::::: \cr \, \cr ::::: K ::::::: } \! \! \matrix{ ::::::: A === \cr ~~~~ \varepsilon {\small \longrightarrow} \cr ::::::: P :::::::: } \! \! \matrix{ ===== M ====== \cr \, \cr :::::::::::::::::::::::::::::::::::: } \! \! \matrix{ == B ::: \cr ~~~\varepsilon \cr :::::: Q ::: } \! \! \matrix{ ::::::: F :::::: \cr \, \cr :::: \hphantom{ :: F :::::: } }$$
$$ \left( \frac{AB}{AF} \right) + \left( \frac{BF}{AF} \right) = 1, $$
$$ \left( \frac{KP}{KQ} \right) + \left( \frac{PQ}{KQ} \right) = 1 \tag{1a} $$
$$\matrix{ ::::::: A :::::::: \cr \varepsilon ~~ \cr :::::: K === } \! \! \matrix{ ::::::::::::::: \cr \, \cr == P == } \! \! \matrix{ :::::::::::::::::::::::::::::: B :: \cr \, \cr === N ======== } \! \! \matrix{ :::::::: F ::::: \cr {\small \longrightarrow} \varepsilon ~ \cr === Q :::: } \! \! \matrix{ ::: \cr \, \, \cr ::: }$$
$$\beta := \beta_{ABF}[ \, Q \, ] = \left( \frac{BF}{AF} \right), $$
$$ \beta := \beta_{KPQ}[ \, A \, ] = \left( \frac{KP}{PQ} \right) = \left( \frac{KP}{KQ} \right) / \left( \frac{PQ}{KQ} \right). \tag{1b} $$
$$ \left( \frac{AB}{AF} \right) = (1 – \beta), $$
$$ \left( \frac{PQ}{KQ} \right) = 1 – \left( \frac{KP}{KQ} \right) = 1 – \left( \frac{KP}{PQ} \right) \, \left( \frac{PQ}{KQ} \right) = 1 – \beta \, \left( \frac{PQ}{KQ} \right) = \frac{1}{(1 + \beta)}. \tag{1c} $$
The decisive requirement of reciprocity of corresponding distance ratios, ensuring mutual comprehensibility and agreeability of the comparison:
$$ \left( \frac{AB}{PQ} \right) = \left( \frac{KQ}{AF} \right). \tag{1d} $$
$$ \left( \frac{AB}{PQ} \right) = \sqrt{ \left( \frac{AB}{PQ} \right) \, \left( \frac{KQ}{AF} \right) } = \sqrt{ \left( \frac{AB}{AF} \right) \, \left( \frac{KQ}{PQ} \right) } = \sqrt{ (1 – \beta) \, (1 + \beta) } = \sqrt{ 1 - \beta^2 }. %\tag{1e} $$
2. Comparison of durations between inertial systems
$$\matrix{ \hphantom{ ::::: J :::::::::: } \cr \, \cr ::::: J :::::::::: } \! \! \matrix{ ::::: A ::::::: \cr \hphantom{ \rightarrow } \varepsilon {\small \longrightarrow} \cr ::::: P ::::::: } \! \! \matrix{ ::::: B ::::::: \cr \, \cr :::::::::::::::: } \! \! \matrix{ :::::::::::::::::::: G ::::: \cr \, \cr ::::::::: U ::::::\hphantom{ ::::::::: } }$$
$$ \left( \frac{AB}{AG} \right) + \left( \frac{BG}{AG} \right) = 1, $$
$$\left( \frac{JP}{JU} \right) + \left( \frac{PU}{JU} \right) = 1. \tag{2a}$$
$$\matrix{ \hphantom{ :::::::: } ::::: A :::::::::::: B :::::::::::::::::::: \cr \, \cr ::::: J ::::::::::::::: P ::::::::::::::::::::::::::: } \! \! \matrix{ ::::::: G :::::::: \cr \leftarrow \! \! \! \! \leftarrow \! \! \! \rightarrow \varepsilon \hphantom{ \longrightarrow ~ } \cr ::::::: U ::::::: }$$
Subsequently:
$$\matrix{ \hphantom{:::} ::::: A ::::: \cr \, \cr ::::: J ::::::::: } \! \! \matrix{ ::::::: B :::::::: \cr \leftarrow \! \! \! \! \leftarrow \! \! \! \! \leftarrow \varepsilon \leftarrow \! \! \! \! \leftarrow ~\cr ::::::: P :::::::: } \! \! \matrix{ :::::::::::::::::: G :::::: \hphantom{ ::::::::: } \cr \, \cr ::::::::::::::::::::::::::: U ::::::: }$$
And finally:
$$\matrix{ ::::: A ::::::::: \cr \phantom{ ~~~~ } \varepsilon \leftarrow \! \! \! \! \leftarrow \! \! \! \! \leftarrow \cr ::::: J ::::::::: } \! \! \matrix{ ::: B :::::::::::: \cr \, \cr :::::::: P ::::::: } \! \! \matrix{ :::::::::::::: G :::::::::: \hphantom{ ::::::::: } \cr \, \cr ::::::::::::::::::::::::::: U ::::::: }$$
$$\beta := \beta_{ABG}[ \, P \, ] := \frac{1}{ \left( \frac{AG}{AB} \right) + \left( \frac{BG}{AB} \right) } = \frac{\left( \frac{AB}{AG} \right)}{1 + \left( \frac{AB}{AG} \right) \left( \frac{BG}{AB} \right) } = \frac{1 – \left( \frac{BG}{AG} \right)}{1 + \left( \frac{BG}{AG} \right)},$$
$$\beta := \beta_{JPU}[ \, A \, ] := \frac{1}{ \left( \frac{PU}{JP} \right) + \left( \frac{JU}{JP} \right) } = \frac{\left( \frac{JP}{JU} \right)}{\left( \frac{JP}{JU} \right) \left( \frac{PU}{JP} \right) + 1 } = \frac{1 – \left( \frac{PU}{JU} \right)}{1 + \left( \frac{PU}{JU} \right)} = \frac{\left( \frac{JU}{PU} \right) – 1}{\left( \frac{JU}{PU} \right) + 1}.$$
The decisive requirement of reciprocity of corresponding duration ratios, ensuring mutual comprehensibility and agreeability of the comparison:
$$\large \left( \frac{\tau^P_{[ \, \CoincidentWith A, \, \CoincidentWith B \, ]}}{\tau^A_{[ \, \CoincidentWith P, \, \SimultaneousTo B \CoincidentWith P \, ]}} \right) = \left( \frac{\tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith J \, ]}}{\tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo J \CoincidentWith A \, ]}} \right). \tag{2b} $$
$$\left( \frac{\tau^P_{[ \, \CoincidentWith A, \, \CoincidentWith B \, ]}}{\tau^A_{[ \, \CoincidentWith P, \, \SimultaneousTo B \CoincidentWith P \, ]}} \right) := \frac{2 \, PU / c}{(AG + BG) / c} := \frac{2}{ \left( \frac{AG}{PU} \right) + \left( \frac{BG}{PU} \right) },$$
$$\left( \frac{\tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith J \, ]}}{\tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo J \CoincidentWith A \, ]}} \right) := \frac{2 \, AG / c}{(PU + JU) / c} := \frac{2}{ \left( \frac{PU}{AG} \right) + \left( \frac{JU}{AG} \right) }.$$
$$\eqalign{ \frac{2}{ \left( \frac{AG}{PU} \right) + \left( \frac{BG}{PU} \right) } &= \frac{2}{ \left( \frac{PU}{AG} \right) + \left( \frac{JU}{AG} \right) } = \left( \frac{AG}{PU} \right) \, \left( \frac{2}{ 1 + \left( \frac{JU}{PU} \right) } \right) \\ &= \left( \frac{AG}{PU} \right) \, \left( \frac{ 1 + \left( \frac{BG}{AG} \right) }{2} \right) \, \left( \frac{2}{ 1 + \left( \frac{BG}{AG} \right) } \right) \, \left( \frac{2}{ 1 + \left( \frac{JU}{PU} \right) } \right) \\ &= \left( \frac{ \left( \frac{AG}{PU} \right) + \left( \frac{BG}{PU} \right) }{2} \right) \, \left( 1 + \beta \right) \, \left( 1 – \beta \right) \\ &= \left( \frac{ \left( \frac{AG}{PU} \right) + \left( \frac{BG}{PU} \right) }{2} \right) \, \left( 1 – \beta^2 \right) \\ &= \sqrt{ 1 – \beta^2 }. }$$
$$\large \left( \frac{\tau^P_{[ \, \CoincidentWith A, \, \CoincidentWith B \, ]}}{\tau^A_{[ \, \CoincidentWith P, \, \SimultaneousTo B \CoincidentWith P \, ]}} \right) = \left( \frac{\tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith J \, ]}}{\tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo J \CoincidentWith A \, ]}} \right) = \sqrt{ 1 – \beta^2 }. \tag{2c}$$
3. Invariance of spacelike intervals
$$ s^2[ \, \varepsilon_{AP}, \varepsilon_{BQ} \, ] := AB^2 \, \overset{?}{=} \, PQ^2 - \left( c \, \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo Q \CoincidentWith B \, ]} \right)^2. \tag{3a} $$
$$ \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo Q \CoincidentWith B \, ]} = \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo Q \CoincidentWith F \, ]} - \tau^Q_{[ \, \CoincidentWith B, \, \CoincidentWith F \, ]} = \left( \frac{1}{2} \right) \, \tau^P_{[ \, \CoincidentWith A, \, \ReceivedAndReflected Q \ReceivedAndReflected P \CoincidentWith A \, ]} - \sqrt{ 1 - \beta^2 } \, \tau^B_{[ \, \CoincidentWith Q, \, \SimultaneousTo F \CoincidentWith Q \, ]},$$
$$ c \, \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo Q \CoincidentWith B \, ]} = \left( \frac{c}{2} \right) \, \tau^P_{[ \, \CoincidentWith A, \, \ReceivedAndReflected Q \ReceivedAndReflected P \CoincidentWith A \, ]} - c \, \sqrt{ 1 - \beta^2 } \, \tau^B_{[ \, \CoincidentWith Q, \, \SimultaneousTo F \CoincidentWith Q \, ]} = PQ - \sqrt{ 1 - \beta^2 } \, \frac{BF}{\beta}. %\tag{3b} $$
$$ AB^2 \, \overset{?}{=} \, PQ^2 - \left( PQ - \sqrt{ 1 - \beta^2 } \, \frac{BF}{\beta} \right)^2 = \frac{BF^2}{\beta^2} - BF^2 - 2 \, PQ \, \sqrt{ 1 - \beta^2 } \, \frac{BF}{\beta}. \tag{3c} $$
$$PQ = \frac{AB}{\sqrt{1 - \beta^2}},$$
$$ AB^2 \, \overset{?}{=} \, \frac{BF^2}{\beta^2} - BF^2 - \frac{2 \, AB \, BF}{\beta}. \tag{3d}$$
$$\frac{1}{\beta} \overset{?}{=} \left( \frac{AB}{BF} \right) + 1 = \left( \frac{AB + BF}{BF} \right) = \left( \frac{AF}{BF} \right) \tag{3e} $$
which holds according to eq. (1a). Therefore eq. (3a) holds as well: the invariance of spacelike interval $s^2[ \, \varepsilon_{AP}, \varepsilon_{BQ} \, ]$, and consequently invariance of spacelike intervals in general.
4. Invariance of timelike intervals
$$ s^2[ \, \varepsilon_{AP}, \varepsilon_{AK} \, ] := -\left( c \, \tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith K \, ]} \right)^2 \, \overset{?}{=} \, KP^2 - \left( c \, \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo K \CoincidentWith A \, ]} \right)^2. \tag{4a} $$
$$ \tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith K \, ]} = \sqrt{ 1 - \beta^2 } \, \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo K \CoincidentWith A \, ]}, $$
$$ \beta = \frac{KP}{c \, \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo K \CoincidentWith A \, ]}} := \beta_{KPQ}[ \, A \, ] = \frac{KP}{PQ} = \frac{KP}{c \, \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo Q \CoincidentWith F \, ]}}. $$
Accordingly substituting $\tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith K \, ]}$ and $KP$, eq. (4a) is seen to be manifestly satisfied:
$$ -\left( c \, \tau^A_{[ \, \CoincidentWith P, \, \CoincidentWith K \, ]} \right)^2 = (\beta^2 - 1) \, \left( c \, \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo K \CoincidentWith A \, ]} \right)^2 = KP^2 - \left( c \, \tau^P_{[ \, \CoincidentWith A, \, \SimultaneousTo K \CoincidentWith A \, ]} \right)^2. $$
Therefore the timelike interval $s^2[ \, \varepsilon_{AP}, \varepsilon_{AK} \, ]$ is invariant; and so are timelike intervals in general.