Not a physicist: but want to use the temporal ordering of events in special relativity as an example for something, and need to answer the following question to do so.
Suppose we have three events, a, b and c such that
(i) a occurs prior to b in all inertial reference frames.
(ii) b occurs prior to c in some (but not all) inertial reference frames.
Do (i) and (ii) jointly entail (iii) below?
(iii) a occurs prior to c in all inertial reference frames.
No, (i) and (ii) together do not imply (iii).
For an explicit counter-example, consider an (unprimed) inertial frame using $(t,x)$ coordinates where $c=1$. Consider in that frame the following events $$a=(0,0)$$$$b=(1,0)$$$$c=(1,2)$$Now, transform to an arbitrary (primed) inertial frame moving at $v$ relative to the unprimed frame. Then we have $$a'=(0,0)$$$$b'=\left( \frac{1}{\sqrt{1-v^2}}, \frac{v}{\sqrt{1-v^2}}\right)$$$$c'=\left( \frac{1+2v}{\sqrt{1-v^2}},\frac{2+v}{\sqrt{1-v^2}} \right)$$
Now, $$t'_a=0<\frac{1}{\sqrt{1-v^2}}=t'_b$$ for all $-1<v<1$. So (i) holds.
And, $$t'_b = \frac{1}{\sqrt{1-v^2}} < \frac{1+2v}{\sqrt{1-v^2}} = t'_c$$ for $0<v<1$ but not for $-1<v<0$. So (ii) holds.
But, $$t'_a = 0 < \frac{1+2v}{\sqrt{1-v^2}} = t'_c$$ does not hold for $-1<v<-\frac{1}{2}$. So (iii) does not hold.
A simpler way to reason about this is to realize that (i) implies that $a$ and $b$ are timelike separated, while (ii) implies that $b$ and $c$ are spacelike separated. But spacetime separation is not transitive so we cannot infer (iii) that $a$ and $c$ are timelike separated.
It can be useful in such questions to use a light cone diagram. If you place event (a) in the origin of your light cone, you are basically saying that event (b) is also within (a)'s light cone. Now, since the ordering of events (b) and (c) can be reversed by moving between inertial frames, it means that (c) is outside of a light cone that has (b) as its origin, but with regards to (a)'s light cone we now have two cases, see red and blue options in the diagram:
So, it turns out that given the information you have provided statement (iii) cannot always be deduced, because as you can see, (c) can be either inside or outside of (a)'s light cone, despite being outside of (b)'s light cone in both cases, which is given by assumption.
If the red option takes place, then statement (iii) happens to be correct, but for the blue option conclusion (iii) is false.
In summary, it is not always true that given (i) and (ii) conclusion (iii) will hold, and the counterexample to show that is the blue option for event (c).
A bit more background about the meaning of such diagrams: we say that two events are timelike separated if they can be connected by a light ray. You can imagine two such events, by thinking of an imaginary ray that has at most speed $c$ (light speed) emitted at one event being able to reach the second event. This emitting of light rays is represented by the light cone: the cone's boundary represents the paths that actual light rays travel, hence any event within such a light cone is causally connected to the event at its origin, and indeed will remain so for all inertial frames (otherwise some interesting paradoxes can be concocted).
Note that not all points within a light cone are causally related, this can be slightly confusing, but you need to consider each pair of events separately and see whether or not a light cone originating in one of them contains the other event.
Having understood that, it's easy also to understand spacelike separated events, and those are all events that can't be connected by anything with speed equal or less than the speed of light. You won't be able to find a light cone originating in one and enveloping the other. You can think of such pairs of events as not causally connected, and hence inertial observers may disagree on which happened first. Since by definition such events are not causally related, the disagreement on which happened first does not lead to any paradox.
There is also the "inbetween" case of two events being exactly related by a light ray, we call them lightlike separated. Only light himself presumably is able to be in two such 'meetings' on time :)
