Relativity of simultaneity subtle example imagine three persons, Person A is standing next to a device these device is an AND gate if two photons(one from right and one from left) reaches the AND gate at the same time the device explodes and Person A dies otherwise if there is a difference in time between the two photons reaching the gate it does not explode , person B and person C are moving to relative to each other and there is photons source , imagine i worked out the numbers to fit that the two photons reach at the same time for one observer, the other will see that the device did not explode and person A does not die what really happened is ?
Note : by seeing here i do n't mean seeing by eye and i ignore the light it takes to reach the human eyes
 A: For simplicity of problem, let us assume that the two photons are emitted by sources at equal distances on either side of a line joining the two sources and the AND gate in A's frame, which is our x axis from now on. Let one source have x coordinate as -L (source 1) and the other have x coordinate as +L (source 2), with the AND gate at 0. Now lets take the observer B who is moving along the x axis with some velocity v. Now in A's frame, the person explodes as the photons arrive simultaneously. Now suppose the photons were released at time t = 0 from the sources and A exploded at t = L/c in A's frame.
Taking a Lorentz transformation of velocities from A's frame to B's frame gives the position and time of 1st source as
$$ x_1' = \gamma (-L )$$
$$ t_1' = \gamma (+Lv/c^2)$$
Similarly, taking a Lorentz transformation of velocities from A's frame to B's frame gives the position and time of 2nd source as
$$ x_2' = \gamma ( L )$$
$$ t_2' = \gamma (-Lv/c^2)$$
As you can see, the photons were not released simultaneously from 1 and 2.
Now the position of the AND gate in B's frame is given by
$$ x_0' = \gamma ( -Lv/c^2 )$$
Our problem now is that we need to find at what time the light from both sources reaches the AND gate in B's frame.
Using the speed of light, we get
$$ c = \dfrac{\Delta x'_{1 \to 0}} {\Delta t'}$$
where 
$$ \Delta x' = \gamma (-Lv/c + L)$$
and 
$$ \Delta t' = t'_0 - \gamma (Lv/c^2)$$
and so $t'_0$ is found out to be $\gamma (L/c)$.
Similarly we can get by using
$$ c = \dfrac{\Delta x'_{2 \to 0}} {\Delta t'}$$
and solve for $t'_0$ to get $\gamma (L/c)$.
So A is dead in the reference frame of of B too as the photons hit the detector simultaneously. Notice that even though the the photons were not released simultaneously, they hit the detector simultaneously. Thats a problem you should try to find out why.
Now instead of telling this whole long story there is an apparent shortcut too. By Einstein's 1st law of relativity, physical laws remain same in every frame of reference. So if an event, by some conclusion is predicted to happen in one frame, then it has to happen in all other frames too. 
