Special relativity postulate says that speed of light is constant for all inertial observers, meaning that measures of time and distance must accommodate in a way to show that speed of light is the same for all.
I know about length contraction and time dilation. Under Lorentz transformations, the length measured by an observer in rest frame is shorter than length measured by an observer who's moving along with the thing he measures (for example, a rod of length $L$), moving relative to the resting observer. And the time between two physical events happening in a moving frame is greater for a resting observer than it is for the other one.
What bothers me are equations for these phenomena:
$$\Delta t = \gamma\Delta t'$$ $\Delta t'$ here means proper time, in other words time measured between two events happening in a moving frame of reference, measured by an observer in that frame. So that means, from this equation that $\Delta t'$ is smaller than $\Delta t$. For a guy on the ground, time between two events is longer.
$$\Delta L = \Delta L' / \gamma$$ In the same way, $\Delta L'$ is the length of, let's say a rod, measured by an observer for which the rod is at rest. And let's say they are moving with a velocity relative to the observer in a resting frame. The latter will measure, according to the equation, a shorter length.
Now I am finally getting to the problem that bothers me: If speed of light is the same for both observers, and speed is distance divided by time, meaning that
c = distance/time,
it means that for a moving observer $c = \Delta L'/ \Delta t'$ and for an observer at rest $c = \Delta L / \Delta t$
So if $\Delta L' > \Delta L$, shouldn't it be that also $\Delta t'> \Delta t$ since $\Delta L'/ \Delta t'$ must be equal to $\Delta L/ \Delta t$? From these two equations it follows that two observers measure different magnitudes for speed of light, and they should measure the same.
What am I doing wrong here?