The point is that if w_{in} = 4, f = 3 and s = 2 with no padding, then the first stride is off the end, right? So you can only do one application of the filter and you get w_{out} = 1. You can only apply the filter in positions where the complete filter fits within the input space. Draw the picture and it will make sense.

With everything the same but s = 1 in that case, then w_{out} would be 2.

Or if you have everything the same but w_{in} = 5, then w_{out} would also be 2. But that’s not what you have.