Understanding why the Solution to Silver 2021 Feburary Problem 3 is correct

#1

http://usaco.org/current/data/sol_prob3_silver_feb21.html

In the above editorial, the following is stated under the second provided solution:

"Alternatively, note that all_onesi,j[k]=all_onesi,j−1[k]&ok[j][k]all_ones�,�[�]=all_ones�,�−1[�]&��[�][�]. So let’s fix i� and compute

all_onesi,i,all_onesi,i+1,…,all_onesi,N−1all_ones�,�,all_ones�,�+1,…,all_ones�,�−1

in order."
I’ve drawn out TCs to see that this is true, but I’m having trouble intuitively understanding why. Would someone mind outlining a “proof”?
Thanks.

#2

To be more specific, my confusion lies in understanding why and how we can transition from the 1D case to cover the 2D case in the way the problem is doing that