The Boolean function $g$ of the variables $x_1 , . . . , x_5 , y_1 , . . . , y_5$ , is given by the formula: $$ \bigwedge_{i=1}^{5} (x_i \ \vee \ y_i) $$ Mission: find the dnf($g$). (the minimum number of conjuncts in the DNF of function $g$)
Some clarifications:
The smallest number of conjuncts in the DNF representing the function $g$ is denoted by dnf($g$).
My thinking:
At first, I just expanded this distributivity formula from the task condition and got a disjunction of 32 conjuncts, each conjunct contains a conjunction of 5 boolean variables (some 5 of $x_1, ..., x_5, y_1, ... y_5$). In fact, this is the answer to the problem, but I can't prove that there won't be less than 32 conjuncts. I tried to prove the opposite: let's assume that there are fewer than 32 conjuncts, etc., but no contradiction was be deduced by me...
Then I tried to consider subtasks when we are given not 10 boolean variables, but for example 4 boolean variables: $(x_1, x_2, y_1, y_2)$. And then you can see that $3^2=9$ combinations of input data $(x_1, x_2, y_1, y_2)$ will give the value 1 for the function $g$, the total values of the function $g$ from 4 variables takes $2^4=16$. Then we can essentially consider a Boolean cube with 16 vertices (vertex = $(x_1,x_2,y_1,y_2)$), and dnf($g$) will be equal to the minimum number of faces from the Boolean cube that cover all vertices $(x_1,x_2,y_1,y_2)$ from whose coordinates the function $g$ takes the value 1, and in fact it will be $2^2 = 4$ faces that cover all 9 vertices on which the function takes the value 1. (all this is very easy to get by drawing a 4-dimensional Boolean cube on paper). But I don't quite understand how to improve this idea for 10 variables already, by induction doing this is probably a very dubious idea. Do you have any suggestions on how to solve this problem with this approach?
