I'm trying to gather my thoughts for proving the following claim:
For $ fRg \Leftrightarrow fg = f$ on $B^{n}$ to $B$, show that $ fRg $ if and only if for any input values $ b_1,...,b_n $, we have that $ f(b_1,...,b_n) \leq g(b_1,...,b_n) $.
I am new to Boolean algebra, so my intuition is shaky at best. Here's a sketch of what I'm thinking of saying so far. I know it is incomplete, but I would like suggestions on how to proceed / anything to revise / how to make things more rigorous / if I'm even on the right track.
Since this is a Boolean relation, all values must be either 0 or 1. And we know that all $ g(b_1,...,b_n) = 1 $, since $ f * 1 = f $ but it is not the case that $ f * 0 = f $ .
The problem is to show that $fg = f$ on $B^n$ iff $f \leq g$ on $B^n$. $\Rightarrow$, if $f \gt g$ fs. $x \in B^n$, then $(fg)(x) = f(x) \gt g(x)$, so $(f\cdot f)(x) \gt f(x)$. But trying the two Boolean values for $f(x)$ gives $1 \gt 1$ or $0 \gt 0$. $\Leftarrow$, $f\cdot f = f$ for Boolean functions so we have given $f \leq g$ that $f \leq f\cdot g$. Similarly and'ing by $g$ gives the other inequality. QED