Can covering be done on two elements?

Question

The covering rule is: $$B \bullet (B+C) = B$$ and $$B+(B \bullet C)=B$$ So does it follow from this rule that: $$B \bullet A \bullet \bar{C} + B \bullet D \bullet\bar{F} = B \bullet (A\bullet\bar{C}+D\bullet\bar{F}) = B? $$ Or does the covering rule apply only to terms with one variable?

Answer 1

No, the equation at the end of your post is not valid. For example, if $A = D = \bot$ and $B\neq \bot$ ($\bot$, also denoted $0$ or $\emptyset$, is the bottom element of the boolean algebra), then $B\cdot A \cdot \bar{C} + B \cdot D \cdot \bar{F} = \bot + \bot = \bot$.

The first equality, $B\cdot A \cdot \bar{C} + B \cdot D \cdot \bar{F} = B\cdot (A\cdot \bar{C} + D \cdot \bar{F})$ is valid by distributivity. But there's no $B$ inside the parentheses, so there's no way to apply the covering rule at this point.

As for your question about terms with one variable, here's an example where you can apply the covering rule: $(x+y)\cdot (x+y+z\cdot w+z\cdot x) = x+y$. We can take $B = x+y$ and $C = z\cdot w+z\cdot x$ in the statement of the covering rule.