Simplify $$ab+\bar ac+bc$$ where $a,b,c$ are Boolean numbers, that is, $0$ or $1$
$$ab+\bar ac+bc$$ $$=ab+\bar ac+bc\cdot1$$ $$=ab+\bar ac+bc(a+\bar a)$$ $$=ab+bca+\bar ac+bc\bar a$$ $$=ab(1+c)+\bar ac(1+b)$$ $$=ab+\bar ac$$
Can I simplify this further using some tricks like I did before$?$ And what is the insight behind these tricks$?$ Can we evaluate it without using these tricks$?$ I already know that this can be done using K maps but what about algebraically.
Any help is greatly appreciated. In my opinion, this question shouldn't be closed under need more focus as the three questions asked are interrelated, though my main question is existing in the title itself.
You already know K maps ... so from that it should be clear that the resulting expression cannot be further simplified.
As to how you would know that algebraically ... that just takes some more practice and experience, just as with algebraically simplifying expressions in the domain of real numbers.
But yeah, you'll have to know your algebra principles, including some advanced ones. In this particular case, it helps to know the Consensus Theorem. But again, with plenty of experience, you just know that no further simplification principles can be applied.