I need to simplify the following expression $$(a+b+\bar c)(a+c+d)(\bar a +b+c) $$
This is what I've done: $$=(a+(b+\bar c)(c+d))(\bar a+b+c)$$ $$=(a+bc+bd+\bar cc+\bar cd)(\bar a+b+c)$$ $$=(a+bc+bd+\bar cd)(\bar a+b+c)$$ $$=a \bar a +ab+ac+\bar abc+bc+bc+ \bar abd+bd+bdc+\bar a \bar cd+b \bar cd+c \bar cd$$ $$=ab+ac+bc+bd+\bar a \bar cd$$
My textbook has given the final answer as $$ab+ac+bc+\bar a \bar cd$$ but I haven't been able to figure out which theorem was used to simplify bd. How was bd removed from their final answer?
If $b$ is true and $d$ is true, then one of
$(ab), (bc),$ or $(\overline{a}~\overline{c}~d)$ must be true.
Therefore, the $(bd)$ term is redundant, and thus eliminated.