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I'm currently simplifying the current equation $$a.b+a.b.\bar{c}+a.\bar{b}+\bar{a}.b.\bar{c}+\bar{a}.\bar{b}.\bar{c}.$$ These are the steps I've taken so far: $$\bar{c}(a.b+\bar{a}.b+\bar{a}.\bar{b})+a(b+\bar{b})$$ and $b+\bar{b}$ simplifies to one. Factoring out $\bar{a}$ we have $$\bar{c}(\bar{a}(b+\bar{b})+a.b)+a$$ and simplifying with same identity we obtain $$\bar{c}(\bar{a}+a.b)+a.$$ I don't know exactly how I simplified to what I have below, but I know its correct. If you know that would be a great bonus! $$\bar{c}(\bar{a}+b)+a$$ This is where I'm confused. Intuitively, I feel $\bar{a}+b$ should somehow simplify to 1, since according to every tool I use, this should just simplify to $\bar{c}+a$. But I just don't see how this is possible. Any idea how to continue from here?

Thank you.

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Ryan_DS
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1 Answers1

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We have \begin{align}ab+ab\bar c+a\bar b+\bar ab\overline c+\bar a\bar b\bar c&=a(b+\bar b)+ab\bar c+\bar ab\overline c+\bar a\bar b\bar c\tag1\\&=a+a(b\bar c)+\bar a\bar c(b+\bar b)\tag2\\&=a+\bar a\bar c\tag3\\&=(a+a\bar c)+\bar a\bar c\tag4\\&=a+(a+\bar a)\bar c\tag5\\&=a+\bar c\tag6\\&=\bar c+a\tag7\end{align}

where

$(1)$: Distributive Law $ab+a\bar b=a(b+\bar b)$

$(2)$: Complement Law $b+\bar b=1$, Commutative Law $b\bar c=\bar cb$ and Distributive Law

$(3)$: Redundancy Law $a+a(b\bar c)=a$, Complement Law

$(4)$: Redundancy Law

$(5)$: Associative Law $(a+a\bar c)+\bar a\bar c=a+(a\bar c+\bar a\bar c)$, Distributive Law

$(6)$: Complement Law

$(7)$: Commutative Law

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