I can't simplify this A’B’C + A’BC + A’BC’ + AB’C + ABC boolean expression to A'B+C

Question

I have to get this expression A’B’C + A’BC + A’BC’ + AB’C + ABC to A'B+C. I did this but I can't finish it, I don't know how to.

A’B’C + A’BC + A’BC’ + AB’C + ABC

A'B(C+C')+C(A'B'+AB'+AB)

A'B+C(A'B'+AB'+AB)

That's it, I don't know how to solve that.

Answer 1

There is this super-weird absorption laws in boolean algebra: $X+XY = X$ and $X(X+Y) = X$.

From your last line, as you "prematurely grouped" the $A'BC$ term into $A'B$, the absorption law can "recreate" an $A'BC$ out of $A'B$:

$$\begin{align*} p &= A'B+C(A'B'+AB'+AB)\\ &= (A'B + A'BC) + C(A'B ' + AB' + AB)\\ &= A'B + C(A'B + A'B' + AB' + AB)\\ &= A'B + C(A' + A)(B'+B)\\ &= A'B + C \end{align*}$$

Also try using Karnaugh map which helps a lot.

Answer 2

\begin{align} E&=A'B+C(A'B'+AB'+AB)\\&=A'B+C((A'+A)B'+AB)\\ &=A'B+C(B'+AB) \end{align}

For boolean expressions $+$ also distributes over product, by example: $$ B'+AB = (B'+A)(B'+B)=B'+A $$ thus

\begin{align} E &=A'B+C(B'+A)\\ &=A'B+CB'+CA \\ &=(A'B+C)(A'B+B')+CA \\ &=(A'B+C)(A'+B')+CA\\ &=A'B+CA'+CB'+CA \\ &=A'B+C+CB' \\ &=A'B+C \end{align}

Answer 3

Terms can be duplicated in boolean algebra, because or and and are idempotent: $X = X+X$ and $X = XX$.

$$\begin{align*} p &= A’B’C + \underline {A’BC} + A’BC’ + AB’C + ABC\\ &= (\underline {A’BC} + A'BC') + (A'B'C + \underline {A’BC} + AB'C + ABC)\\ &= A'B(C+C') + (A'+A)(B'+B)C\\ &= A'B + C \end{align*}$$

(Underlined terms were duplicated, no special meaning otherwise)

Also try using Karnaugh map which helps a lot.

Answer 4

$$A'B'C'+A'BC'+A'BC+ABC=$$

$$=A'C(B'+B)+A'BC+ABC' , \{B'+B=1\}$$

$$=A'C'+B(A'C+AC')$$

$$=A'C' +B$$