Simplifying the Boolean binary equation

Question

A'B + A'B'C + ABC'C' + AB' + AB'C'

I need to simplify this boolean equation in simplest form but im not sure of my answer.

Answer 1

Hint: One possibility is: $A'C + BC' + AB'$. Try drawing a Karnaugh map. Otherwise, here's a quick sketch of a proof:

• $ABC'C'=ABC'$.
• $AB'C'$ can be absorbed into $AB'$.
• $A'B$ can be decomposed into $A'BC$ and $A'BC'$.
• $A'B'C$ and $A'BC$ can be combined to form $A'C$.
• $ABC'$ and $A'BC'$ can be combined to form $BC'$

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EDIT: Here are the details.

$$ \begin{align*} &A'B + A'B'C + ABC'C' + AB' + AB'C'\\ &= A'B + A'B'C + ABC' + AB' + AB'C' & \text{Idempotent Law} \\ &= A'B(1) + A'B'C + ABC' + AB' + AB'C' & \text{Identity Law} \\ &= A'B(C + C') + A'B'C + ABC' + AB' + AB'C' & \text{Inverse Law} \\ &= A'BC + A'BC' + A'B'C + ABC' + AB' + AB'C' & \text{Distributive Law} \\ &= (A'B'C + A'BC) + (ABC' + A'BC') + (AB' + AB'C') & \text{Commutative/Associative Law} \\ &= A'C(B' + B) + BC'(A + A') + AB'(1 + C') & \text{Distributive Law} \\ &= A'C(1) + BC'(1) + AB'(1 + C') & \text{Inverse Law} \\ &= A'C(1) + BC'(1) + AB'(1) & \text{Domination Law} \\ &= A'C + BC' + AB' & \text{Identity Law} \\ \end{align*} $$