How do i simplify this SOP expression?
You can verify that the two expressions are equivalent by using some logical equivalences. Below I'm using 1 to represent true.
$$\begin{align}
a'bc'+a'bc + ab'c'+ab'c &amp;= (a'bc'+a'bc) + (ab'c'+ab'c) &amp;\text{associativity of } +\\
 &amp;= a'b(c'+c) + ab'(c'+c) &amp;\text{distributivity, twice}\\
 &amp;= a'b(1) + ab'(1) &amp;\text{inverse for }+\\
 &amp;= a'b + ab' &amp;\text{identity for }\cdot
\end{align}$$