How do i simplify this SOP expression? Hi i have derived the following SoP (Sum of Products) expression , by analyzing the truth table of a 3 bit , binary to gray code converter. I ask for verification, because i feel as though this answer may not be correct or complete.


  X = a'bc' + a'bc + ab'c' + ab'c


which, using k-maps, was simplified to 


  X = ab' + a'b


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