# 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 &= (a'bc'+a'bc) + (ab'c'+ab'c) &\text{associativity of } +\\ &= a'b(c'+c) + ab'(c'+c) &\text{distributivity, twice}\\ &= a'b(1) + ab'(1) &\text{inverse for }+\\ &= a'b + ab' &\text{identity for }\cdot \end{align}