$$f(x_1, x_2, x_3) := \sum m(2, 3, 4, 5, 6, 7)$$
With the normal SOP expression for this function, it must be, with the use of minterm:
$$f = m_2 + m_3 + m_4 + m _6 + m_7 = x_1'x_2x_3'+x_1'x_2x_3+x_1x_2'x_3'+x_1x_2x_3'+x_1x_2x_3$$
How can this function be simplified?
(I copied this exercise from a book. I think there should be a mistake because something is missing. Specifically, i looked author's solution for his exercise and i see that he added a term that you cannnot understand where it comes from. So, probably it's a mistake.)
This is the author's solution:
$$f = x_1'x_2(x_3'+x_3)+x_1'(x_2'+x_2)x_3'+x_1x_2(x_3'+x_3)\\ \qquad \quad = x_1'x_2+x_1x_3'+x_1x_2=(x_1'+x_1)x_2+x_1x_3'=x_2+x_1x_3'$$
So, is his solution correct or did he make a mistake?