Boolean Algebra / Digital Logic

I am trying to figure out how to simply a canonical sum of products expression that is from this expression: $$f_1(x_1,x_2,x_3) = \sum m (2,3,4,6,7)$$ where m is canonical minterms

I got: $$\bar{x}_{1}x_2\bar{x}_3 + \bar{x}_{1}x_2x_3+x_1\bar{x}_{2}\bar{x}_{3}+x_1x_2\bar{x}_{3} + x_1x_2x_3$$

then I simplify and get: $$x_2+x_1x_2$$

which is incorrect. I the answer in the book is: $$x_2+x_1\bar{x}_{3}$$

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What is $m$? And I think the word you want to use is "simplify", not "simply". –  TMM Sep 22 '11 at 15:12
Sorry, I edited question for further clarity. –  Nick Sep 22 '11 at 15:15

Let $a = x_{1}$, $b = x_{2}$, and $c = x_{3}$.

Then to simplify $a'bc' + a'bc + ab'c' + abc' + abc$

$a'(bc' + bc) + a(b'c' + bc') + abc$

$a'(b(c' + c)) + a(c'(b' + b)) + abc$

$a'b + ac' + abc$

$a'b + a(c' + bc)$

$a'b + a(c' + b)$

$a'b + ac' + ab$

$b(a' + a) + ac'$

$b + ac'$

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genius! Thank you. I was doing fuzzy math, haha. –  Nick Sep 22 '11 at 16:46