Boolean simplification gives two different results

Question

Here is the question I'm trying to solve:

Use algebraic manipulation to find the minimum sum-of-products expression for the function $f=x_1x_3+x_1\overline{x}_2+\overline{x}_1x_2x_3+\overline{x}_1\overline{x}_2\overline{x}_3$

The way I attempted it was as follows:

f = ac + ab' + a'bc + a'b'c'
  = ac + b'(a + a'c') + a'bc   [distributive]
  = c(a + a'b) + b'(a + a'c')  [distributive]
  = c(a + b) + b'(a + c')      [using x + x'y = x + y]
  = ac + bc + ab' + b'c'

where a = x1, b = x2, c = x3

The correct answer, however is:

$f=x_1x_3+x_2x_3+\overline{x}_2\overline{x}_3$

Can anyone tell me where I went wrong?

Answer 1

$\begin{align} &ac + bc + ab' + b'c'\\=~&bc +~ ac + ab' + b'c'\\ =~& bc+~(abc+ab'c)+(ab'c+ab'c')+(ab'c'+a'b'c')\\\vdots~~& \\ =~& bc+~ac+b'c'\\=~& ac+bc+b'c' \end{align} $

Complete the missing steps, and you are done.

Answer 2

Karnough maps are more standard and consistent than algebra, but algebra works too. You are almost there. Observe that bc + b'c' is true iff b$\equiv$c. If this is the case, then ac + ab'$\iff$a(c+b')$\iff$a. On the other hand, if bc + b'c' is false, then b$\equiv$c' and ac + ab'$\iff$a(c+b')$\iff$ac. So either the bc + b'c' term makes the expression true, or ac + ab' is equivalent to just ac.

Answer 3

You haven't simplified it completely: $$ x_1\bar{x_2}=x_1\bar{x_2}(x_3+\bar{x_3})=x_1\bar{x_2}x_3+x_1\bar{x_2}\bar{x_3}. $$ The first term contains $x_1x_3$ and the second term contains $\bar{x_2}\bar{x_3}$. So $x_1\bar{x_2}$ is redundant in your $f$.