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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?

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  • $\begingroup$ Next time, put the question and answer here directly, or at least a summary, rather than using links to external images $\endgroup$ – HackerBoss Oct 10 '18 at 23:49
  • $\begingroup$ You did't quite go wrong. You just missed one simplification. The $x_1\bar{x}_2$ term is redundant. $\endgroup$ – Fabio Somenzi Oct 10 '18 at 23:57
  • $\begingroup$ You're right, but how is x1x2' redundant? I don't quite see it... $\endgroup$ – StaticCrazee Oct 11 '18 at 0:07
  • $\begingroup$ One way to see it is in user10354138's answer. Another way is to note that it is the consensus term of the first and third terms. Consensus terms are redundant. $\endgroup$ – Fabio Somenzi Oct 11 '18 at 5:16
  • $\begingroup$ @FabioSomenzi Yes, I've clearly missed that! A little reordering and consensus was the key. Thanks! $\endgroup$ – StaticCrazee Oct 11 '18 at 16:23
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$\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.

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  • $\begingroup$ Thanks for answering. The missing steps is where I'm stuck. For instance, where'd the ab' disappear to? $\endgroup$ – StaticCrazee Oct 11 '18 at 2:46
  • $\begingroup$ Look carefully at the three statements in brackets. Do you notice any thing? [Hint: idempotence] @StaticCrazee $\endgroup$ – Graham Kemp Oct 11 '18 at 3:50
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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 $f$.

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  • $\begingroup$ Not the most complete answer $\endgroup$ – HackerBoss Oct 11 '18 at 0:02
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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.

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  • $\begingroup$ Thanks for answering. But that doesn't match the answer? $\endgroup$ – StaticCrazee Oct 11 '18 at 0:06

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