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An exercise says : Use algebraic manipulation to find the mimimum SOP expression for the function $$f = x_1x_3 + x_1x_2' + x_1'x_2x_3 + x_1'x_2'x_3'$$

The given solution says:

$f = x_1x_3 + x_1x_2' + x_1'x_2x_3 + x_1'x_2'x_3'\\ = x_1(x_2' + x_2)x_3 + x_1x_2'(x_3' + x_3) + x_1'x_2x_3 + x_1'x_2'x_3'\\ = x_1x_2'x_3 + x_1x_2x_3 + x_1x_2'x_3' + x_1'x_2x_3 + x_1'x_2'x_3'\\ = x_1x_3 + (x_1 + x_1')x_2x_3 + (x_1 + x_1')x_2'x_3'\\ = x_1x_3 + x_2x_3 + x_2'x_3'$

In the last two equalities, how $x_1x_3$ showed up?

Shubham Johri
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Κωνσταντίνος Κορναράκης
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1 Answers1

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$\color{red}{x_1x_2'x_3+x_1x_2x_3}+\color{blue}{x_1x_2'x_3'}+\color{green}{x_1'x_2x_3}+\color{blue}{x_1'x_2'x_3'}\\=\color{red}{x_1x_2'x_3+x_1x_2x_3}+\color{green}{x_1x_2x_3}(*)+\color{blue}{x_1x_2'x_3'}+\color{green}{x_1'x_2x_3}+\color{blue}{x_1'x_2'x_3'}\\=\color{red}{x_1(x_2+x_2')x_3}+\color{blue}{(x_1+x_1')x_2'x_3'}+\color{green}{(x_1+x_1')x_2x_3}\\=x_1x_3+x_2'x_3'+x_2x_3$

$(*)$ due to the Idempotent Law, $A=A+A\implies x_1x_2x_3=x_1x_2x_3+x_1x_2x_3$.

Shubham Johri
  • 17,659