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I have started a discrete math module for my computer science course and I'm having a little trouble using the identity, idempotent and complement laws to convert a boolean expression into sum of products (SOP) form.

$F(x,y,z) = (x+y')*z$
$=(z*x)+(z*y')$
$=(z+z)*(z+y')*(z+x)*(x+y')$

This is where I'm stuck. I could simplify it but I am not sure how to get to $(x'y'z)+(xy'z)+(xyz)$ that i got with a truth table.

Any guidance would be greatly appreciated.

Kind regards, Luke

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    $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial, and equation editing how-to. $\endgroup$
    – mark
    Commented Dec 17, 2021 at 10:27
  • $\begingroup$ Your first step, distributing $z$ over $x+y'$, gives you a sum of products. I'm not sure what to make of the next line. Is it a new problem? It doesn't seem to have any relationship to the line above it. $\endgroup$
    – hardmath
    Commented Dec 18, 2021 at 23:41
  • $\begingroup$ @hardmath using De Morgan's law twice and distributive law once, $zx+zy'=(zx+zy')''=((z'+x')(z'+y))'=(z'z'+z'y+x'z'+x'y)'=(z+z)(z+y')(x+z)(x+y')$ $\endgroup$
    – cineel
    Commented Dec 19, 2021 at 4:14
  • $\begingroup$ @hardmath it's the distributive law where the 'plus' distributes over the 'times'. $\endgroup$
    – cineel
    Commented Dec 19, 2021 at 4:29

2 Answers 2

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After your first step you have $(z*x)+(z*y')$. That is in SOP, so you are done!

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first find $x+y'$ in SOP form.
$x+y'=x(y+y')+y'(x+x')$
$=xy+xy'+xy'+x'y'$
$=xy+xy'+x'y'$

multiply both sides by $z$.
$(x+y')z=(xy+xy'+x'y')z$
$=xyz+xy'z+x'y'z$

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  • $\begingroup$ Thanks @cineel. As im still learning the laws, could i ask if im right in thinking what xou did was use distributive with complement law to expand the (x+y'), then idempotent to remove one of the + xy'. Then used distriutive factorising to re introduce the z? im not sure if that is right and where i could use the identity law? $\endgroup$
    – luke mason
    Commented Dec 17, 2021 at 12:08
  • $\begingroup$ @lukemason x=x*1=x*(y+y') the first step uses identity law and the second step uses complement law. $\endgroup$
    – cineel
    Commented Dec 17, 2021 at 15:29

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