# Rewriting a boolean expression in SOP form

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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– mark
Commented Dec 17, 2021 at 10:27
• 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. Commented Dec 18, 2021 at 23:41
• @hardmath using De Morgan's law twice and distributive law once, $zx+zy'=(zx+zy')''=((z'+x')(z'+y))'=(z'z'+z'y+x'z'+x'y)'=(z+z)(z+y')(x+z)(x+y')$ Commented Dec 19, 2021 at 4:14
• @hardmath it's the distributive law where the 'plus' distributes over the 'times'. Commented Dec 19, 2021 at 4:29

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

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 $$z$$.
$$(x+y')z=(xy+xy'+x'y')z$$
$$=xyz+xy'z+x'y'z$$

• 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? Commented Dec 17, 2021 at 12:08
• @lukemason x=x*1=x*(y+y') the first step uses identity law and the second step uses complement law. Commented Dec 17, 2021 at 15:29