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$$\overline{X}\overline{Y} + YZ + \overline{X}Y\overline{Z}$$

I'm having a lot of trouble simplifying this boolean expression.

I used commutative property and re-arranged it as my first step: $$\overline{X}Y\overline{Z} + \overline{X}\overline{Y} + YZ$$ 2nd Step: Factored out the common literal, $\overline{X}$. $$\overline{X}(\overline{Y} + Y\overline{Z}) + YZ$$ 3rd Step: ? I first used one of the identities to turn $\overline{Y} + Y$ into 1; I wasn't too sure what to do here and I believed what I did was wrong, so I then used the distributive property for the sum term in the parentheses instead. But, I don't know if this aforementioned step is correct.

S. Coughing
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1 Answers1

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$\overline{X}\overline{Y}+YZ+\overline{X}Y\overline{Z}=$

$\overline{X}\overline{Y}+\color{blue}{YZ}+\overline{X}Y\overline{Z}=$

$=\overline{X}\overline{Y}+\color{blue}{YZ\left(1+\overline{X}\right)}+\overline{X}Y\overline{Z}=$

$=\overline{X}\overline{Y}+YZ+\overline{X}YZ+\overline{X}Y\overline{Z}=$

$=\overline{X}\overline{Y}+YZ+\overline{X}Y\left(Z+\overline{Z}\right)=$

$=\overline{X}\overline{Y}+YZ+\overline{X}Y=$

$=\overline{X}\left(\overline{Y}+Y\right)+YZ=$

$=\overline{X}+YZ$

More details about the steps:

$\color{blue}{YZ=YZ\cdot1=YZ\left(1+\overline{X}\right)}$

because $\;1+\overline{X}=1\;$ whatever is the value of $\;X\;,$

indeed, if $\;X=0\;,\;$ then $\;1+\overline{X}=1+1=1\;,$

whereas if $\;X=1\;,\;$ then $\;1+\overline{X}=1+0=1\;.$

Angelo
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  • Your first step doesn't make sense. As a counterexample, let $X=0, Y=Z=1$. Your final expression is $0$ whereas the original expression is $1$. The actual final answer should be $\overline{X} + XYZ$. – Joshua P. Swanson Feb 05 '21 at 02:16
  • If $X=0,Y=Z=1$, my final expression is $1$, not $0$, moreover $\overline{X}+XYZ=\overline{X}+YZ$ is true in any case. My first step is correct because $1+A=1$ for any boolean variable A. – Angelo Feb 05 '21 at 08:47
  • Hm, I was interpreting $+$ as xor, whereas you're interpreting it as or. Is this a standard convention? I'm used to boolean algebra being modeled on arithmetic in the field with two elements, which uses xor for addition. – Joshua P. Swanson Feb 05 '21 at 10:13
  • Usually in boolean algebra the symbol of ‘xor’ is ‘$⊕$’, whereas the symbol of ‘or’ is ‘$+$’. Boolean algebra is not the field $\mathbb{Z}_2$. If you use ‘$+$’ for ‘xor’, which symbol will you use for ‘or’? ‘$\lor$‘? Usually it is not so. – Angelo Feb 05 '21 at 12:30
  • @JoshuaP.Swanson, if you have any doubt, why do not you ask the original poster what he means with the symbol ‘$+$’? – Angelo Feb 05 '21 at 12:31
  • Angelo is correct: ⊕ is XOR. I've seen people on YouTube introduce new terms to simplify Boolean expressions, but I didn't understand how I would use that to my advantage. If this answer had more details about the steps taken, I would approve of it. I also want to know if my way of doing it would yield a correct answer? – S. Coughing Feb 05 '21 at 17:31
  • @S.Coughing, do you need more details? Look at my answer now. I think your way would not yield a correct answer because the final expression you get, could be simplified further. – Angelo Feb 05 '21 at 18:05
  • It's all good, thank you! – S. Coughing Feb 05 '21 at 19:28
  • Thanks for clarifying, apologies for the digression. – Joshua P. Swanson Feb 05 '21 at 22:06