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$$(xy'+xy)+xz'$$

Using boolean algebra I achieve $x + xz'$, which is pretty obvious by just looking at the problem, however I can't find another way to go after there in order to cancel the $z',$ since there is no $xz$ or similar term to cancel the product.

The below image shows the properties I am allowed to use, highlighted in blue. The properties that are not highlighted, like $1 = a + 1$, or so I am not allowed to use.

EDIT: reloaded the website and I think there was a problem with the website, I got an email from the book company and I was given a new question, since this one seemed misleading, nonetheless I appreciate those who took the time to read the question and gave their input.

user376343
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Miguel Rodriguez
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  • Do you mean you cannot use a rule of your image when only its left side is in blue but not its right? [all the left sides are blue... more clear not to make either side blue if you can't use the rule.] – coffeemath Oct 23 '21 at 19:48
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    I think you used (xy+xy')=x(y+y') which is allowed, and then (y+y'=1) that is not allowed. If the latest is allowed, then also (x+xz')=x(1+z')=x.1=x is ok. – user376343 Oct 23 '21 at 19:48
  • I can use all of the left sides, sorry for the inconvenience, the website set which rules I can and cannot use, I can't really change the colors but basically I can only use the left side so if i have something in the form of a + 1 for example which is on the left side I can convert to 1, but if I had 1 I cannot convert 1 to a + 1 since one is on the right – Miguel Rodriguez Oct 23 '21 at 19:50
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    @MiguelRodriguez Equality ought to be a symmetric relation! You are allowed to write $1=1+z'$. – Berci Oct 23 '21 at 20:09
  • Why don't you use Karnaugh diagrams ? It's so simple ! – Jean Marie Oct 23 '21 at 21:58

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EDIT $$\begin{aligned} (xy'+xy)+xz'&=x(y'+y)+xz'\qquad &\text{distributivity}\\&=x\cdot 1+xz' \qquad &\text{complement}\\ &=x(1+z') \qquad & \text{distributivity}\\&=x(z'+1) \qquad & \text{commutativity}\\ &=x\cdot 1 \qquad &\text{null element}\\&=x \qquad &\text{identity} \end{aligned}$$

user376343
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  • You don't need the step called 'identity'. In fact, you introduce the $1$ again in the following step. (+1) just the same, but I'd edit to remove that step. – amrsa Oct 25 '21 at 18:27
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    Yes, but my point is that you could just make $x1+xz'=x(1+z')$. From $x+xz'$ we would get $x$ by absorption, which the OP apparently cannot use, but what you did was actually reintroduce the element $1$ to apply distributivity. – amrsa Oct 25 '21 at 19:33
  • Thank you @amrsa , edited. – user376343 Oct 25 '21 at 20:34