# Boolean Algebra - $ABC+B'=AC+B'$?

I'm doing a bit of homework, and it says to prove or disprove the statement

$XZ+X'Y'+Y'Z'=XZ+Y'$

If you do a truth table and take the sum-of-products, you can eventually simplify the equation down to $XYZ+Y'$. Now that was way too close for me to stop there, and I couldn't think of anything beyond that. I plugged it into one of the WolframAlpha Boolean Algebra equators and it came out with $XYZ+Y'=XZ+Y'$. I'm perplexed as to how it got that, or if it's just a bug in the program.

The only thing I can think of is that it is an extension on the $A+A'B=A+B$ law, but I'm not entirely sure that can be used on $3$ different variables like this. Can someone explain if this is the legitimate answer, and if so can you please prove it so I can see how that's true? Thanks a bunch.

XZ + Y' = XZY +XZY' + Y' = XYZ + (XZY' +Y')
XZY' is a subset of Y'. So,
=XYZ + Y'

Best thing to do would be draw three intersecting circles as X,Y and Z and try to visualize LHS and RHS

Note that $\bar{Y} \implies X\bar{Y}Z$ and so $\bar{Y} = X\bar{Y}Z + \bar{Y}$.

Then $XYZ+ \bar{Y} = XYZ+ X\bar{Y}Z + \bar{Y}= X(Y+\bar{Y})Z + \bar{Y} = XZ + \bar{Y}$.

ABC + B’ = AC + B’ ?

The expression on the left means if B then AC, if not B then true. Obviously, the repeated mentioning of B is redundant, so the expression to the right, which reads if not B then true, else AC is its simplified equivalent. You can verify this by writing $B'+ABC=B'+BAC=$ $=(B'+B)(B'+AC)=1\cdot(B'+AC)=AC+B'.$