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In a Boolean algebra $\mathcal B$, we know that $$x+\bar{x}y=x+y\text{ for all } x, y\in \mathcal B.$$ By following the above identity, we can also write $$xy+\bar{x}yz=xy+yz.$$ Can we write $$\bar{y}xz+yp=xz+yp,\text{ where $p$ is distinct from $x$ and $z$}?$$ Or is there any alternative way to simplify the expression on the left side of the last equation further?

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  • $\begingroup$ It would be good if you include in your question the steps to get what you wrote after "By following the above identity". $\endgroup$
    – coffeemath
    Jun 24, 2021 at 4:25
  • $\begingroup$ @JMP How, it is not true? In that case, both sides of the equation give you $1$. $\endgroup$
    – gete
    Jun 24, 2021 at 5:50
  • $\begingroup$ @JMP These are logical operations and symbols, but not usual addition, multiplication or real number $1$ and $0$. $0$ and $1$ are just symbols. $\endgroup$
    – gete
    Jun 24, 2021 at 6:07
  • $\begingroup$ @JMP You can't remove $z$ from the equation $xyz+x'yz=xyz+yz$ unless the structure is cancelative. However, such an equation will hold according to the second identity. $\endgroup$
    – gete
    Jun 24, 2021 at 9:32
  • $\begingroup$ @JMP His answer is a counter example showing that the last identity doesn't hold. Thereby giving a good hint that the expression $\bar{y}xz+yp$ is in the most simplified form. The second identity which you are mentioning here is an already proven identity. $\endgroup$
    – gete
    Jun 24, 2021 at 9:44

1 Answer 1

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No. Taking $x = y = z = 1$ and $p = 0$, you get $\bar yxz +yp = 0$ but $xz + yp = 1$.

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  • $\begingroup$ Thank you for the answer. By the ways, do you mean that the expression $\bar{y}xz+yp$ cannot be further simplified? $\endgroup$
    – gete
    Jun 24, 2021 at 4:30
  • $\begingroup$ I don't think that $\bar yxz + yp$ can be simplified. $\endgroup$
    – J.-E. Pin
    Jun 24, 2021 at 4:37
  • $\begingroup$ Well. That's fine and it got my problem fixed. Thank you. $\endgroup$
    – gete
    Jun 24, 2021 at 4:55

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