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?
Asked
Active
Viewed 111 times
3
-
It would be good if you include in your question the steps to get what you wrote after "By following the above identity". – coffeemath Jun 24 '21 at 04:25
-
@JMP How, it is not true? In that case, both sides of the equation give you $1$. – gete Jun 24 '21 at 05:50
-
@JMP These are logical operations and symbols, but not usual addition, multiplication or real number $1$ and $0$. $0$ and $1$ are just symbols. – gete Jun 24 '21 at 06:07
-
@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. – gete Jun 24 '21 at 09:32
-
@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. – gete Jun 24 '21 at 09:44
1 Answers
4
No. Taking $x = y = z = 1$ and $p = 0$, you get $\bar yxz +yp = 0$ but $xz + yp = 1$.
J.-E. Pin
- 42,871
-
Thank you for the answer. By the ways, do you mean that the expression $\bar{y}xz+yp$ cannot be further simplified? – gete Jun 24 '21 at 04:30
-
-