I wanna know if $X'(X+Y)$ means $X'.X+Y.X'$?

Does it have an AND gate after $X'$?


  • $X'$ : NOT $X$
  • $X + Y$: $X$ OR $Y$ (OR gate)
  • $X.Y$ : $X$ AND $Y$ (AND gate)

New to boolean, can't seem to understand this question.

  • $\begingroup$ What do ' and . mean? $\endgroup$
    – nir shahar
    May 30 '21 at 14:03
  • $\begingroup$ Perhaps you should add a bit of context to help understanding your question? $\endgroup$
    – Nathaniel
    May 30 '21 at 14:04
  • $\begingroup$ ' = prime . = and thanks for the reply. $\endgroup$ May 30 '21 at 14:06
  • $\begingroup$ The title and the body of the question do not seem to match... Are you only interested in learning if $X'(X + Y) = X'.X+Y.X'$ is correct or are you trying to ask something (simplification?) about the formula in the title? Also what is "prime" supposed to mean? Is it like a NOT gate? $\endgroup$
    – phan801
    May 30 '21 at 15:35
  • $\begingroup$ X′(X+Y)=X′.X+Y.X′ I'm interested in whether is this simplification is correct. meaning like is there a AND gate in between X'? like is this X'(X OR Y) simplify gonna be X'AND(X OR Y)? $\endgroup$ May 30 '21 at 17:34

I am not aware of any different conventinos so it is probably implied that your expression is $X'.(X + Y)$

By definition in Boolean Algebra, $+$ and $.$ are distributive over one another. This means that $x.(y+z)=xy+xz$ and $x+(y.z)=(x+y).(x+z)$.

Obviously what you are asking is the first case, so let's take $x.(y+z)=xy+xz$. This is not a proof, I am writing it like this to help you understand what is going on. In natural language, it says that the expression holds if $x$ and ($y$ OR $z$) is true. Or equivalently that $x$ and simultaneously at least one of $y$, $z$ is true. Or in other words, if $x$ and $y$ are true or if $x$ and $z$ are true. Or to return to the math notation $x.y + x.z$

  • $\begingroup$ Okay. Thanks for the reply. Will try to understand what you mean. Need time to digest. :D $\endgroup$ May 30 '21 at 19:53
  • $\begingroup$ So if i want to simplify my title expression, may i know how to do it? $\endgroup$ May 30 '21 at 20:15

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