# Simplify $X'(X+Y) + (Y+X.X) ( X + Y') + Z + X.Z$

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

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

Notation:

• $$X'$$ : 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.

• What do ' and . mean? May 30 '21 at 14:03
• Perhaps you should add a bit of context to help understanding your question? May 30 '21 at 14:04
• ' = prime . = and thanks for the reply. May 30 '21 at 14:06
• 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? May 30 '21 at 15:35
• 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)? May 30 '21 at 17:34

## 1 Answer

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 $$x.y + x.z$$

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