Use Boolean Logic Rules To Show The Expressions Are Valid?

Question

$(X + Y) + (X' Y') = 1 $

I know I am doing this wrong but when I look at this all I can see to do is combine the two, and then they would cancel out and just be $0 = 1$. I am not sure how to go about doing this one.

The other example given is $(X + Y) (X +Y') = x$. I am honestly very confused with these types of problems, and don't even know where to start.

Answer 1

For the first one note that $X'Y'=(X+Y)'$. Hence $$(X + Y) + (X' Y') =(X + Y)+(X+Y)'=1.$$ For the second one $$(X + Y) (X +Y')=XX+XY'+XY+YY'=X+X(Y'+Y)+0=X+X=X.$$

Answer 2

In the case $X'$ denotes $\neg X$ (negation of $X$), using the de morgan rules, you need to find terms containing $X+X'$ or $Y+Y'$, which is equal to 1, or $X'X$ which is equal to $0$. This simplifies a lot the expressions.

Here, you can do $(X+Y)+(X'Y')=(X+X'+Y)(X+Y+Y')=1 \cdot 1 = 1$

But decompose this a little more, and look for the specific rule you apply at each step.