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X'Y'Z' + XYZ

I have the equation above (Boolean Algebra truth table), and as I know, if I get x' and the value of x is 0, the value will change to 1. But Y' with the top bar being higher, what happens if I replace Y wit 0? Will it also become a 1? Please mind my noob question, as I am still a noob.

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There is no equation in what you have written. Please try again. – Dilip Sarwate Sep 25 '12 at 18:35
What do you mean by "the top bar being higher"? – Henning Makholm Sep 25 '12 at 18:39

X'Y'Z' + XYZ reads as "not X, not Y, and not Z, OR X, Y, and Z". By simple evaluation, we see that if you plug in 0 for X, X' becomes true and the expression simplifies to Y'Z'. If you plug in 1, X is true and the expression simplifies to YZ.

A similar thing happens if you only substitute for Y: Y=0 $\to$ X'Z', Y=1 $\to$ XZ. If Y=X', the expression evaluates to zero since XY=0 and X'Y'=0, regardless of Z. If Y=X, then the expression evaluates either to Z if X=Y=1 or Z' if X=Y=0.

In other words, it's all or nothing; the expression evaluates to 1 if X=Y=Z, regardless of whether the value of all three is 1 or 0. If any one of the three does not equal the others, the expression evaluates to zero.

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Suppose two variables take different values, then Without assume $X=1$ and $Z=0$, then

$X'Y'Z'=0$ (since $X=1\implies X'=0$) and

$XYZ=0$ (since $Z=0 $)

Thus, $X'Y'Z'+XYZ=0$ whenever any two variables take different values.

If all variables take same value $0$ then,

$X'Y'Z'=1\implies X'Y'Z'+XYZ=1$

and if they all take same value $1$ then,

$XYZ=1\implies X'Y'Z'+XYZ=1$

Thus, $X'Y'Z'+XYZ=1$ whenever all the variables take same value.

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