Why does A'B'C' + ABC not equal 1 due to the boolean algebra law X' + X = 1?

Question

I am learning boolean algebra, but this is confusing me.

How does A'B'C' + ABC not equal 1? If you do the truth table for ABC then the output will not always be 1, for example 101 gives an output of 0. Why is this? Why can't we use the law X' + X = 1?

Answer 1

It's because $(ABC)' = A'+B'+C'$, rather than $A'B'C'$

Answer 2

It might help to look at the two variable case, and to consider it in terms of the truth values of some statements. Let

$$A = \textrm{Today is Monday} \\ B = \textrm{It is raining today}$$

Then if we look at the possible combinations of $A$ and $A'$ with $B$ and $B'$, they correspond to:

$$\begin{array}{|c|c|c|} \hline & A & A' \\ \hline B & \textrm{Today is Monday and it is raining} & \textrm{Today is not Monday and it is raining} \\ \hline B' & \textrm{Today is Monday and it is not raining} & \textrm{Today is not Monday and it is not raining} \\ \hline\end{array}$$

So you can see that $AB + A'B' \neq 1$, because it only covers rainy Mondays and non-rainy non-Mondays, missing out fine Mondays and wet Thursdays entirely.

You can state that $AB + (AB)' = 1$, but notice that $(AB)'$ corresponds to "It is not a rainy Monday", which holds true if either today is not Monday or it isn't raining, i.e. $(AB)' = A'B + AB' + A'B' = A' + B'$. This is one of De Morgan's Laws, which is the correct way to deal with the interaction of negation with multiplication or addition in Boolean algebra.

Answer 3

Just take $A=1$, $B = 1$ and $C = 0$. Then $A'B'C' = 0$ and $ABC = 0$ whence $A'B'C' + ABC = 0$.