0

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?

Trev347
  • 103
  • 1
    Note $(1-a)(1-b)(1-c)\neq 1-abc$. – user10354138 Oct 18 '23 at 02:05
  • What happens to $ABC'$ or $A'BC'$ or several other possibilities? – Henry Oct 18 '23 at 02:10
  • 1
    What is true is that $ABC +(ABC)'=1$ but $A'B'C'$ need not be and usually is not $(ABC)'$ – Henry Oct 18 '23 at 02:12
  • @Henry Can you help explain to my brain why this does not work? If I saw this in a problem I would automatically assume it would work, but I am having a hard time understanding why it does not. The only way I know how it doesn't work is because I did the truth table. – Trev347 Oct 18 '23 at 02:16
  • 1
    The truth table shows that $A'B'C' \ne (ABC)'$. Instead, De Morgan's laws apply. – peterwhy Oct 18 '23 at 02:17
  • 1
    $(ABC)'$ is every possibility except $ABC$ so is the sum of $2^3-1=7$ terms $ABC'+AB'C+AB'C'+A'BC+A'BC'+A'B'C+A'B'C'$, not just one of them. – Henry Oct 18 '23 at 02:19
  • @Henry What about A'B' + AB'C. Why can't I factor out the B' and obtain B'(A' + AC) then use the law "X + X'Y = X + Y" to get B'(A' + C)? These are the kinds of examples that confuse me cause it seems as if it should work, cause it has the same format as these laws, but they do not work. Does anybody have a good explanation for this that makes sense? – Trev347 Oct 18 '23 at 02:24
  • 2
    $A'B' + AB'C=B'(A' + C)$ is correct in Boolean algebra – Henry Oct 18 '23 at 02:28
  • 2
    Because $A^\prime B^\prime C^\prime$ is actually not equal to $(ABC)^\prime$. – Rócherz Oct 18 '23 at 07:21

3 Answers3

1

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.

ConMan
  • 24,300
1

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

Bram28
  • 100,612
  • 6
  • 70
  • 118
0

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

J.-E. Pin
  • 40,163