# Representing Boolean expressions in a truth table.

Right so I'm trying to understand truth tables in the context of digital logic. And paticularly with lettered boolean expresssions.

Now I do understand truth tables, you have either true or false as the value, and you have some operators that determine that true or false answer. And, Or, Not etc.

But the set of questions I've bene set, I can't get my head around. It's asking for things like ABC + ~A~B~C. I'm not sure what this expression actually means.

Can anyone explain to me what the questions in the below image are asking me to do? These are the questions i've been set.

I do actually have the answers but I honestly can't figure out why it works like that.

Image of set questions

• Authentication is required for your links, so most of us cannot actually see what you are talking about. – Graham Kemp Feb 5 '16 at 13:06
• + means or and ~ means not. Multiplication usually means and, so ~A~B means not A and not B – Gregor de Cillia Feb 5 '16 at 13:10

Authentication is required for your links, so most of us cannot actually see what you are talking about, but for this bit:

It's asking for things like ABC + ~A~B~C. I'm not sure what this expression actually means.

In boolean algebra, multiplication is "and", addition (+) is "or" and the tilda (~) is negation ("not").   Thus we read "(A and B and C) or (not A and not B and not C)"

This says "Either $A,B,C$ are all true or they are all false."   Which is another way of saying "Their boolean values are all the same."

So let's fill out the truth table

$$\begin{array}{c:c:c|c:c|c} A & B & C & A~B~C & \bar A~\bar B~\bar C & A~B~C + \bar A~\bar B~\bar C \\ \hline 0 & 0 & 0 & 0 & 1 & 1 & (0\cdot 0\cdot 0)+(1\cdot 1\cdot 1) = 0+1=1 \\ 0 & 0 & 1 & 0 & 0 & 0 & (0\cdot 0\cdot 1)+(1\cdot 1\cdot 0) = 0+0=0 \\ 0 & 1 & 0 & 0 & 0 & 0 & (0\cdot 1\cdot 0)+(1\cdot 0\cdot 1) = 0+0=0 \\ 0 & 1 & 1 & 0 & 0 & 0 & (0\cdot 1\cdot 1)+(1\cdot 0\cdot 0) = 0+0=0 \\ 1 & 0 & 0 & 0 & 0 & 0 & (1\cdot 0\cdot 0)+(0\cdot 1\cdot 1) = 0+0=0 \\ 1 & 0 & 1 & 0 & 0 & 0 & (1\cdot 0\cdot 1)+(0\cdot 1\cdot 0) = 0+0=0 \\ 1 & 1 & 0 & 0 & 0 & 0 & (1\cdot 1\cdot 0)+(0\cdot 0\cdot 1) = 0+0=0 \\ 1 & 1 & 1 & 1 & 0 & 1 & (1\cdot 1\cdot 1)+(0\cdot 0\cdot 0) = 1+0=1 \end{array}$$

• Right i get what the expressions means but i still don't get the logic at all. Also these are the questions: gyazo.com/a2a5f8336feedcd4cd0cefc58be09659 – user310998 Feb 5 '16 at 13:28
• It's unclear what your problem is @GR412 If you "get what the expressions mean", then you should "get the logic"; that's what the expressions are. Where exactly are you having difficulty? – Graham Kemp Feb 5 '16 at 13:44
• Right I get it now, I think the issue was because I didn't know ABC meant A and B and C. And the question didn't tell me to write out the entire table like this. Thanks for your help. I'm really bad at Maths yet i'm trying to do a comp sci degree, lol. – user310998 Feb 5 '16 at 14:36
• @Grahem Kemp Could you also explain what the 1's are when simplifiyng boolean expressions? I don't understand where the 1's come from. These are the two questions i'm talking about: gyazo.com/a583eb9a0efba7695b7042ca7afbaa31 – user310998 Feb 5 '16 at 15:50
• @GR412 In Boolean Algebra, $1$ is the symbol for the constant value of 'true' and $0$ is that of 'false', with the logical operators defined as $$\begin{array}{c|cc}+ & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 1\end{array}\quad\begin{array}{c|cc}\cdot & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1\end{array}\quad\begin{array}{c|cc}\bar{~} & 0 & 1 \\ \hline ~ & 1 & 0\end{array}$$ – Graham Kemp Feb 5 '16 at 22:44