# Thread: Truth table - Find formula

1. ## Truth table - Find formula

Okey, I have been given this truth table.

$\begin{array}{|c|c|c|c|c|}
\hline \bold{P} & \bold{Q} & \bold{ F} \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline \end{array}$

I can't seem to see the combination of which forumla represents this table. eg. ~(p v q)
Any help in the right direction would be greatly appriciated!

2. let me try to help you this. although its bit difficult for me to explain it with paper and pen.

when ever you have any truth tableand you want to derive the formula, take all thos rows in which output F is one.
Here F is 1 for three rors
a) P = 0 and Q = 0
b) p = 0 an Q = 1
c) p= 1 and Q = 0

now take each term in binary and add them to get expression f\of F.
like

a) P = 0 and Q = 0 means $\overline P . \overline Q$
b) p = 0 an Q = 1 means $\overline P . Q$

Let me know if teh concept is still not clear. I will explain you taking more examples.

Arpit
c) p= 1 and Q = 0 means $P . \overline Q$

So expression for F would be F = $\overline P . \overline Q$ + $\overline P . Q$ + $P . \overline Q$

which on simplification gives $F = \overline P + \overline Q$ = $\overline {P.Q}$

3. Hello, jokke22!

I have been given this truth table.

$\begin{array}{|c|c||c|} \hline
p & q & f \\ \hline \hline
0 & 0 & 1 \\ \hline
0 & 1 & 1 \\ \hline
1 & 0 & 1 \\ \hline
1 & 1 & 0 \\ \hline \end{array}$

I can't seem to see the formula that represents this table.
e.g. . $\sim (p \vee q)$

Where did you get that example? . . . It's correct!

. . $\begin{array}{|c|c||c|c|} \hline
p & q & p \vee q & \sim(p \vee q) \\ \hline \hline
0 &0&0&{\color{blue}1} \\ \hline 0&1&0&{\color{blue}1}\\ \hline 1&0&0&{\color{blue}1} \\ \hline 1&1&1&{\color{blue}0} \\ \hline \end{array}$

4. Originally Posted by Soroban
Hello, jokke22!

Where did you get that example? . . . It's correct!

. . $\begin{array}{|c|c||c|c|} \hline
p & q & p \vee q & \sim(p \vee q) \\ \hline \hline
0 &0&0&{\color{blue}1} \\ \hline 0&1&0&{\color{blue}1}\\ \hline 1&0&0&{\color{blue}1} \\ \hline 1&1&1&{\color{blue}0} \\ \hline \end{array}$

It's what I've seen as the only logical answere, but it was kinda the "wrong" way so I was unsure if it's correct!

Thanks though for confirming!

I will have to look further into this regaring reply 2, arpitagarwal82, thanks for your help as well though so far!

5. There are two things in my solution.
1) writing the expression by seeing the table
2) simplifying the expression urther.

Are you clear with both the parts. If not let me know I can give further examples and explanation.