# Constructing a truth table from a giving statement

• Jan 21st 2010, 07:39 PM
ipatch
Constructing a truth table from a giving statement
Hello, I am currently trying to construct a truth table for the giving statement. p v ¬ q

I am not sure how to format it in this post but this is what I have so far.

p q | p v ¬ q

T T | F
F T | F
T F | T
F F | F

The "T" means true and the "F" means False.

Is this truth table correct or am I entirely wrong?

cheers.
• Jan 21st 2010, 07:44 PM
VonNemo19
Quote:

Originally Posted by ipatch
Hello, I am currently trying to construct a truth table for the giving statement. p v ¬ q

I am not sure how to format it in this post but this is what I have so far.

p q | p v ¬ q How come there is no operator between pq ? do you mean that both p and q are true? Or what? and what do mean by |? Can you try to put forth the argument in english?

T T | F
F T | F
T F | T
F F | F

The "T" means true and the "F" means False.

Is this truth table correct or am I entirely wrong?

cheers.

.
• Jan 21st 2010, 07:49 PM
ipatch
Problem
All the problem in the text says is "construct a truth table for the given statement.

"p v ¬ q" This is is the statement it gives me, and it wants a truth

table from it.
• Jan 21st 2010, 07:59 PM
VonNemo19
Quote:

Originally Posted by ipatch
All the problem in the text says is "construct a truth table for the given statement.

"p v ¬ q" This is is the statement it gives me, and it wants a truth

table from it.

So the excercise is to see when the weddge holds true. Note that when we have $n$ simple statements (in this case we have 2: $p$ and $q$) that the number of lines that are necessary in the truth table are $L=2^n$. Also note that the $\neg$ reverses the truth value. So,

$p\vee\neg{q}$

$TTFT$
$TTTF$
$FFFT$
$FTTF$
• Jan 22nd 2010, 01:05 AM
emakarov
I understand the OP's approach, but I am not sure I understand VonNemo's remarks. The truth table for $p\lor\neg q$ must indeed have the following form:

Code:

p q | p \/ (not q) ----+------------- T T |  ? F T |  ? T F |  ? F F |  ? 
(Though I prefer the following order of the first two columns going down: F F, F T, T F, T T. If T is replaced by 1 and F by 0, we get 00, 01, 10, 11, which in binary notation means 0, 1, 2, 3. Following this systems helps prevent missing or repeated rows.)

However, the OP's truth values in the last column are incorrect. For example, when p = T and q = T, $p\lor\neg q$ is T because disjunction is true if at least one of the arguments (in this case, p) is true.
• Jan 22nd 2010, 07:05 AM
ipatch
Quote:

Originally Posted by VonNemo19
So the excercise is to see when the weddge holds true. Note that when we have $n$ simple statements (in this case we have 2: $p$ and $q$) that the number of lines that are necessary in the truth table are $L=2^n$. Also note that the $\neg$ reverses the truth value. So,

$p\vee\neg{q}$

$TTFT$
$TTTF$
$FFFT$
$FTTF$

I get what your saying about the ¬ representing negation in all, but I just realized today that there are going to be four columns setup for this statement.

p \/ ¬ q

is going to be setup like this

p q ¬q p \/ ¬ q

T T F T
T F T T
F T F F
F F T T

I believe this is what I put down for the problem.

Also here is a good primer example for the problem stated in the thread.

p \/ q

p q p \/ q

T T T
T F T
F T T
F F F

Basically the "or" ( \/ ) can include either or, and both.

Cheers.
• Jan 22nd 2010, 08:11 AM
Soroban
Hello, ipatch!

Quote:

Construct a truth table for: . $p\: \vee \sim q$

. . $\begin{array}{c|c||ccc}
p & q & p & \vee & \sim q \\ \hline
T & T & T & {\color{red}T} & F \\
F & T & F & {\color{red}F} & F \\\
T & F & T & {\color{red}T} & T \\
F & F & F & {\color{red}T} & T
\end{array}$

• Jan 22nd 2010, 09:27 AM
VonNemo19
Quote:

Originally Posted by emakarov
I understand the OP's approach, but I am not sure I understand VonNemo's remarks. .

What is it that you do not understand?
• Jan 22nd 2010, 10:22 AM
ipatch
Quote:

Originally Posted by Soroban
Hello, ipatch!

. . $\begin{array}{c|c||ccc}
p & q & p & \vee & \sim q \\ \hline
T & T & T & {\color{red}T} & F \\
F & T & F & {\color{red}F} & F \\\
T & F & T & {\color{red}T} & T \\
F & F & F & {\color{red}T} & T
\end{array}$

not sure what you mean by using the ~

i know ¬ means take the negation of a statement.
• Jan 22nd 2010, 10:27 AM
VonNemo19
Quote:

Originally Posted by ipatch
not sure what you mean by using the ~

i know ¬ means take the negation of a statement.

They mean the same thing. Are you OK with how to construct a truth table now, or do you have other questions?
• Jan 22nd 2010, 12:30 PM
ipatch
yeah i am getting better with the truth tables now. thanks for the help.
• Jan 23rd 2010, 07:59 AM
Hello ipatch

You might find these notes that I wrote for Wikibooks useful. There's a step-by-step guide to creating truth tables, with worked examples and exercises with answers. If you have any questions about them, post them on this Forum.

• Jan 23rd 2010, 10:26 AM
HallsofIvy
Quote:

Originally Posted by VonNemo19
What is it that you do not understand?

Quote:

p q | p v ¬ q How come there is no operator between pq ? do you mean that both p and q are true?
after mistaking the p and q heading the first two columns in the truth table for part of the formula.
• Jan 23rd 2010, 10:35 AM
VonNemo19
Quote:

Originally Posted by HallsofIvy
Your remarks after mistaking the p and q heading the first two columns in the truth table for part of the formula.

(Giggle) I don't feel that I made a mistake anywhere. How can asking for clarification be considered as a mistake? The notation in logic varies very much. This is evidenced by post nine.