Constructing a truth table from a giving statement

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January 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.
January 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.

.
January 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.
January 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,

$p\vee\neg{q}$

$TTFT$
$TTTF$
$FFFT$
$FTTF$
January 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.
January 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.
January 22nd 2010, 08:11 AM
Soroban

Hello, ipatch!

Quote:

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

. . $\begin{array}{c|c||ccc} p & q & p & \vee & \sim q \\ \hline T & T & T & {\color{red}T} & F \\ F & T & F & {\color{red}F} & F \\\ T & F & T & {\color{red}T} & T \\ F & F & F & {\color{red}T} & T \end{array}$
January 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?
January 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.
January 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?
January 22nd 2010, 12:30 PM
ipatch

yeah i am getting better with the truth tables now. thanks for the help.
January 23rd 2010, 07:59 AM
Grandad

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.

Grandad
January 23rd 2010, 10:26 AM
HallsofIvy

Quote:

Originally Posted by VonNemo19

What is it that you do not understand?

Your remarks

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.
January 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.