Question

Can anybody help me construct a truth table????

Answer 1

I can... but I am limited by the capacity of drawing truth tables here...

Answer 2

we could do one row at a time if that would be easier because i'm not sure how to draw one on here.

Answer 3

I'll give you the basic framework. First, you know your logical operators right? The "not", "and", "or", "implies" etc?

Answer 4

yes

Answer 5

Okay... first thing you gotta do is identify the main operator.
Now... let's start with the easiest.
\[\large \left[\begin{matrix}\neg & P \\ ? & ? \\ ? &?\end{matrix}\right]\]
The main operator here is the negation (NOT) operator, right?

Answer 6

yes

Answer 7

Right. First, pick one of your variables. You only have one, so let's pick P. Under it, write down all possible "truth-values" of P. Truth values just mean either a true or false, 1 or 0, you be the judge. I'll just use T and F. Just tell me if you'd be more comfortable using 1 and 0.

Answer 8

T and F are fine.

Answer 9

\[\large \left[\begin{matrix}\neg & P \\ ? & \color{green}T \\ ? & \color{red}F\end{matrix}\right]\]

Answer 10

Right? Because P can only be either true or false. Catch me so far?

Answer 11

yeah!

Answer 12

Now... what does the "not" operator do to "true" values? What about to "false" values?

Answer 13

makes the true values false, and the false values true?

Answer 14

do you have any expression for which you have to draw a truth table?

Answer 15

Correct. So, write down the effect of the operator... accordingly.
\[\large \left[\begin{matrix}\neg & P \\ \color{red}F & \color{green}T \\ \color{green}T & \color{red}F\end{matrix}\right]\]
\[\large \left[\begin{matrix}\neg & P \\ \color{red}F &T \\ \color{green}T & F\end{matrix}\right]\]

The column underneath the main operator, that is the result of your truth table.

Answer 16

oh okay!

Answer 17

Now, let's do another one... which one do you want, AND, OR, or IMPLIES, or what? :)

Answer 18

so if i give you an example can you help me with it?

Answer 19

Sure. But it might take a while.

Answer 20

that's fine

Answer 21

Hit me~

Answer 22

You're scaring me... this seems lengthy o.O

Answer 23

okay, so i can't get the graph to work but in my solution is pvq...

Answer 24

Just draw it. Or type it out to the best of your abilities :D

Answer 25

the first row is T,T,T,?,?
second row is T,T,F,?,?
third row is T,F,T,?,?

Answer 26

i have more rows after this

Answer 27

What was given? I mean, the question, what was it?

Answer 28

Was it just p v q ?

Answer 29

all those rows are possible values, but you have to give main question..

Answer 30

probably its not just p v q since 3 columns have T/F values :D

Answer 31

The way I'd work out p v q
would involve 3 columns though. Go figure :/

Answer 32

But there are two ?'s on each row, implying five columns... the question, @lexiporter12 ?

Answer 33

here is what im looking at..

Answer 34

i think he has 3 variables

Answer 35

can you see it okay?

Answer 36

okay... it's just p v q after all...
Let's put that down...
\[\large \left[\begin{matrix}p & \lor & q\\ ? & ? & ? \\ ? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{matrix}\right]\]

Answer 37

what is compound statement? and why do you have r?

Answer 38

(pvq)vr
a compound statement is what is in the parenthesis

Answer 39

tsk...\[\large \left[\begin{matrix}(p & \lor & q) & \lor & r \\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\end{matrix}\right]\]

Answer 40

Ready to proceed, @lexiporter12 ?

Answer 41

yes

Answer 42

Now, the main operator of your compound statement is this
\[\large \left[\begin{matrix}(p & \boxed\lor & q) & \lor & r \\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\end{matrix}\right]\]

Answer 43

right!

Answer 44

Scratch that, I think you misunderstood your own question @lexiporter12
The main operator of p v q is this...\[\large \left[\begin{matrix}(p & \boxed\lor & q) & \lor & r \\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\end{matrix}\right]\]

While the main operator of the compound statement is this\[\large \left[\begin{matrix}(p & \lor & q) & \boxed\lor & r \\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\\ ? & ? & ? & ? & ?\end{matrix}\right]\]

Answer 45

ohh okay

Answer 46

so pick a variable... you can pick any variable you want, as long as it's p

Answer 47

well my chart already tells me what the first 3 columns are, so i just need to find the last two

Answer 48

So under the variable p, divide the column into two halves, the first half being TRUE, the second half being FALSE \[\large \left[\begin{matrix}(p & \lor & q) & \lor & r \\ \color{green}T & ? & ? & ? & ?\\ \color{green}T & ? & ? & ? & ?\\ \color{green}T & ? & ? & ? & ?\\ \color{green}T & ? & ? & ? & ?\\ \color{red}F & ? & ? & ? & ?\\ \color{red}F & ? & ? & ? & ?\\ \color{red}F & ? & ? & ? & ?\\ \color{red}F & ? & ? & ? & ?\end{matrix}\right]\]

Answer 49

okay got that

Answer 50

No... you will learn to do this from scratch, @lexiporter12 for your own good :D

Answer 51

haha okay

Answer 52

the 2nd column would be TTFFTTFF

Answer 53

second colum or the column under q.

Answer 54

under q

Answer 55

\[\large \left[\begin{matrix}(p & \lor & q) & \lor & r \\ \color{green}T & ? & \color{green}T & ? & ?\\ \color{green}T & ? & \color{green}T & ? & ?\\ \color{green}T & ? & \color{red}F & ? & ?\\ \color{green}T & ? & \color{red}F & ? & ?\\ \color{red}F & ? & \color{green}T & ? & ?\\ \color{red}F & ? & \color{green}T & ? & ?\\ \color{red}F & ? & \color{red}F & ? & ?\\ \color{red}F & ? & \color{red}F & ? & ?\end{matrix}\right]\]

Answer 56

You can also put the values for r... what would they look like?

Answer 57

TFTFTFTF for r

Answer 58

You see the pattern, right? Alternating T's and F's, but faster alternations every time.\[\large \left[\begin{matrix}(p & \lor & q) & \lor & r \\ \color{green}T & ? & \color{green}T & ? & \color{green}T\\ \color{green}T & ? & \color{green}T & ? & \color{red}F\\ \color{green}T & ? & \color{red}F & ? & \color{green}T\\ \color{green}T & ? & \color{red}F & ? & \color{red}F\\ \color{red}F & ? & \color{green}T & ? & \color{green}T\\ \color{red}F & ? & \color{green}T & ? & \color{red}F\\ \color{red}F & ? & \color{red}F & ? & \color{green}T\\ \color{red}F & ? & \color{red}F & ? & \color{red}F\end{matrix}\right]\]

Answer 59

right

Answer 60

So... what does the v (OR) operator do to two truth values?

Answer 61

make them F

Answer 62

No. The v (OR) operator would be true if AT LEAST ONE of the two values it works with are true. That's why I needed to draw another truth table...
\[\large \left[\begin{matrix}P & \boxed\lor &Q\\ \color{blue}T & \boxed{\color{green}T} & \color{blue}T \\ \color{blue}T & \boxed{\color{green}T} & \color{orange}F\\ \color{orange}F & \boxed{\color{green}T} & \color{blue}T\\ \color{orange}F & \boxed{\color{red}F} & \color{orange}F\end{matrix}\right]\]

Answer 63

So... can you finish up the second column?

Answer 64

Remember, under the v (OR) operator, you would put a T if there is a T on the left OR a T on the right.

Answer 65

FFTTT

Answer 66

There are 8 of them...

Answer 67

TTTFFFTT

Answer 68

NO. I told you, if there is at least ONE true value between the two variables (in this case p and q) it would make the value of the v (OR) operator true. For example, in the first row, you have both p and q as T. Therefore, the value of the v (OR) operator on that row is T.

Answer 69

TTTTTTFF

Answer 70

That's much better. And those are the truth values of your statement p v q.

Answer 71

okay so what about the compound statements

Answer 72

\[\large \left[\begin{matrix}(p & \lor & q) & \lor & r \\ \color{green}T & \color{green}T & \color{green}T & ? & \color{green}T\\ \color{green}T & \color{green}T& \color{green}T & ? & \color{red}F\\ \color{green}T & \color{green}T & \color{red}F & ? & \color{green}T\\ \color{green}T & \color{green}T & \color{red}F & ? & \color{red}F\\ \color{red}F & \color{green}T & \color{green}T & ? & \color{green}T\\ \color{red}F & \color{green}T & \color{green}T & ? & \color{red}F\\ \color{red}F & \color{red}F & \color{red}F & ? & \color{green}T\\ \color{red}F & \color{red}F & \color{red}F & ? & \color{red}F\end{matrix}\right]\]

Answer 73

It's the same logic, as with the v (OR) operator, but this time, you compare the truth values of the r-variable to the truth values of the entire pvq statement to the left.

Answer 74

so if at least one of them is true then it makes the compound statement true?

Answer 75

Yes.

Answer 76

TTTTTTTF

Answer 77

And that's your answer. Nicely played.

Answer 78

THANK YUOU

Answer 79

No problem.

Answer 80

I have to go to school now... until the next, guys...
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Terence out