# Thread: truth table check =]

1. ## truth table check =]

I just unsure on the truth table ive done and was wondering if someone was able to see if it looks right or not lol

Heres my forumla:

P=(p0^p1) V ¬p2

Heres the table:

po|p1|p2|P
1 | 1 |0 | 1
0 | 0 |1 | 0
1 | 0 |1 | 0
1 | 1 |1 | 0
0 | 1 |1 | 0
0 | 1 |0 | 0
1 | 0 |0 | 0

If ive got it wrong which i think i have, if you could point me in the right direction that would be great!

Thanks,

~Crag

You're missing one row: the p0 = 0, p1 = 0, p2 = 0 row.

Your fourth row is incorrect, I think, as are your sixth and seventh rows.

I would recommend, in general, building up your expression by bits and pieces. Here's an example:

$\begin{tabular}{c|c|c|c|c}
p_{0} &p_{1} &p_{2} &p_{0}\land p_{1} &\neg p_{2}\\ \hline
0 &0 &0 &0 &1\\
0 &0 &1 &0 &0\\
0 &1 &0 &0 &1\\
0 &1 &1 &0 &0\\
1 &0 &0 &0 &1\\
1 &0 &1 &0 &0\\
1 &1 &0 &1 &1\\
1 &1 &1 &1 &0
\end{tabular}$

Then finish. What do you get?

Note the organized way in which I have arranged the rows. It's a good idea to do that as well (I'm basically counting in binary), so that you don't miss a row. If you have $n$ variables, then the truth table must have $2^{n}$ rows, in this case $2^{3}=8$ rows.

3. $\begin{tabular}{c|c|c|c|c|c}
p_{0} &p_{1} &p_{2} &p_{0}\land p_{1} &\neg p_{2} &P \\ \hline
0 &0 &0 &0 &1 &1\\
0 &0 &1 &0 &0 &0\\
0 &1 &0 &0 &1 &1\\
0 &1 &1 &0 &0 &0\\
1 &0 &0 &0 &1 &1\\
1 &0 &1 &0 &0 &0\\
1 &1 &0 &1 &1 &0\\
1 &1 &1 &1 &0 &1
\end{tabular}$

Is that looking right? wasn't sure about the one on row seven as it was both, so im guessing it has to be one or the other to equal P due to the or.

Is this right?

4. Technically, what you've done is sort of mix up the inclusive or and the exclusive or. You're asked to compute the inclusive or, which is true if either of the disjuncts is true, or if both of them are true. It is false only when both disjuncts are false. The exclusive or is true if the disjuncts have the opposite truth value, and false if they have the same truth value. Make sense?

5. $\begin{tabular}{c|c|c|c|c|c}
p_{0} &p_{1} &p_{2} &p_{0}\land p_{1} &\neg p_{2} &P \\ \hline
0 &0 &0 &0 &1 &1\\
0 &0 &1 &0 &0 &0\\
0 &1 &0 &0 &1 &1\\
0 &1 &1 &0 &0 &0\\
1 &0 &0 &0 &1 &1\\
1 &0 &1 &0 &0 &0\\
1 &1 &0 &1 &1 &1\\
1 &1 &1 &1 &0 &1
\end{tabular}$

Right i think i get that, is the table right now?

6. There you go.

7. Right i thinking im getting this now

i have also been asked to draw truth tables for the formulas Q and R,

Q= ¬p2
R= ¬p0 V ¬p1

So im guessing i can add this onto the current table ive got already, like so:

$\begin{tabular}{c|c|c|c|c|c|c|c}
p_{0} &p_{1} &p_{2} &p_{0}\land p_{1} &\neg p_{2} &P &Q &R \\ \hline
0 &0 &0 &0 &1 &1 &0 &1\\
0 &0 &1 &0 &0 &0 &1 &1\\
0 &1 &0 &0 &1 &1 &0 &0\\
0 &1 &1 &0 &0 &0 &1 &0\\
1 &0 &0 &0 &1 &1 &0 &0\\
1 &0 &1 &0 &0 &0 &1 &0\\
1 &1 &0 &1 &1 &1 &0 &0\\
1 &1 &1 &1 &0 &1 &1 &0
\end{tabular}$

Have i got the or mixed up on formula R again? So for example should it be - R when it is either p0 or p1 but it is not R when they both p0 and p1 are 1. Or like how i have it in my table?

Also going back to the question ive been asked
i have also been asked to draw truth tables for the formulas Q and R,
Does this mean i have to draw a table for each formula or can i do it like i have above?

Thanks so much for the help!

8. Both your Q and R are incorrect. Your Q should be precisely the same as the fifth column you already computed! I would normally build up R from the constituent pieces. Do the negations, and then the disjunct.

9. ahh right i see what ive done lol attempt 39 lol XD

$\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
p_{0} &p_{1} &p_{2} &p_{0}\land p_{1} &\neg p_{2} &\neg p_{0} &\neg p_{1} &P &Q &R \\ \hline
0 &0 &0 &0 &1 &1 &1 &1 &1 &1\\
0 &0 &1 &0 &0 &1 &1 &0 &0 &1\\
0 &1 &0 &0 &1 &1 &0 &1 &1 &1\\
0 &1 &1 &0 &0 &1 &0 &0 &0 &1\\
1 &0 &0 &0 &1 &0 &1 &1 &1 &1\\
1 &0 &1 &0 &0 &0 &1 &0 &0 &0\\
1 &1 &0 &1 &1 &0 &0 &1 &1 &1\\
1 &1 &1 &1 &0 &0 &0 &1 &0 &0
\end{tabular}$

Hows that looking? im just a bit unsure on R due to the or

Thanks

10. Q is correct. Double check rows 6 and 7 for R. Like Plato said, you can use WolframAlpha to check your work.

11. $\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
p_{0} &p_{1} &p_{2} &p_{0}\land p_{1} &\neg p_{2} &\neg p_{0} &\neg p_{1} &P &Q &R \\ \hline
0 &0 &0 &0 &1 &1 &1 &1 &1 &1\\
0 &0 &1 &0 &0 &1 &1 &0 &0 &1\\
0 &1 &0 &0 &1 &1 &0 &1 &1 &1\\
0 &1 &1 &0 &0 &1 &0 &0 &0 &1\\
1 &0 &0 &0 &1 &0 &1 &1 &1 &1\\
1 &0 &1 &0 &0 &0 &1 &0 &0 &1\\
1 &1 &0 &1 &1 &0 &0 &1 &1 &0\\
1 &1 &1 &1 &0 &0 &0 &1 &0 &0
\end{tabular}$

Right im guessing that table is correct now, i checked it on WolfFram and it looks right

Now thats sorted i got a few questions to answer, im gonna post them here and my answers and would just like someone to tell me if they sound right or not please

Heres the first one:

Which of the following are true?
• {P} |= Q
• {Q} |= P
• {P,R} |= Q

Is the correct answer {Q} |= P due to the fact when ever Q is true so is P, this shows us that Q follows from P.

Thanks for the help

12. Hmm. Couple of questions.

1. How are you interpreting the symbol |= ? I've usually seen it mean "semantically implies", but I'm not sure how that would work here.
2. What does the symbol {P,R} mean? Does that mean P and R |= Q? Or does it mean P or R |= Q?

13. 1. the symbol means follow from
2. it means and

Cheers,

~Crag

14. Originally Posted by Ackbeet
1. How are you interpreting the symbol |= ? I've usually seen it mean "semantically implies", but I'm not sure how that would work here.
That's what it means here as well.
Originally Posted by Ackbeet
2. What does the symbol {P,R} mean? Does that mean P and R |= Q? Or does it mean P or R |= Q?
It's just emphasizing that what is left of |= is a set of formulas, which is interpreted as a conjunction. Usually the set notation is omitted.

Is the correct answer {Q} |= P due to the fact when ever Q is true so is P, this shows us that Q follows from P.
Yes, though in words this says that P follows from Q, not the other way around. Also, {P,R} |= Q holds.

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