1. ## Truth table problem

Is the following valid or invalid? (show by using a truth table).

If Rick gains employment, then Rick will earn a paycheck. p->q
If Rick earns a paycheck, then Rick will be excited. q->r

Rick gains employment. p

____________________________
Rick will be excited. r

what is my final statement? Which is correct before completing my truth table?
[p^(p->q)]^(q->r) or is it [p^(p->q)]^r

2. Originally Posted by CPR
Is the following valid or invalid? (show by using a truth table).

If Rick gains employment, then Rick will earn a paycheck. p->q
If Rick earns a paycheck, then Rick will be excited. q->r

Rick gains employment. p

____________________________
Rick will be excited. r

what is my final statement? Which is correct before completing my truth table?
[p^(p->q)]^(q->r) or is it [p^(p->q)]^r
P => R

you want to know if gaining employment makes Rick excited in every instance

3. Hello, CPR!

Is the following valid or invalid? (show by using a truth table).

If Rick gains employment, then Rick will earn a paycheck. . $p \to q$
If Rick earns a paycheck, then Rick will be excited. . $q \to r$
Rick gains employment. . $p$
-----------------------------------------------
Therefore: Rick will be excited. . $r$

The argument is: . $\bigg[(p \to q) \wedge (q \to r) \wedge p\bigg] \to r$

(The argument is valid; the final column should be eight T's.)

4. ## Thanks!

Thanks for the help. I got it. I was on the right page. Perhaps you can help me with my post under Urgent Help. It is an Induction problem.
It is 1+2+3....(4n-2)= n(3n-1)

The basic step is n=1. 4(1)-2 = 1(3(1)-1)
Induction step is:
1+2+3....(4k-2) = k(3k-1)

1+2+3....(4k-2)+(k+1) = k(3k-1) = (K+1)

5. Originally Posted by CPR
Thanks for the help. I got it. I was on the right page. Perhaps you can help me with my post under Urgent Help. It is an Induction problem.
It is 1+2+3....(4n-2)= n(3n-1)

The basic step is n=1. 4(1)-2 = 1(3(1)-1)
Induction step is:
1+2+3....(4k-2) = k(3k-1)

1+2+3....(4k-2)+(k+1) = k(3k-1) = (K+1)

I think your k+1 is in the wrong place...if i'm correct, you would assume that
1+2+3....(4k-2) = k(3k-1) for all n >= 1

Then you want to prove:

1+2+3....(4k-1) = k(3k)

Can someone confirm i'm correct?

6. Instead of the last equal sign I meant to type =k(3k-1)+ (k+1)

Yes that is what I got on my first attempt but where do I go from there.
So after =(3k-K) + (k+1)
= 3k +1
now what? I need your help!

7. Originally Posted by CPR
Instead of the last equal sign I meant to type =k(3k-1)+ (k+1)

Yes that is what I got on my first attempt but where do I go from there.
So after =(3k-K) + (k+1)
= 3k +1
now what? I need your help!
you want to prove:

1+2+3....(4k-1) = (k+1)(3k)
which equals
1+2+3....(4k-1) = (3k^2 + 3k)

So now rather than try to simplify it any further (like you did above), you need to prove that (4k-1) = (3k^2+3k)...so start from the left, and try to get the right. Don't start with what you want to prove..You will need to suppose that (4k-2) = k(3k-1) (This is your IH) and use it to prove k+1, so in your algebra, you want to get 4k-2, so you can say "I know that 4k-2 = k(3k-1) and therefore...blah blah blah"

Use that and do the algebra to get (4k-1) = (3k^2+3k), don't forget about your IH which you'll need to complete the proof.

8. okay, I'm trying but its not working. I'll continue to try for awhile, but Im not getting it.

9. Originally Posted by CPR
okay, I'm trying but its not working. I'll continue to try for awhile, but Im not getting it.
Try re-writing it using the summation notation, maybe that will help you understand it better?

10. What is summation notations?

11. Originally Posted by CPR
What is summation notations?
writing a sum using the $\sum$ symbol

i need to constuct a truth table...but i absolutely cannot remember how!

(pv~r)^~(qvr)

13. Originally Posted by malia
i need to constuct a truth table...but i absolutely cannot remember how!

(pv~r)^~(qvr)

-Dan

14. Hello, malia!

I can't give you a review on truth tables,
. . but this is what you should have . . .

$\begin{array}{c|c|c|c|c|c|c|c|c|c|c}
p & q & r & (p & \vee & \sim r) & \wedge & \sim & (q & \vee & r) \\ \hline
T & T & T & T & T & F & {\color{blue}F} & F & T & T & T \\
T & T & F & T & T & T & {\color{blue}F} & F & T & T & F \\
T & F & T & T & T & F & {\color{blue}F} & F & F & T & T \\
T & F & F & T & T & T & {\color{blue}T} & T & F & F & F \end{array}$

. $\begin{array}{c|c|c|c|c|c|c|c|c|c|c}
F & T & T & F & F & F & {\color{blue}F} & F & T & T & T \\
F & T & F & F\, & T & \;\;T\:\; & {\color{blue}F} & F & T \,& T \,& F \\
F & F & T & F & F & F & {\color{blue}F} & F & F & T & T \\
F & F & F & F & T & T & {\color{blue}T} & T & F & F & F \\
& & & ^1 & ^2 & ^1 & ^4 & ^3 & ^1 & ^2 & ^1
\end{array}$

15. Originally Posted by Soroban
Hello, malia!

I can't give you a review on truth tables,
. . but this is what you should have . . .

$\begin{array}{c|c|c|c|c|c|c|c|c|c|c}
p & q & r & (p & \vee & \sim r) & \wedge & \sim & (q & \vee & r) \\ \hline
T & T & T & T & T & F & {\color{blue}F} & F & T & T & T \\
T & T & F & T & T & T & {\color{blue}F} & F & T & T & F \\
T & F & T & T & T & F & {\color{blue}F} & F & F & T & T \\
T & F & F & T & T & T & {\color{blue}T} & T & F & F & F \end{array}$

. $\begin{array}{c|c|c|c|c|c|c|c|c|c|c}
F & T & T & F & F & F & {\color{blue}F} & F & T & T & T \\
F & T & F & F\, & T & \;\;T\:\; & {\color{blue}F} & F & T \,& T \,& F \\
F & F & T & F & F & F & {\color{blue}F} & F & F & T & T \\
F & F & F & F & T & T & {\color{blue}T} & T & F & F & F \\
& & & ^1 & ^2 & ^1 & ^4 & ^3 & ^1 & ^2 & ^1
\end{array}$

nice use of LaTeX, that must have taken you forever!