Truth table problem

• Feb 26th 2008, 11:04 AM
CPR
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
• Feb 26th 2008, 11:17 AM
Jhevon
Quote:

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
• Feb 26th 2008, 01:54 PM
Soroban
Hello, CPR!

Quote:

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

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

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

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

• Feb 26th 2008, 03:56 PM
CPR
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)

• Feb 26th 2008, 04:54 PM
shawn
Quote:

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?
• Feb 26th 2008, 05:05 PM
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!
• Feb 26th 2008, 05:13 PM
shawn
Quote:

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.
• Feb 26th 2008, 05:42 PM
CPR
okay, I'm trying but its not working. I'll continue to try for awhile, but Im not getting it.
• Feb 26th 2008, 06:16 PM
shawn
Quote:

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?
• Feb 27th 2008, 09:50 AM
CPR
What is summation notations?
• Feb 27th 2008, 01:46 PM
Jhevon
Quote:

Originally Posted by CPR
What is summation notations?

writing a sum using the $\displaystyle \sum$ symbol
• Mar 23rd 2008, 03:53 PM
malia
i need to constuct a truth table...but i absolutely cannot remember how!

(pv~r)^~(qvr)
• Mar 23rd 2008, 05:41 PM
topsquark
Quote:

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

(pv~r)^~(qvr)

-Dan
• Mar 24th 2008, 08:59 AM
Soroban
Hello, malia!

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

$\displaystyle \begin{array}{c|c|c|c|c|c|c|c|c|c|c} p & q & r & (p & \vee & \sim r) & \wedge & \sim & (q & \vee & r) \\ \hline T & T & T & T & T & F & {\color{blue}F} & F & T & T & T \\ T & T & F & T & T & T & {\color{blue}F} & F & T & T & F \\ T & F & T & T & T & F & {\color{blue}F} & F & F & T & T \\ T & F & F & T & T & T & {\color{blue}T} & T & F & F & F \end{array}$
.$\displaystyle \begin{array}{c|c|c|c|c|c|c|c|c|c|c} F & T & T & F & F & F & {\color{blue}F} & F & T & T & T \\ F & T & F & F\, & T & \;\;T\:\; & {\color{blue}F} & F & T \,& T \,& F \\ F & F & T & F & F & F & {\color{blue}F} & F & F & T & T \\ F & F & F & F & T & T & {\color{blue}T} & T & F & F & F \\ & & & ^1 & ^2 & ^1 & ^4 & ^3 & ^1 & ^2 & ^1 \end{array}$

• Mar 24th 2008, 09:02 AM
Jhevon
Quote:

Originally Posted by Soroban
Hello, malia!

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

$\displaystyle \begin{array}{c|c|c|c|c|c|c|c|c|c|c} p & q & r & (p & \vee & \sim r) & \wedge & \sim & (q & \vee & r) \\ \hline T & T & T & T & T & F & {\color{blue}F} & F & T & T & T \\ T & T & F & T & T & T & {\color{blue}F} & F & T & T & F \\ T & F & T & T & T & F & {\color{blue}F} & F & F & T & T \\ T & F & F & T & T & T & {\color{blue}T} & T & F & F & F \end{array}$
.$\displaystyle \begin{array}{c|c|c|c|c|c|c|c|c|c|c} F & T & T & F & F & F & {\color{blue}F} & F & T & T & T \\ F & T & F & F\, & T & \;\;T\:\; & {\color{blue}F} & F & T \,& T \,& F \\ F & F & T & F & F & F & {\color{blue}F} & F & F & T & T \\ F & F & F & F & T & T & {\color{blue}T} & T & F & F & F \\ & & & ^1 & ^2 & ^1 & ^4 & ^3 & ^1 & ^2 & ^1 \end{array}$

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