Show that argument form with premises

(p ^ t) --> (r v s),

q --> ( u ^ t),

u --> p,

negation ~s

end up with conclusion q --> r

by using rules of inference

Thank you

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- Sep 26th 2008, 09:37 AM #1

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- Sep 26th 2008, 09:46 AM #2

- Sep 26th 2008, 09:55 AM #3

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i just want to know how would you form these arguments, and end up with

q --> r

I know how to do it with the basic ones with 1 principle say " p --> q, q, end up with p"

However, I am little confused with 2 or more principles within an arguments say " (p ^ t) --> ( r v s)"

How do i simplify something like this?

- Sep 26th 2008, 10:16 AM #4
you have to do a simplification.

you know how to set up these problems right?

$\displaystyle \begin{array}{lll} 1. & (P \wedge T) \implies (R \vee S) & \\

2. & Q \implies (U \wedge T) & \\

3. & U \implies P & \\

4. & \neg S & \backslash \therefore Q \implies R \\

5. & Q & \text{assumption} \\

6. & U \wedge T & \text{2,5 M.P.} \end{array}$

.

.

.

.

now how do we move on?

- Sep 26th 2008, 10:38 AM #5

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- Sep 26th 2008, 11:48 AM #6

- Sep 26th 2008, 03:52 PM #7

- Sep 26th 2008, 04:21 PM #8

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- Sep 26th 2008, 04:24 PM #9
you can think of it in terms of compound statements. it is like the rule for Modus Ponens except you have (p ^ t) for p and (r v s) for q. get to (p ^ t) which will imply (r v s) by M.P. then try to get to r from (r v s). once you get r, you are done

but we are no where near that as yet. for now, we know 6 statements. what can you do with them to get to (p ^ t) ?

- Sep 26th 2008, 05:17 PM #10

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- Sep 26th 2008, 05:26 PM #11

- Sep 26th 2008, 06:23 PM #12

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- Sep 26th 2008, 06:26 PM #13
why assume (r v s)?

first of all, we want to get to r, going from (r v s) to (p ^ t) makes no sense, you are going in the wrong direction. if you have r, stick with it. but it is not enough to assume you have r. you must show that you have r by assuming q. remember, we want to show q => r, so we want to start at q and get r. that is your objective. stay focused. secondly, logically you are going in the wrong direction. we have (p ^ t) => (r v s) NOT (r v s) => (p ^ t), so your line 8 does not follow by M.P. it violates the implication

- Sep 26th 2008, 06:54 PM #14

- Sep 29th 2008, 12:18 PM #15
$\displaystyle \begin{array}{lll} 1. & (P \wedge T) \implies (R \vee S) & \\

2. & Q \implies (U \wedge T) & \\

3. & U \implies P & \\

4. & \neg S & \backslash \therefore Q \implies R \\

5. & Q & \text{assumption} \\

6. & U \wedge T & \text{2,5 M.P.} \\

7. & U & \text{6, Simp.} \end{array}$

$\displaystyle

\begin{array}{lll}

8. & T & \text{6, Comm., Simp.} \\

9. & P & \text{3,7 M.P.} \\

10. & P \wedge T & \text{8,9 Conj.} \\

11. & R \vee S & \text{1, 10 M.P.} \\

12. & R & \text{4,11 Comm., D.S.} \\

\hline 13. & Q \implies R &

\end{array}$