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Math Help - please help

  1. #1
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    please help

    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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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by justin016 View Post
    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
    do you know the rules of inference?

    assume q and show that it implies r. what rules will you need?
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  3. #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?
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    Quote Originally Posted by justin016 View Post
    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?
    you have to do a simplification.

    you know how to set up these problems right?

    \begin{array}{lll} 1. & (P \wedge T) \implies (R \vee S) & \\<br />
2. & Q \implies (U \wedge T) & \\<br />
3. & U \implies P & \\<br />
4. & \neg S & \backslash \therefore Q \implies R \\<br />
5. & Q & \text{assumption} \\<br />
6. & U \wedge T & \text{2,5 M.P.} \end{array}
    .
    .
    .
    .

    now how do we move on?
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  5. #5
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    how did you get from

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

    to

    q --> ( u ^ t)

    i know this is the hypothetical syllogism, but it doesnt make any sense that

    ( r v s) turn into q
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  6. #6
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    Quote Originally Posted by justin016 View Post
    how did you get from
    i know this is the hypothetical syllogism, but it doesnt make any sense that ( r v s) turn into q
    Where in the world did you come up with that?
    He used statement 2 and statement 5 with modus ponens to get statement 6.
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    Quote Originally Posted by justin016 View Post
    how did you get from

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

    to

    q --> ( u ^ t)

    i know this is the hypothetical syllogism, but it doesnt make any sense that

    ( r v s) turn into q
    evidently you still don't have a good grasp of the rules of inference. please review them. then we can talk about how to proceed.

    be aware, your objective is to get to R
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  8. #8
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    Pardon my ignorant of this stuff. This is my first time doing these

    Like i said, I understand the basic function of inference rules, but

    something like this (p ^ t) --> ( r v s) confuses me . What should i do with 1?
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  9. #9
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    Quote Originally Posted by justin016 View Post
    Pardon my ignorant of this stuff. This is my first time doing these

    Like i said, I understand the basic function of inference rules, but

    something like this (p ^ t) --> ( r v s) confuses me . What should i do with 1?
    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) ?
    Last edited by Jhevon; September 29th 2008 at 12:08 PM. Reason: wrote Pollens instead of Ponens :D
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  10. #10
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    oh okay, thank you that cleared up a lot. All i have to do is to get ( r v s ) be itself by using Modus ponens on ( p ^ t), then use disjunctive syllogism of

    ( r v s)
    --s
    to get r
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  11. #11
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    Quote Originally Posted by justin016 View Post
    oh okay, thank you that cleared up a lot. All i have to do is to get ( r v s ) be itself by using Modus ponens on ( p ^ t), then use disjunctive syllogism of

    ( r v s)
    --s
    to get r
    um, looking at the 6 lines we have so far, we don't have (p ^ t)

    please write out the steps of your argument clearly using the format i showed you

    say what rule of inference you used and with what lines you are using them
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  12. #12
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    why do you want to get ( p ^ t)?

    Can i use M.P implying ( r v s) to get (p ^ t), but isnt that illegal?
    Last edited by justin016; September 26th 2008 at 06:34 PM.
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  13. #13
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    Quote Originally Posted by justin016 View Post
    7. ( r v s) assumption
    8. ( p ^ t) 1,7 M.P

    why do you want to get ( p ^ t)?
    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
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  14. #14
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    Quote Originally Posted by justin016 View Post
    why do you want to get ( p ^ t)?
    because i want to get to r. as i said, getting r is our objective. if i can get to (p ^ t) i can get to r since (p ^ t) => (r v s)

    Can i use M.P implying ( r v s) to get (p ^ t), but isnt that illegal?
    i told you you couldn't do that
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  15. #15
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    \begin{array}{lll} 1. & (P \wedge T) \implies (R \vee S) & \\<br />
2. & Q \implies (U \wedge T) & \\<br />
3. & U \implies P & \\<br />
4. & \neg S & \backslash \therefore Q \implies R \\<br />
5. & Q & \text{assumption} \\<br />
6. & U \wedge T & \text{2,5 M.P.} \\<br />
7. & U & \text{6, Simp.} \end{array}
    <br />
\begin{array}{lll}<br />
8. & T & \text{6, Comm., Simp.} \\<br />
9. & P & \text{3,7 M.P.} \\<br />
10. & P \wedge T & \text{8,9 Conj.} \\<br />
11. & R \vee S & \text{1, 10 M.P.} \\<br />
12. & R & \text{4,11 Comm., D.S.} \\<br />
\hline 13. & Q \implies R & <br /> <br />
\end{array}
    Last edited by Jhevon; September 29th 2008 at 03:15 PM.
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