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Math Help - modus ponens

  1. #1
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    modus ponens

    premise is -p

    -r -> p therefore -r, modus ponens

    is this assumption correct?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    this is not using modus ponens, and it's not true anyway. Remember, \neg R \implies P is the same as saying \neg P \implies R. Since \neg P is true, R follows by modus ponens, not \neg R
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  3. #3
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    thanks
    how about
    premise -q
    -p or q, therefore p, disjunctive syllogism

    if q is false then p must be true, otherwise the whole condition is false??

    or is it,
    if -q(i.e. q is false)
    then -p must be true for -p or q.

    ??
    so
    premise -q
    -p or q, therefore -p, disjunctive syllogism

    ??
    Last edited by dunsta; June 30th 2010 at 09:33 PM. Reason: more thinking/confusion
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dunsta View Post
    so
    premise -q
    -p or q, therefore -p, disjunctive syllogism

    ??
    correct
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  5. #5
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    thanks for the help

    the question I am working on is:

    -r --> p, (-p) or q, -s -->(-p)and (-r), (-p) and r --> (-s) or t, -q, therefore t.

    So far I have
    1: -p, -p or q therefore -p conjunctive syllogism
    2: -p, -r --> p therefore r modus ponens
    3: -p, r, therefore -p and r conjunctive addition
    (the next 2 lines I am unsure of)
    4: -p and r, -s --> (-p) and(-r) /*-p is true from 1. but r is true from 2, so -r is not true. therefore (-p) and(-r) is not true so - (-p) and(-r) = -(p or r) demorgans.??? I don't know where to go with this? */
    therefore s, because -s --> (-p) and(-r) is false.
    5: (-p) and r (from 3), -s is false, (-s) or t therefore t disjunctive syllogism

    am I close to showing the argument is valid??
    thanks for any input and help
    so -s --> (-p) and(-r)
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