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Need Help

  1. Jan 13, 2008 #1
    Perchance someone here has some background in Philosophy, specifically symbolic logic? For the life of me I can't figure out the following:

    A v D
    (~B ^ ~C) = D
    B => ~(C=> A)
    ~B

    What I have so far is this:

    1| A v D ............................P
    2| (~B ^ ~C) = D .............P
    3| B => ~(C =>A) ............P
    4| |B ................................A
    5| | |C .............................A
    r| | |?
    s| | |?
    t| | |?
    u| | |?
    v| | |A ..............................1, ?-?, ?-?, vE
    w| | C=>A ........................5-v, =>I
    x| | B => ~(C =>A) ..........2, R
    y| | ~(C =>A) ..................w, x, =>E
    z| ~B ...............................4, w, y, ~I

    What I can't figure out is how to get the disjunction elimination for A in the second sub-derivation. Anyone have a clue?

    [​IMG]
     
  2. jcsd
  3. Jan 13, 2008 #2
    I'm not sure about what you're doing as I have limited experience with this kind of thing, but as far as I can tell, if you assume B you get a contradiction:

    B => C ^ ~A (equivalent to proposition 3)
    ~A => ~B ^ ~C (props 1 and 2)
    So, you get B => C ^ ~C (or B => ~B...either way).

    Or is that way off?
     
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