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Logic Proof

  1. Mar 12, 2005 #1
    1. ~(A v B) It is neither A nor B.
    2. ~(C v D) It is neither C nor D.
    3. E > (C v B) If E, then C or B
    Therefore, ~(~E > A)
    --------------------------------------------
    This is a logic proof I cannot seem to resolve.

    I've tried working backwards from the conclusion.

    http://www.hu.mtu.edu/~tlockha/h2701n10.s02.doc
    Here are rules of replacement and rules of implication.

    I know that ~ (~E > A) is a true statement.
    I know that because I know ~E > A is false
    Because ~E > ~ A (proved below) is true so ~E > A cannot be true.
    A is not true because ~A is true. ~E is also true.

    ~(~E > A) cannot be unpacked because it is a horseshoe.
    I can introduce (~E > A) as an addition (trying to work backwards):

    1. ~A v (~E > A) add on ~A
    2. A > (~E > A) 1, impl
    We don't have a rule that says not P therefore not Q

    3 (~E > A) v ~A 1 comm
    4. ~ (~E >A) > ~A 3 impl

    We don't have a rule that says Q therefore P


    The proof as far as I can get it
    > 1. ~(A v B)
    > 2. ~(C v D)
    > 3. E > (C v B)
    > Therefore ~( ~E > A)
    > 4. ~A and ~B .............1, DeMorgan
    > 5. ~C and ~D ............2. DeMorgan
    > 6. ~C...................... 5. simplification
    > 7. ~B.......................4. simplification
    > 8. ~C and ~B............6,7 conjunction
    > 9. ~(C v B)...............8, DeMorgan
    > 10. ~E.....................3,9, Modus Tolens
    11. ~A.......................4 simpl




    Additional attempts:
    > 12. ~A v ~A.............11 tautology
    > 13. A > ~A..............12. implication
    > 14 ~A v E...............11 addition
    > 15. A> E..................14. impl
    > 16. ~E > ~A.............15. contraposition

    17 ~A v ~E.................11, add
    18 A > ~E...................17 impl
    19 E > ~A...................18, contra
     
    Last edited: Mar 12, 2005
  2. jcsd
  3. Mar 12, 2005 #2

    AKG

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    In your first attempt, you get ~E (line 10) and ~A (line 11). This gives
    12. ~E & ~A (conjunction introduction, 10, 11)
    13. ~(~(~E & ~A)) (double negation, 12)
    14. ~(~E > A) (implication, 13)
     
  4. Mar 12, 2005 #3
    I agree with 12.

    According to Demorgan
    ~E*~A :: ~(E v A)

    Implication
    (P>Q) :: (~P v Q)
    no conjunctions are involved.
     
  5. Mar 12, 2005 #4
    I think you have skipped at least one step.
    Also, can you really operate on things within a parentheses with a negation outside?
    ~(E v A ) becomes ~ (~E >A) with DN and impl
    I don't see that in the rules anywhere.
    Thanks for your help.
     
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