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Logic again

  1. Jun 18, 2006 #1
    Is there a way to formally prove:
    A -> (A -> B) equivalent to (A -> B)

    do I just assume A and then use modus ponenes (by deduction method)? It just looks a bit odd asserting a premise out of the blue....
    I checked the truth table for those, it seems to be true, however, truth table is not going to work as the answer, I need to do this formally.
    Thanks in advance.

    ps: sorry for trashing forum with my logic threads... from now on, i'll just put my questions (if i have any) here :redface:
  2. jcsd
  3. Jun 18, 2006 #2


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    More than just "seems"! Any logical equivalence provable via truth table can be deduced with the rules of inference you're using, and vice versa.
  4. Jun 18, 2006 #3


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    Proofs by exhaustion (like truth tables) are perfectly rigorous.
  5. Jun 18, 2006 #4
    well that is the thing.... I check my "inference" steps with truth tables, however I need a rule or transformation of some sort and I can't see anything unless I assert the premise (A).
    Ok, so here's the full thing:
    prove Y -> W
    given these premises:
    1. Y -> Z'
    2. X' -> Y
    3. Y -> (X -> W)
    4. Y -> Z
    -----my proof----
    5. X' -> Z' (2,1, hypoth. syllogism)
    6. Z -> X (5, contraposition)
    7. Y -> X (4,6, hypothet. syllogism)
    8. (X' v W)' -> Y' (3, contraposition)
    9. (X ^ W') -> Y' (DeMorgan's)
    10. X -> (W' ->Y') (9, exportation)
    11. X -> (Y ->W) (contraposition)

    12. Y -> (Y -> W) (7,11, hypothet. syllogism)
    13. Y (?)
    14. Y -> W (?)

    12-14 is where I need something or can I just assert Y?
    Last edited: Jun 18, 2006
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