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Thread: Isn't (A or B) and (not A or C) the same as (B or C)?

  1. #1
    Feb 2010
    New Jersey

    Isn't (A or B) and (not A or C) the same as (B or C)?

    When doing symbolic logic deductive proof, I have the following two statements
    $\displaystyle A \lor B$
    $\displaystyle \neg A \lor C$.

    Since either A is true, or A is false, it seems to me the above two statements are similar to
    $\displaystyle B \lor C$.

    Is that right? It seems so to me. But I can't figure out how to prove it, and when I make a truth table out "((A v B) & (~A v C)) <=> (B v C)" (using Truth Table Constructor), it doesn't show that they're equal. What gives? Thanks

    Edit: OK, note that what I said is not equivalent to the truth table I created. The correct statement would be "((A v B) & (~A v C)) => (B v C)" (the major operator is one way implication, not two way). The truth table for that comes up entirely true. So the title of this post is not true, but it would be if I replaced "the same as" with "implies that".
    Last edited by MSUMathStdnt; Jul 28th 2012 at 05:41 PM.
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