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Math Help - Showing equality via a simple calculation

  1. #1
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    Showing equality via a simple calculation

    I'm studying for an introductory course on logic, I'm at the point where I've to show the equality of certain formulas using a calculation with some basic standard equivalences. Like Inversion, True/False elimination, DeMorgan, distributivity et cetera. I'ce come across a formula that's intuitively very clear, I just don't understand how to work it from A to B.

    I'm asked to provide a calculation showing that: P \wedge (P \lor Q) = P. Intuitively I can see this to be true, drawing the truthtables confirms this, but how can I show this via a simple calculation?

    I know I can rewrite this as follows: by distribution I get: (P \wedge P) \lor  (P \wedge Q) = P, from which by idempotence follows that: P \lor  (P \wedge Q) = P. Which actually leads me to exercise b . To show that is equivalent to P.

    The big question is: how do I eliminate the Q? What rule can I apply here to show the equivalence to P?
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  2. #2
    MHF Contributor
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    Re: Showing equality via a simple calculation

    P\lor(P\land Q)=(P\land 1)\lor(P\land Q)=P\land(1\lor Q)=P\land1=P.
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  3. #3
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    Re: Showing equality via a simple calculation

    Ah I see now, thanks! I haven't seen distributive and true/false elimination being used like that
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