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Discrete Math: Proving something is logically equivalent

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that (p ∧ q) → r and (p → r) ∧ (q → r) are not
    logically equivalent.


    2. Relevant equations
    a → b = [itex]\neg[/itex]a v b



    3. The attempt at a solution
    I'm sorry. I'm completely stumped on how to go about this problem. I'm not asking for the solution since I want to know how to do this instead of just getting the answer. Any help would be appreciated. Thank you.
    Here is what I had just so no one thinks I didn't try.

    (p ∧ q) → r
    => [itex]\neg[/itex] ( p [itex]\wedge[/itex] q ) [itex]\vee[/itex] r
    => ([itex]\neg[/itex]p [itex]\wedge[/itex] [itex]\neg[/itex]q ) [itex]\vee[/itex] r
    => (switched it around) r [itex]\vee[/itex] ([itex]\neg[/itex]p [itex]\wedge[/itex] [itex]\neg[/itex]q )
    => (distributed) (r [itex]\vee[/itex] [itex]\neg[/itex]p ) [itex]\wedge[/itex] ( r v [itex]\neg[/itex] q)
    => ([itex]\neg[/itex]p v r ) [itex]\wedge[/itex] ([itex]\neg[/itex]q v r )
    => (p -> r ) [itex]\wedge[/itex] (q -> r)

    It said disprove but somehow I'm getting that they are L.E.
     
    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    Dick

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    If you want to prove they are not equivalent then just figure out how you can assign true and false values to p, q and r so that the two sides give you different values.
     
  4. Mar 10, 2012 #3
    That makes sense. Here is what I got, please correct me if I'm wrong.

    Let
    p = true
    q = false
    r = false
    then (p ∧ q) → r is true.
    and (p → r) ∧ (q → r) is false.
    Hence it's not L.E.

    Any mistakes?
     
  5. Mar 10, 2012 #4

    Dick

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    Looks ok to me. true → false is false. false → false is true.
     
    Last edited: Mar 10, 2012
  6. Mar 10, 2012 #5

    Office_Shredder

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    As far as the original logic goes...

    Your problem is here.
    not (p and q) = (not p OR not q)
     
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