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Homework Help: Truth table and proving a tautology

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to prove that this statement is a tautology (with truth table):

    (p -> ((p OR q) AND ~q)) = (~p OR ~q) AND p


    2. Relevant equations



    3. The attempt at a solution

    Code (Text):

    p | q | ~p | ~q | p OR q | (p OR q) AND ~q | p -> ((p OR q) AND ~q) | (~p OR ~q) | (~p OR ~q) AND p
    ----------------------------------------------------------------------------------------------------
    1   1    0    0      1            0                      0                 0               0
    1   0    0    1      1            1                      1                 1               1
    0   1    1    0      1            0                      1                 1               0
    0   0    1    1      0            0                      1                 1               0


     
    But here cleary is that it is not tautology, because left side of equation gets 0 1 1 1 and right side 0 1 0 0

    So where did i make mistake, because i cant see it
     
  2. jcsd
  3. May 11, 2010 #2

    D H

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    Staff Emeritus
    Science Advisor

    Are you sure you have the parentheses right in the expression that is to be proven?
     
  4. May 11, 2010 #3
    yes, it is just like that in the excercise paper
     
  5. May 11, 2010 #4
    so do you think that I have done it right and the excercise is badly worded?
     
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