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Homework Help: Tautology or contradiction

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Wrie out the truth table for the statement form P -> ~(Q ^ ~P). Is it a tautology or a contradiction?

    2. Relevant equations



    3. The attempt at a solution

    First off is it true to say that P -> ~(Q ^ ~P) and P -> (~Q v P) are equal.

    P | Q | ~P | ~Q | P -> (~Q v P)
    T T F F F
    T F F T F
    F T T F F
    F F T T F

    I believe it is a contradiction. Is this correct
     
  2. jcsd
  3. Aug 30, 2010 #2

    Mark44

    Staff: Mentor

    Technicality: ~(Q ^ ~P) is equivalent to (not equal to) (~Q v P), so P ==> ~(Q ^ ~P) and P ==> (~Q v P) are also equivalent.

    If I were doing the truth table I would have columns for P and Q (which you do), plus one column for ~Q v P.
     
  4. Aug 30, 2010 #3
    P | Q | (~Q v P) | P -> (~Q v P)
    T__T_____F_________F
    T__F_____T_________T
    F__T_____F__________F
    F__F_____F__________F

    is the second row correct. If P, then not Q or P. It sounds like a contradiction but P is true ~QvP.
     
  5. Aug 30, 2010 #4
    how come F->F is false?
     
  6. Aug 30, 2010 #5
    and there's something wrong in the (~Q v P) column, in the first row.

    FvT is false?
     
  7. Aug 30, 2010 #6
    not followings u
     
  8. Aug 30, 2010 #7
    hmmm

    P--->(~QvP)
    T_()___T
    T_()___T
    F_()___F
    F_()___F

    fill in the blank ()

    ;P
     
  9. Aug 30, 2010 #8
    for ~QvP, it is only true when it Q is false and P is true.
     
  10. Aug 30, 2010 #9

    Mark44

    Staff: Mentor

    No, that's not right. ~QvP is true in all cases other than when P is false and Q is true.
     
  11. Aug 30, 2010 #10
    "v" is or. it's only true when either ~Q or P is true

    your statement here "for ~QvP, it is only true when it Q is false and P is true."

    suppose to be "for ~Q[tex]\wedge[/tex]P, it is only true when it Q is false and P is true."
     
  12. Aug 30, 2010 #11
    P | Q | (~Q v P)
    T__T___T
    T__F___F
    F__T___T
    F__F___T

    ok i got this part. but when u pu P->(~Q v P) you get True only when (~Q v P) is True and when P is True.

    P | Q | (~Q v P) | P->(~Q v P)
    T__T___T___________T
    T__F___F___________F
    F__T___T___________F
    F__F___T___________F
     
  13. Aug 30, 2010 #12
    in second row is wrong

    and also whenever P is false, it must be true too

    because if the premises is already false, either the consequence is false or true, it doesn't matter. The statement must be true
     
  14. Aug 30, 2010 #13
    P | Q | (~Q v P) | P->(~Q v P)
    T__T___T___________T
    T__F___T___________T
    F__T___F___________T
    F__F___T___________T
     
  15. Aug 30, 2010 #14
    yea, that's correct, i hope you understand each of them. and btw, you can also proof it algebraically. using all those law, assiosiative, identity, commut, distributive and so on,
     
  16. Aug 30, 2010 #15
    im having a hard time grasping the abstractness of this. do you know of any good sites i can use for added material.
     
  17. Aug 30, 2010 #16
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