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Pv (Qv~P)

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that Pv(Qv~P) is always true whatever the values of p and q.


    Attempt

    Pv(Qv~P)
    (PvQ) v (Pv~P)
    (PvQ) v T
     
  2. jcsd
  3. Oct 2, 2012 #2

    Mark44

    Staff: Mentor

    Re: Pv(Qv~P)

    The step above is incorrect. The logical disjunction operator (V) is associative, with P V (Q V R) <==> (P V Q) V R
     
  4. Oct 2, 2012 #3
    Re: Pv(Qv~P)

    So.. it should it be P V (Q V ~P) = (~P V P) V Q?
     
  5. Oct 2, 2012 #4
    Re: Pv(Qv~P)

    So is it T V Q?
     
  6. Oct 2, 2012 #5

    Mark44

    Staff: Mentor

    Re: Pv(Qv~P)

    Yes. Can you give reasons for each step? (There are a couple.)

    That's not the final, simplified expression. What does that simplify to?
     
  7. Oct 2, 2012 #6
    Re: Pv(Qv~P)

    Oh Umm because (~P V P) = T

    hence T V Q

    I think there is an indentity for P = T.. so p V q?
     
  8. Oct 2, 2012 #7

    Mark44

    Staff: Mentor

    Re: Pv(Qv~P)

    You're not done.

    The part in red should have been a clue as to what you should conclude.


    T V <whatever> ⇔ T
     
    Last edited by a moderator: Oct 3, 2012
  9. Oct 3, 2012 #8
    Re: Pv(Qv~P)

    I don't see what's in red. I'm sorry :(
     
  10. Oct 3, 2012 #9
    Re: Pv(Qv~P)

    but isn't it P V Q = T ? Since it's "OR" it's either one or the other or both?
     
  11. Oct 3, 2012 #10

    vela

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    Staff Emeritus
    Science Advisor
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    Re: Pv(Qv~P)

    I fixed the tag in Mark's post, so you can see now what he intended.

    No, look at the truth table for ⋁. If both P and Q are false, then P⋁Q is false.
     
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