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Prove: ~p ^ (~q ^ r) v (q ^ r) v (p ^ r) ≡ r

  1. Sep 22, 2012 #1


    show that :
    ~p ^ (~q ^ r) v (q ^ r) v (p ^ r) ≡ r

    when i made the truth table of course they were equivalent..

    but i don't know what i'm doing wrong ! the end is always T "true" when i try to prove it logically...


    Please help !


     
  2. jcsd
  3. Sep 22, 2012 #2

    Mark44

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    There's a lot of simplification that you can do. For example, how can you rewrite (~q ^ r) v (q ^ r)?
     
  4. Sep 22, 2012 #3
    What exactly do you mean by 'the end is always T "true" when i try to prove it logically...'?

    Do you mean that the LHS becomes True once you've simplified it? If so, show us step by step how you got to that and someone will spot the error in your working.
     
  5. Sep 22, 2012 #4

    micromass

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    Indeed. Please post your work or this thread will be deleted.
     
  6. Sep 23, 2012 #5

    Thank you all , but i have solved it ^^

    ~p ^ (~q ^ r) v (q ^ r) v (p ^ r) ≡ r


    1- rearrange it:

    [~p ^ (p ^ r) ] v (~q ^ r) v (q ^ r) ≡ r

    [ (~p ^ p) ^ (~p ^ r) ] v (~q ^ r) v (q ^ r) ≡ r

    [F ^ (~p ^ r) ] v (~q ^ r) v (q ^ r) ≡ r

    F v (~q ^ r) v (q ^ r) ≡ r

    (~q ^ r) v (q ^ r) ≡ r

    [ (~q ^ r) v q) ] ^ [ (~q ^ r) v r) ] ≡ r

    (~q v q ) ^ ( r v q ) ^ ( ~q v r ) ^ ( r v r ) ≡ r

    Put braces :

    (~q v q ) ^ [ ( r v q ) ^ ( ~q v r ) ] ^ ( r v r ) ≡ r

    * [ ( r v q ) ^ ( ~q v r ) ] ==> r v (q ^ ~q)

    * ( r v r ) => r

    T ^ [ r v (q ^ ~q) ] ^ r ≡ r

    * (q ^ ~q) => F

    T ^ r ^ r ≡ r

    *( r ^ r) => r

    T ^ r ≡ r

    ^^



    Al7amdulillah <3
     
  7. Sep 23, 2012 #6
    I think you got lucky! :smile:

    The first line in your "rearrangement" seems invalid to me but almost trivially causes your LHS to be equivalent to r.

    The original LHS is:
    ~p ^ (~q ^ r) v (q ^ r) v (p ^ r)

    and your rearrangement is:
    [~p ^ (p ^ r)] v (~q ^ r) v (q ^ r)

    As (q ^ r) is common to both of these expressions, you are effectively saying that:
    ~p ^ (~q ^ r) v (p ^ r) ≡ [~p ^ (p ^ r)] v (~q ^ r)

    which is NOT the case.

    I'm actually having a hard time trying to object to your reasoning as each step is logically correct and equivalent to the previous one; it's just your initial rearrangement where I can't understand how you got to it!

    It looks like you saw A ^ B v C v D and thought you could rearrange it to be [A ^ D] v B v C, which in general is not true. :smile:

    (I may be making a complete fool of myself here if I've missed something obvious!)

    The original LHS can actually be simplified to r in about 3 steps as Mark was hinting at earlier.
     
  8. Sep 24, 2012 #7

    HallsofIvy

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    The first part of your statement, "~p" says that p is false. That means that "p^r" is false so that statement reduces to "(~q^r)v(q^r)". If q is false, "q^r" is false so we must have "~q^r" and so r is true. If q is true, "~q^r" is false so we must have "q^r" and so r is true. In any case, r is true.
     
  9. Sep 24, 2012 #8
    Maybe I'm being stupid, but this post makes no sense to me.

    r isn't necessarily true or false. It's not a question of finding whether r is true or false; it's about proving that the LHS is equivalent to r.
     
  10. Sep 26, 2012 #9
    Anyone?
     
  11. Oct 1, 2012 #10
    A pity , you fail to fathom still after such excellent hints. Do you know what <=> in Proposition means ? If I say , that p<=>q , then p<=>q = (p=>q)^(q=>p) = (~p+q)^(~q+p).

    Use this theorem in your question and prove !! Well truth is easiest method , but if you are not satisfied then use conditional elimination , as I stated. Lets see what you get.

    Do you know about tautology , contrapositive , converse , etc.. in propositional logic ?
     
  12. Oct 1, 2012 #11

    vela

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    You're right. Nuha22's solution is wrong for the reason you noted.
     
  13. Oct 1, 2012 #12
    I think you may be confusing me with the OP (Nuha22), who hasn't responded for over a week, possibly still believing that he's "solved" it.

    Thanks. :smile:

    But I'm still not sure what Ivy was getting at...
     
  14. Oct 2, 2012 #13
    Yeah !! I mistaken you for OP , who was Nuha22 !! Silly me !! I apologize for that though. :smile:

    And yes , you're right. Ivy's hint was slightly confusing and little hard to fathom. But he was right. And that's what I stated , in my previous post. Ivy was using "conditional elimination." If you prove the statement to be a tautology , not a contradiction , then indeed , the inference from the premises is derived.
     
  15. Oct 3, 2012 #14
    I'm quite happy to be corrected, but I think Ivy is female.

    I still don't understand what she or you mean. I don't mean to be any mathematical genius.

    I'm sure I can solve the OP's problem, but Ivy's reasoning is a bit confusing to me.
     
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