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Symbolic Logic (quick easy question)

  1. Apr 5, 2010 #1
    I have ~(P&Q) and I want to get ~P v ~Q

    (& means and)
    (v means or)

    Am I right in thinking that I can:
    1.~(P&Q) Main assumption
    Sub derivation
    2..Assume P&Q
    3.Reiterate ~(P&Q)
    End sub derivation
    4.Then do 1-2, ~I (negation elimination) to get ~P
    5.then do vI (disjunction introduction) (to number 4) to get ~P v ~Q

    I'm confused on how exactly to extract ~P from the negation ~(P&Q). My text book completely ignores it.

    Is my method correct?


    (Mods I didn't know where to put this. I'm thinking it isn't here)
     
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2
    I don't claim to me very knowledgble in this area, and if you are looking for some sort of rigorous derivation I cannot help, but you can get from ~(P&Q) to ~Pv~Q by DeMorgan's law's which state exactly that, you can also construct a truth table for the two statements and show their equivelency. If P="team 1 won the game" and Q="team 2 won the game" than ~(P&Q) means "both team 1 and 2 didn't win" which conveys the same meaning as "Either team 1 lost or team 2"
     
  4. Apr 5, 2010 #3
    Hey thanks for the reply.

    I can't use truth tables it has to be in derivation form.

    I get the demorgan law I just need to figure out how to turn that into a negation elimination or introduction.
     
  5. Apr 5, 2010 #4
    Looks contradictory to me. You should be able to derive anything you want from that, but I don't know if it'll earn too many points :smile:. I'm rusty though, and I don't have my logic book handy on my lunch break.
     
  6. Apr 5, 2010 #5
    Hey thanks for the reply.

    oh right so can I derive ~P from the steps below? Would it go in the main scope line? I'm thinking now maybe you can't because step 1 is a negation and step 2 is a conjunction.

    1.~(P&Q) Main assumption
    Sub derivation
    2..Assume P&Q

    Which logic textbook are you using? When you get home can you check if this is possible lol.
     
  7. Apr 5, 2010 #6
    What I meant was that your two assumptions look contradictory. You can derive a contradiction to show that an assumption is impossible, but if you flat out assume something that is a contradiction of another assumption, I don't think you can get anywhere useful. You can literally derive anything from assuming P & ~P, or (P&Q) & ~(P&Q) in your case. The problem is you end up with logic that isn't valid.

    Unless I missed something, which is quite possible, I suggest trying another approach. I'm too rusty at the moment to be of much more help though. Seeing the equivalent of P and ~P as assumptions is throwing off red flags for me is all. If you can't use DeMorgan's, I'm not sure what rules you are supposed to be playing by.

    We used Copi (Symbolic Logic)... 5 years ago?
     
    Last edited: Apr 5, 2010
  8. Apr 6, 2010 #7

    in3

    User Avatar

    It was quite a while since I took logic, but shouldn't this work?

    Code (Text):
    1      (1)  ~(P & Q)      Premise
           (2)  P v ~P        LEM
    3      (3)  P             Assumption
    4      (4)  Q             Assumption
    3,4    (5)  P & Q         3,4 &I
    1,3,4  (6)  absurd        1,5 ~E
    1,3    (7)  ~Q            4,6 ~I
    1,3    (8)  ~P v ~Q       7 vI
    9      (9)  ~P            Assumption
    9     (10)  ~P v ~Q       9 vI
    1     (11)  ~P v ~Q       2,3,8,9,10 vE
    The leftmost column denotes dependencies and the rightmost column what rule is invoked. I'm using the notation found in Forbes' "Modern Logic", it seems you're using it too.

    One shaky thing here is the use of the law of excluded middle which is only valid as long as you can use reductio ad absurdum, so it is not valid in intuitionistic logic. Otherwise LEM is not difficult to prove (assume ~(A v ~A) and go from there).
     
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