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Truth table help... did I do it correctly

  1. Jan 28, 2016 #1
    1. The problem statement, all variables and given/known data
    The question I'm asking about is number two. Should've made it clearer.
    20160128_181508.jpg
    2. Relevant equations
    N/A

    3. The attempt at a solution
    My proof table, I'm not sure but it seems that PΞQ is not true.
    20160128_181518.jpg
     
  2. jcsd
  3. Jan 28, 2016 #2
    Hi Kingyou:

    Your calculation looks right to me.

    Regards,
    Buzz
     
  4. Jan 28, 2016 #3
    A few hints:
    1. Set out your truth table methodically - this makes it easier for you to check and easier for a marker to give you partial marks
      Code (Text):
      p q r
      -----
      T T T
      T T F
      T F T
      T F F
      F T T
      F T F
      F F T
      F F F
    2. Read the question carefully - why have you calculated p → r ?
    3. Learn the truth table for implies
      Code (Text):
      a b | a→b
      ----------
      T T | T
      T F | F
      F T | T
      F F | T
     
  5. Jan 28, 2016 #4
    Awesome, thank you
     
  6. Jan 29, 2016 #5

    WWGD

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    You can also use an argument:
    Assume ## p\rightarrow## ~ r , and assume p. Then you have ##p\rightarrow q ##, from which q follows ,from which r follows.
     
  7. Jan 29, 2016 #6
    How does that show that there are values of (p, q, r) for which the identity does not hold)?
     
  8. Jan 29, 2016 #7

    WWGD

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    It shows there are none, since this shows that the wff is a theorem.
     
  9. Jan 29, 2016 #8
    I don't understand which wff you are referring to, or any steps of your outline proof I am afraid. In any case I'm pretty sure the question setter is looking for a truth table and is going to mark a deductive proof harshly unless it is presented flawlessly - why take the risk?
     
  10. Jan 29, 2016 #9

    WWGD

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    Sorry, I was unclear, I was aiming for a proof by contradiction, but I agree, might as well go for the truth table argument.
     
  11. Jan 29, 2016 #10
    To be clear, I meant why have you calculated p → r (incorrectly) in the first calculated column of the TT?
     
  12. Jan 29, 2016 #11

    WWGD

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    I assumed ## p \rightarrow -r ## together with the given premises and concluded r , so I showed by contradiction that ## p \rightarrow r ##. Unfortunately I don't know well the Latex for logic symbols to do
    the actual derivation of ## p \rightarrow r## and ##p \rightarrow - r ##..
     
    Last edited: Jan 29, 2016
  13. Jan 30, 2016 #12
    Oh this has got very confusing, my post #10 was intended to clarify my post #3 pointing out that the OP had misread the question which asks for (p → q) → (q → r) ≡ (p → r) and attempted to show a TT for (p → r) → (q → r) ≡ (p → r) instead.

    As for your post #11 WWGD, note that the negation of ## (p \to q) \to (q \to r) \equiv (p \to r) ## is not ## (p \to q) \to (q \to r) \equiv (p \to \neg r) ## it is ##(p \to q) \to (q \to r) \neq (p \to r) ##
     
  14. Jan 30, 2016 #13

    WWGD

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    I know (and you're right that I only proved one side of the equivalence), I don't mean to be impolite, but it is getting too confusing; let's just drop it if you don't mind, sorry for the dead end.
     
  15. Jan 30, 2016 #14
    Agreed - the OP seems to have gone anyway (unfortunately I think with the impression that his workings were OK but we tried...)
     
  16. Feb 2, 2016 #15
    What is wrong with my work?
     
  17. Feb 2, 2016 #16
    Look at what you have written at the top of your fourth and sixth columns and read the question again.

    And then apply the truth table I showed you correctly.
     
  18. Feb 2, 2016 #17
    The easiest way to do this is to create the truth table for the expression

    ## (p \to q) \to (q \to r) ##

    Then look for two things. First, check to see if the truth value (for any fixed values of p and r) depends on q. If so, the equivalence cannot hold. If not, then check if the truth table for the above expression produces the same values for a fixed p and r as ## (p \to r) ##.
     
  19. Feb 3, 2016 #18

    SammyS

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    You have the wrong result for a ⇒ b . The only instance in which a ⇒ b is false, is the case of a is false and b is true.
     
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