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No-data

How can I simplify this: \[\left(P\wedge\neg R\right)\vee\left(P\wedge\neg Q\right)\vee\left(Q\wedge\neg R\right)\]

  • one year ago
  • one year ago

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  1. No-data
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    I've been working on this but it seem I can't do anymore. =/

    • one year ago
  2. No-data
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    The original expression is this: \[\left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right]\]

    • one year ago
  3. No-data
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    I see that it is like prove the transitivy propertie for the implication operator. I made a truth table for this and it turned out to be false, but I want to get to the same conclussion without using truth tables.

    • one year ago
  4. No-data
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    This is the work I've done so far.

    • one year ago
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  5. abb0t
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    Well, tautology is a proposition that is always true. That's all I know. That's all I know.

    • one year ago
  6. Rohangrr
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    good reasoning @No-data

    • one year ago
  7. No-data
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    Thank you @Rohangrr

    • one year ago
  8. No-data
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    @brinethery Do you have any idea?

    • one year ago
  9. wio
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    Are you sure it's true? It looks like the converse is true.

    • one year ago
  10. wio
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    \[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]Suppose \(P \gets T\quad Q\gets F\quad R\gets T \) \[ (T\implies T)\implies [(T\implies F)\wedge (F \implies T)] \\ T\implies (F\wedge T) \\ T\implies F \\ F \]

    • one year ago
  11. Edutopia
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    I feel like q is irrelvent and its just If P then R

    • one year ago
  12. wio
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    \[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]I think it should be \[ \left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \Rightarrow \left(P\Rightarrow R\right) \]

    • one year ago
  13. Edutopia
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    ^^ makes alot of sense

    • one year ago
  14. wio
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    There is nothing wrong with using a counter example to show something is false. Trying to prove it without a counter example seems pointless to me.

    • one year ago
  15. No-data
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    I did not say it was true

    • one year ago
  16. No-data
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    Mmm I don't understand the expression after "Suppose" what those arrows mean?

    • one year ago
  17. No-data
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    Ok I got the counter example, is a row on my truth table and its correct. Thank you @wio

    • one year ago
  18. No-data
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    I was just trying to show that the expression is not a tautology.

    • one year ago
  19. No-data
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    The expression you wrote saying you thought it should be is, in fact, a tautology.

    • one year ago
  20. wio
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    Imagine if I have you the proposition \(P\) and claimed it was a tautology. How would you prove it is wrong?

    • one year ago
  21. wio
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    There is no boolean algebra to prove it, you just give a counter example, right?

    • one year ago
  22. No-data
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    Do you think that is not possible to find a way to show this using only theorems and definitions?

    • one year ago
  23. wio
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    Which theorems?

    • one year ago
  24. wio
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    How would you show that \(P\) in and of itself is not a tautology?

    • one year ago
  25. No-data
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    There is no way to know. Unless you tell me that P can take the values T and F.

    • one year ago
  26. wio
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    I mean, I have nothing against trying to do things a particular way, but what are the rules really? How would you show \(P\) is not a tautology without a counter example.

    • one year ago
  27. No-data
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    and then, by definition of P. P is not a tautology

    • one year ago
  28. wio
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    So you'd say by definition of proposition, a proposition is not a tautology? Then how about \(P\wedge Q\)?

    • one year ago
  29. No-data
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    You meant that proposition is not a tautology.

    • one year ago
  30. No-data
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    As far as I know a tautology is a predicate that is always true for any value of their parameters.

    • one year ago
  31. No-data
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    \[P\wedge Q\] is a predicate that can be true or false. So it's not a tautology.

    • one year ago
  32. No-data
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    Anyway I appreciate your help.

    • one year ago
  33. wio
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    You're claiming it can be false... The point is to get you to prove it without counterexample.

    • one year ago
  34. wio
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    I'm just curious as to the rules of the game.

    • one year ago
  35. No-data
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    I think I'm not using a counter example and that is a valid proof.

    • one year ago
  36. Edutopia
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    A tautology is a restatement within the same premiss like all trees are made of wood

    • one year ago
  37. wio
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    What if I say "is a predicate that can be true or false.", about the original predicate? Said it about \[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]Would simply saying that be a valid proof that it is not a tautology?

    • one year ago
  38. wio
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    I don't think saying "is a predicate that can be true or false." is valid without a counter example.

    • one year ago
  39. wio
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    Or maybe there is another way, but I'm not sure

    • one year ago
  40. No-data
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    Is not evident that that predicate can be true or false. Its not a valid proof.

    • one year ago
  41. No-data
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    and \[P\wedge Q\]is by definition true or false.

    • one year ago
  42. wio
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    Sure but then what makes something 'evident'?

    • one year ago
  43. wio
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    Is there some standard form you want it to be in?

    • one year ago
  44. No-data
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    I wanted to say that there is a definition by saying "evident"

    • one year ago
  45. wio
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    What about \(P\wedge Q \wedge R\)?

    • one year ago
  46. No-data
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    I accepted you method buy I feel like you're trying to say that there is no other method to prove this problem.

    • one year ago
  47. wio
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    What I am saying is that if there is another way to prove this problem, then there needs to be a way to prove very simple predicates.

    • one year ago
  48. No-data
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    let \[P\wedge Q\Leftrightarrow S \]\[S\wedge R\]

    • one year ago
  49. wio
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    I mean prove without counter example

    • one year ago
  50. No-data
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    so it's not a tautology.

    • one year ago
  51. No-data
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    I know that, I already proved it using other method. I just wanted to know if it was possible to do it the way I'm trying.

    • one year ago
  52. wio
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    Using your logic that \(\wedge\) doesn't give a tautology, then couldn't we say \(\neg P\vee P\) isn't a tautology?

    • one year ago
  53. Edutopia
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    ITs a Syllogism if the expanded form is the antecedent, but because its the consequent it is not tautological

    • one year ago
  54. wio
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    Well it's just an interesting thing to think about. I would like to find a way to prove false without counterexample but it seems hard to understand what is allowed.

    • one year ago
  55. Edutopia
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    q's relation to p and r

    • one year ago
  56. No-data
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    This is not a tautology:\[P\wedge Q\], not just the \[\wedge \] symbol.

    • one year ago
  57. No-data
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    Sorry I did not understand that @Edutopia

    • one year ago
  58. No-data
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    I'm just beginning with this logic things lol.

    • one year ago
  59. Edutopia
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    my comp is sllooow, i ment to say: q's relation to p and r would have to be in the premiss for it to be tautological

    • one year ago
  60. No-data
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    ahh that is right Edutopia.

    • one year ago
  61. wio
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    You know what's interesting about proving something is true, is that you are showing it is true for all cases. If you are proving something is false, you only need to show one case is false. To prove that it is false for all cases can't always be done and doesn't need to be done. I think to even say \(P\) is not a tautology, you are implicitly calling upon the counter example just to claim "\(P\) can be false".

    • one year ago
  62. wio
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    But maybe that is a semantic argument? It's just my thought.

    • one year ago
  63. Edutopia
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    in IF (IF P THEN R) THEN BOTH (IF P THEN Q) AND ( IF Q THEN R) the relation of Q to P and R is not established, imagine the argument where P and R have to do with Physics and Q is someones opinion on abortion.

    • one year ago
  64. No-data
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    Yeah it's interesting @wio and I agree your counterexamples are a good tool. I just think they are not elegant.

    • one year ago
  65. wio
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    When I first saw \[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]The \(Q\) made me think it's likely not a tautology right away. Since it wasn't something inconsequential like \(Q\vee \neg Q\).

    • one year ago
  66. No-data
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    I'm not sure, but I think it's valid in math.If\[x=y\]then\[x+a=y+a\]

    • one year ago
  67. wio
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    That is the addition property of equality. \(Q\) was not being appended to both sides of the implication.

    • one year ago
  68. Edutopia
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    yes, but in math its all a valid argument because you are using numbers

    • one year ago
  69. No-data
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    I know it just look similar. Thank you guys. It is nice to be able to discuss this kind of things with other people. Thank you for you r time.

    • one year ago
  70. Edutopia
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    a=a and so forth

    • one year ago
  71. brinethery
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    @No-data I have no idea :-). I'm so stupid.

    • one year ago
  72. brinethery
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    Is this for linear algebra?

    • one year ago
  73. No-data
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    It is propositional logic. You're not stupid at all @brinethery. I was curious about how databases work so I picked a book about math applied to databases and this is a problem from the first chapter. Logic and Set Theory are the foundation of DB systems.

    • one year ago
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