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No-data

  • one year ago

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

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

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

  3. No-data
    • one year ago
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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.

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

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

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

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

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

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

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

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

  12. wio
    • one year ago
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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 \[ \left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \Rightarrow \left(P\Rightarrow R\right) \]

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

  14. wio
    • one year ago
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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.

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

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

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

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

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

  20. wio
    • one year ago
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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?

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

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

  23. wio
    • one year ago
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    Which theorems?

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

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

  26. wio
    • one year ago
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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.

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

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

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

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

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

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

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

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

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

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

  37. wio
    • one year ago
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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?

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

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

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

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

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

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

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

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

  46. No-data
    • one year ago
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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.

  47. wio
    • one year ago
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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.

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

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

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

  51. No-data
    • one year ago
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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.

  52. wio
    • one year ago
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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?

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

  54. wio
    • one year ago
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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.

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

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

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

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

  59. Edutopia
    • one year ago
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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

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

  61. wio
    • one year ago
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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".

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

  63. Edutopia
    • one year ago
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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.

  64. No-data
    • one year ago
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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.

  65. wio
    • one year ago
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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 \(Q\vee \neg Q\).

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

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

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

  69. No-data
    • one year ago
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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.

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

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

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

  73. No-data
    • one year ago
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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.

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