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No-data

  • 2 years 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
    • 2 years ago
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    I've been working on this but it seem I can't do anymore. =/

  2. No-data
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    This is the work I've done so far.

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  5. abb0t
    • 2 years 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
    • 2 years ago
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    good reasoning @No-data

  7. No-data
    • 2 years ago
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    Thank you @Rohangrr

  8. No-data
    • 2 years ago
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    @brinethery Do you have any idea?

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

  10. wio
    • 2 years 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
    • 2 years ago
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    I feel like q is irrelvent and its just If P then R

  12. wio
    • 2 years 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
    • 2 years ago
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    ^^ makes alot of sense

  14. wio
    • 2 years 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
    • 2 years ago
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    I did not say it was true

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

  17. No-data
    • 2 years 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
    • 2 years ago
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    I was just trying to show that the expression is not a tautology.

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

  20. wio
    • 2 years 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
    • 2 years ago
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    There is no boolean algebra to prove it, you just give a counter example, right?

  22. No-data
    • 2 years 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
    • 2 years ago
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    Which theorems?

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

  25. No-data
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    and then, by definition of P. P is not a tautology

  28. wio
    • 2 years 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
    • 2 years ago
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    You meant that proposition is not a tautology.

  30. No-data
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Anyway I appreciate your help.

  33. wio
    • 2 years 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
    • 2 years ago
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    I'm just curious as to the rules of the game.

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

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

  37. wio
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Or maybe there is another way, but I'm not sure

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

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

  42. wio
    • 2 years ago
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    Sure but then what makes something 'evident'?

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

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

  45. wio
    • 2 years ago
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    What about \(P\wedge Q \wedge R\)?

  46. No-data
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    let \[P\wedge Q\Leftrightarrow S \]\[S\wedge R\]

  49. wio
    • 2 years ago
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    I mean prove without counter example

  50. No-data
    • 2 years ago
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    so it's not a tautology.

  51. No-data
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    q's relation to p and r

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

  57. No-data
    • 2 years ago
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    Sorry I did not understand that @Edutopia

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

  59. Edutopia
    • 2 years 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
    • 2 years ago
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    ahh that is right Edutopia.

  61. wio
    • 2 years 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
    • 2 years ago
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    But maybe that is a semantic argument? It's just my thought.

  63. Edutopia
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    yes, but in math its all a valid argument because you are using numbers

  69. No-data
    • 2 years 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
    • 2 years ago
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    a=a and so forth

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

  72. brinethery
    • 2 years ago
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    Is this for linear algebra?

  73. No-data
    • 2 years 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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