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princesspixie

  • 2 years ago

show that the statement (~p V q) ^ (p^~q) is a logical contradiction

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  1. KingGeorge
    • 2 years ago
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    Have you been taught how to do truth tables?

  2. princesspixie
    • 2 years ago
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    no

  3. KingGeorge
    • 2 years ago
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    Well, they make the problem easier, but you can still do it without them. What we need to do, is break the problem up into 4 different cases. Case 1: p=True, q=True. Case 2: p=True, q=False. Case 3: p=False, q=True. Case 4: p=False, q=False. I'll work through the first case.

  4. princesspixie
    • 2 years ago
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    ok

  5. princesspixie
    • 2 years ago
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    ive seen them in my text book but i am not sure how to do one

  6. KingGeorge
    • 2 years ago
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    Suppose p=True, and q=True. Then (~p V q) is the same as (False OR True), which is True, since q is True. Also, (p ^ ~q) is the same as (True AND False), which is False, since q and p have different truth values (one is true, the other is false). Thus, we end with True AND False, which by the same reasoning, is False. To do the rest, you need to go through each case, and show that you always end with False. Can you try to do Case 2 by yourself?

  7. UnkleRhaukus
    • 2 years ago
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    if you were gong to use a truth table set it up like this: \[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&&T\\T&&F \\F&&T\\T&&F\\\hline\end{array}\]

  8. princesspixie
    • 2 years ago
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    @KingGeorge i am not even sure where to start...

  9. KingGeorge
    • 2 years ago
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    Alright. So start with Case 2: p is True, and q is False. Now let's take it one step at a time. What is ~p?

  10. princesspixie
    • 2 years ago
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    false

  11. KingGeorge
    • 2 years ago
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    right. Now, ~p is False, and q is true. What is ~p OR q?

  12. princesspixie
    • 2 years ago
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    false

  13. KingGeorge
    • 2 years ago
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    Not so fast. Remember that q is True, and since you only need one thing to be true in an OR statement, ~p OR q is actually True.

  14. KingGeorge
    • 2 years ago
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    Let's go to the next step. What is ~q?

  15. princesspixie
    • 2 years ago
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    I dont get it..

  16. princesspixie
    • 2 years ago
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    true

  17. KingGeorge
    • 2 years ago
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    Alright, let's take a step back, and look at ~p OR q again. For (~p OR q) to be true, either ~p is True, or q is True. Since q is True, (~p OR q) is True. Does this make more sense?

  18. princesspixie
    • 2 years ago
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    i guess

  19. KingGeorge
    • 2 years ago
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    Alright, stop me if you start to not understand anything. Now, q is True. So what is ~q?

  20. princesspixie
    • 2 years ago
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    false

  21. KingGeorge
    • 2 years ago
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    Precisely. Now, recall that p is True, and ~q is False. Can you tell me what (p AND ~q) is?

  22. princesspixie
    • 2 years ago
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    false

  23. KingGeorge
    • 2 years ago
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    oops, I screwed up a bit.

  24. KingGeorge
    • 2 years ago
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    Since we're assuming p is true, and q is False, ~q is True. So, (p AND ~q) is actually...?

  25. princesspixie
    • 2 years ago
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    trur

  26. KingGeorge
    • 2 years ago
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    Oh wow. I screwed up earlier as well. You were correct in saying that (~p OR q) is False. Anyways, then we only have (~p OR q) AND (p AND ~q) left. This simplifies to (False AND True). What is the truth value of this?

  27. princesspixie
    • 2 years ago
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    false true

  28. KingGeorge
    • 2 years ago
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    Well, for (False AND True) to be True, either both need to be false, or both need to be true. Since neither is the case, (False AND True) is False. Now, I'm going to fill in the truth table that UnkleRhaukus kindly provided.

  29. KingGeorge
    • 2 years ago
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    \[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F\\T&T&F&T\\\hline\end{array}\] The first two lines correspond to Case 1 and Case 2 respectively. Now we want to do Case 3. So p is False, and q is True. Can you tell me what (~p OR q) is?

  30. princesspixie
    • 2 years ago
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    true

  31. KingGeorge
    • 2 years ago
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    right. what about (p AND ~q)?

  32. princesspixie
    • 2 years ago
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    false since q is true

  33. KingGeorge
    • 2 years ago
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    Bingo. Then, what about (~p OR q) AND (p AND ~q)?

  34. princesspixie
    • 2 years ago
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    true

  35. UnkleRhaukus
    • 2 years ago
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    whops i set up the table with a mistake \[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F\\\color{red}F&T&F&T\\\hline\end{array}\]

  36. KingGeorge
    • 2 years ago
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    Not quite. Remember that (~p OR q) is True, but (q AND ~q) is False. Since they're different truth values, (~p OR q) AND (p AND ~q) must be False.

  37. KingGeorge
    • 2 years ago
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    Good catch^^

  38. KingGeorge
    • 2 years ago
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    Now we just have the final case, where p is False, and q is False. Can you work through that, and tell me what you think (~p OR q) is, and what (p AND ~q) is?

  39. princesspixie
    • 2 years ago
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    false

  40. princesspixie
    • 2 years ago
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    @KingGeorge

  41. KingGeorge
    • 2 years ago
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    What exactly are you saying is false?

  42. princesspixie
    • 2 years ago
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    ~p or q

  43. princesspixie
    • 2 years ago
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    actually i mean that one is true since p is false

  44. KingGeorge
    • 2 years ago
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    Right. what about (p AND ~q)?

  45. princesspixie
    • 2 years ago
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    true

  46. KingGeorge
    • 2 years ago
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    Remember that p is False, and ~q is True, so (p AND ~q) would actually be False.

  47. KingGeorge
    • 2 years ago
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    Now here's the very last step. What is (~p OR q) AND (p AND ~q)?

  48. princesspixie
    • 2 years ago
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    false and false?

  49. KingGeorge
    • 2 years ago
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    Well the statement as a whole is False. Now, if we look at the truth table, we get a table that looks like this. \[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F&T&F&F\\F&T&F&T&T&F&F\\\hline\end{array}\]Since the entire last column is filled with F, we've shown that the original statement is a logical contradiction.

  50. KingGeorge
    • 2 years ago
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    If you want to know some more about truth tables, and how to construct them, feel free to ask.

  51. princesspixie
    • 2 years ago
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    ok great so that truth table is the whole answer?

  52. princesspixie
    • 2 years ago
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    thanks so much! i have other questions but there not about truth tables

  53. KingGeorge
    • 2 years ago
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    The truth table is a great way to show that you have a logical contradiction, and one of the easiest ways to see it directly.

  54. princesspixie
    • 2 years ago
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    |dw:1355892045358:dw|

  55. princesspixie
    • 2 years ago
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    The number of rows in the truth table for the compound statement

  56. princesspixie
    • 2 years ago
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    would the answer be 8?

  57. KingGeorge
    • 2 years ago
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    First off,how many variables do you have?

  58. princesspixie
    • 2 years ago
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    7?

  59. KingGeorge
    • 2 years ago
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    I mean like p and q are variable. So in this context, it's a variable if it can be true or false.

  60. KingGeorge
    • 2 years ago
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    How many of those are there?

  61. princesspixie
    • 2 years ago
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    4?

  62. KingGeorge
    • 2 years ago
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    Right. Then for these truth problems, each variable can be either T or F, and each row in the truth table corresponds to a different combination of these possibilities. Since each variable has 2 possibilities, and there are 4 variables, there are \(2\cdot2\cdot2\cdot2=16\) rows in the truth table. In general, for a truth statement with \(k\) variables, there would be \[2^k\] rows in the truth table.

  63. princesspixie
    • 2 years ago
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    so it would be 16?

  64. KingGeorge
    • 2 years ago
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    yup.

  65. princesspixie
    • 2 years ago
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    great do you now anything about contrapositive statments?

  66. KingGeorge
    • 2 years ago
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    Sure. If you have more questions though, please put it in a new question.

  67. princesspixie
    • 2 years ago
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    ok thanks!!

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