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princesspixie Group Title

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

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

    • 2 years ago
  2. princesspixie Group Title
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    no

    • 2 years ago
  3. KingGeorge Group Title
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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.

    • 2 years ago
  4. princesspixie Group Title
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    ok

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

    • 2 years ago
  6. KingGeorge Group Title
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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?

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

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

    • 2 years ago
  9. KingGeorge Group Title
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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?

    • 2 years ago
  10. princesspixie Group Title
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    false

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

    • 2 years ago
  12. princesspixie Group Title
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    false

    • 2 years ago
  13. KingGeorge Group Title
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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.

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

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

    • 2 years ago
  16. princesspixie Group Title
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    true

    • 2 years ago
  17. KingGeorge Group Title
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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?

    • 2 years ago
  18. princesspixie Group Title
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    i guess

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

    • 2 years ago
  20. princesspixie Group Title
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    false

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

    • 2 years ago
  22. princesspixie Group Title
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    false

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

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

    • 2 years ago
  25. princesspixie Group Title
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    trur

    • 2 years ago
  26. KingGeorge Group Title
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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?

    • 2 years ago
  27. princesspixie Group Title
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    false true

    • 2 years ago
  28. KingGeorge Group Title
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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.

    • 2 years ago
  29. KingGeorge Group Title
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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?

    • 2 years ago
  30. princesspixie Group Title
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    true

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

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

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

    • 2 years ago
  34. princesspixie Group Title
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    true

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

    • 2 years ago
  36. KingGeorge Group Title
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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.

    • 2 years ago
  37. KingGeorge Group Title
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    Good catch^^

    • 2 years ago
  38. KingGeorge Group Title
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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?

    • 2 years ago
  39. princesspixie Group Title
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    false

    • 2 years ago
  40. princesspixie Group Title
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    @KingGeorge

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

    • 2 years ago
  42. princesspixie Group Title
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    ~p or q

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

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

    • 2 years ago
  45. princesspixie Group Title
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    true

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

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

    • 2 years ago
  48. princesspixie Group Title
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    false and false?

    • 2 years ago
  49. KingGeorge Group Title
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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.

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

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

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

    • 2 years ago
  53. KingGeorge Group Title
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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.

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

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

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

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

    • 2 years ago
  58. princesspixie Group Title
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    7?

    • 2 years ago
  59. KingGeorge Group Title
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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.

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

    • 2 years ago
  61. princesspixie Group Title
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    4?

    • 2 years ago
  62. KingGeorge Group Title
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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.

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

    • 2 years ago
  64. KingGeorge Group Title
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    yup.

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

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

    • 2 years ago
  67. princesspixie Group Title
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    ok thanks!!

    • 2 years ago
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