## princesspixie Group Title show that the statement (~p V q) ^ (p^~q) is a logical contradiction one year ago one year ago

1. KingGeorge Group Title

Have you been taught how to do truth tables?

2. princesspixie Group Title

no

3. KingGeorge Group Title

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 Group Title

ok

5. princesspixie Group Title

ive seen them in my text book but i am not sure how to do one

6. KingGeorge Group Title

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 Group Title

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 Group Title

@KingGeorge i am not even sure where to start...

9. KingGeorge Group Title

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 Group Title

false

11. KingGeorge Group Title

right. Now, ~p is False, and q is true. What is ~p OR q?

12. princesspixie Group Title

false

13. KingGeorge Group Title

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 Group Title

Let's go to the next step. What is ~q?

15. princesspixie Group Title

I dont get it..

16. princesspixie Group Title

true

17. KingGeorge Group Title

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 Group Title

i guess

19. KingGeorge Group Title

Alright, stop me if you start to not understand anything. Now, q is True. So what is ~q?

20. princesspixie Group Title

false

21. KingGeorge Group Title

Precisely. Now, recall that p is True, and ~q is False. Can you tell me what (p AND ~q) is?

22. princesspixie Group Title

false

23. KingGeorge Group Title

oops, I screwed up a bit.

24. KingGeorge Group Title

Since we're assuming p is true, and q is False, ~q is True. So, (p AND ~q) is actually...?

25. princesspixie Group Title

trur

26. KingGeorge Group Title

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 Group Title

false true

28. KingGeorge Group Title

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 Group Title

$\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F\\T&T&F&T\\\hline\end{array}$ 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 Group Title

true

31. KingGeorge Group Title

right. what about (p AND ~q)?

32. princesspixie Group Title

false since q is true

33. KingGeorge Group Title

Bingo. Then, what about (~p OR q) AND (p AND ~q)?

34. princesspixie Group Title

true

35. UnkleRhaukus Group Title

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 Group Title

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 Group Title

Good catch^^

38. KingGeorge Group Title

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 Group Title

false

40. princesspixie Group Title

@KingGeorge

41. KingGeorge Group Title

What exactly are you saying is false?

42. princesspixie Group Title

~p or q

43. princesspixie Group Title

actually i mean that one is true since p is false

44. KingGeorge Group Title

Right. what about (p AND ~q)?

45. princesspixie Group Title

true

46. KingGeorge Group Title

Remember that p is False, and ~q is True, so (p AND ~q) would actually be False.

47. KingGeorge Group Title

Now here's the very last step. What is (~p OR q) AND (p AND ~q)?

48. princesspixie Group Title

false and false?

49. KingGeorge Group Title

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 Group Title

If you want to know some more about truth tables, and how to construct them, feel free to ask.

51. princesspixie Group Title

ok great so that truth table is the whole answer?

52. princesspixie Group Title

thanks so much! i have other questions but there not about truth tables

53. KingGeorge Group Title

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 Group Title

|dw:1355892045358:dw|

55. princesspixie Group Title

The number of rows in the truth table for the compound statement

56. princesspixie Group Title

57. KingGeorge Group Title

First off,how many variables do you have?

58. princesspixie Group Title

7?

59. KingGeorge Group Title

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 Group Title

How many of those are there?

61. princesspixie Group Title

4?

62. KingGeorge Group Title

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 Group Title

so it would be 16?

64. KingGeorge Group Title

yup.

65. princesspixie Group Title

great do you now anything about contrapositive statments?

66. KingGeorge Group Title

Sure. If you have more questions though, please put it in a new question.

67. princesspixie Group Title

ok thanks!!