## princesspixie 2 years ago show that the statement (~p V q) ^ (p^~q) is a logical contradiction

1. KingGeorge

Have you been taught how to do truth tables?

2. princesspixie

no

3. KingGeorge

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

ok

5. princesspixie

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

6. KingGeorge

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

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

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

9. KingGeorge

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

false

11. KingGeorge

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

12. princesspixie

false

13. KingGeorge

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

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

15. princesspixie

I dont get it..

16. princesspixie

true

17. KingGeorge

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

i guess

19. KingGeorge

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

20. princesspixie

false

21. KingGeorge

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

22. princesspixie

false

23. KingGeorge

oops, I screwed up a bit.

24. KingGeorge

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

25. princesspixie

trur

26. KingGeorge

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

false true

28. KingGeorge

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

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

true

31. KingGeorge

right. what about (p AND ~q)?

32. princesspixie

false since q is true

33. KingGeorge

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

34. princesspixie

true

35. UnkleRhaukus

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

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

Good catch^^

38. KingGeorge

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

false

40. princesspixie

@KingGeorge

41. KingGeorge

What exactly are you saying is false?

42. princesspixie

~p or q

43. princesspixie

actually i mean that one is true since p is false

44. KingGeorge

Right. what about (p AND ~q)?

45. princesspixie

true

46. KingGeorge

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

47. KingGeorge

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

48. princesspixie

false and false?

49. KingGeorge

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

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

51. princesspixie

ok great so that truth table is the whole answer?

52. princesspixie

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

53. KingGeorge

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

|dw:1355892045358:dw|

55. princesspixie

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

56. princesspixie

57. KingGeorge

First off,how many variables do you have?

58. princesspixie

7?

59. KingGeorge

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

How many of those are there?

61. princesspixie

4?

62. KingGeorge

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

so it would be 16?

64. KingGeorge

yup.

65. princesspixie

great do you now anything about contrapositive statments?

66. KingGeorge

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

67. princesspixie

ok thanks!!