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anonymous
 4 years ago
show that the statement (~p V q) ^ (p^~q) is a logical contradiction
anonymous
 4 years ago
show that the statement (~p V q) ^ (p^~q) is a logical contradiction

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KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Have you been taught how to do truth tables?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Well, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ive seen them in my text book but i am not sure how to do one

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Suppose 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?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1if you were gong to use a truth table set it up like this: \[\begin{array}{ccccc}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&&T\\T&&F \\F&&T\\T&&F\\\hline\end{array}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@KingGeorge i am not even sure where to start...

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Alright. 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?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2right. Now, ~p is False, and q is true. What is ~p OR q?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Not 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.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Let's go to the next step. What is ~q?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Alright, 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?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Alright, stop me if you start to not understand anything. Now, q is True. So what is ~q?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Precisely. Now, recall that p is True, and ~q is False. Can you tell me what (p AND ~q) is?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2oops, I screwed up a bit.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Since we're assuming p is true, and q is False, ~q is True. So, (p AND ~q) is actually...?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Oh 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?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Well, 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.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2\[\begin{array}{ccccc}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F\\T&T&F&T\\\hline\end{array}\] 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?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2right. what about (p AND ~q)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0false since q is true

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Bingo. Then, what about (~p OR q) AND (p AND ~q)?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1whops i set up the table with a mistake \[\begin{array}{ccccc}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F\\\color{red}F&T&F&T\\\hline\end{array}\]

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Not 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.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Now 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?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2What exactly are you saying is false?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually i mean that one is true since p is false

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Right. what about (p AND ~q)?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Remember that p is False, and ~q is True, so (p AND ~q) would actually be False.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Now here's the very last step. What is (~p OR q) AND (p AND ~q)?

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

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2If you want to know some more about truth tables, and how to construct them, feel free to ask.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok great so that truth table is the whole answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks so much! i have other questions but there not about truth tables

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2The truth table is a great way to show that you have a logical contradiction, and one of the easiest ways to see it directly.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355892045358:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The number of rows in the truth table for the compound statement

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would the answer be 8?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2First off,how many variables do you have?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2I mean like p and q are variable. So in this context, it's a variable if it can be true or false.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2How many of those are there?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Right. 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0great do you now anything about contrapositive statments?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.2Sure. If you have more questions though, please put it in a new question.
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