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princesspixie
show that the statement (~p V q) ^ (p^~q) is a logical contradiction
Have you been taught how to do truth tables?
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.
ive seen them in my text book but i am not sure how to do one
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?
if you were gong to use a truth table set it up like this: \[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&&T\\T&&F \\F&&T\\T&&F\\\hline\end{array}\]
@KingGeorge i am not even sure where to start...
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?
right. Now, ~p is False, and q is true. What is ~p OR q?
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.
Let's go to the next step. What is ~q?
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?
Alright, stop me if you start to not understand anything. Now, q is True. So what is ~q?
Precisely. Now, recall that p is True, and ~q is False. Can you tell me what (p AND ~q) is?
oops, I screwed up a bit.
Since we're assuming p is true, and q is False, ~q is True. So, (p AND ~q) is actually...?
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?
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.
\[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F\\T&T&F&T\\\hline\end{array}\] 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?
right. what about (p AND ~q)?
false since q is true
Bingo. Then, what about (~p OR q) AND (p AND ~q)?
whops i set up the table with a mistake \[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F\\\color{red}F&T&F&T\\\hline\end{array}\]
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.
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?
What exactly are you saying is false?
actually i mean that one is true since p is false
Right. what about (p AND ~q)?
Remember that p is False, and ~q is True, so (p AND ~q) would actually be False.
Now here's the very last step. What is (~p OR q) AND (p AND ~q)?
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.
If you want to know some more about truth tables, and how to construct them, feel free to ask.
ok great so that truth table is the whole answer?
thanks so much! i have other questions but there not about truth tables
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.
|dw:1355892045358:dw|
The number of rows in the truth table for the compound statement
would the answer be 8?
First off,how many variables do you have?
I mean like p and q are variable. So in this context, it's a variable if it can be true or false.
How many of those are there?
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.
so it would be 16?
great do you now anything about contrapositive statments?
Sure. If you have more questions though, please put it in a new question.