anonymous
  • anonymous
show that the statement (~p V q) ^ (p^~q) is a logical contradiction
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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KingGeorge
  • KingGeorge
Have you been taught how to do truth tables?
anonymous
  • anonymous
no
KingGeorge
  • 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.

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anonymous
  • anonymous
ok
anonymous
  • anonymous
ive seen them in my text book but i am not sure how to do one
KingGeorge
  • 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?
UnkleRhaukus
  • UnkleRhaukus
if you were gong to use a truth table set it up like this: \[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&&T\\T&&F \\F&&T\\T&&F\\\hline\end{array}\]
anonymous
  • anonymous
@KingGeorge i am not even sure where to start...
KingGeorge
  • 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?
anonymous
  • anonymous
false
KingGeorge
  • KingGeorge
right. Now, ~p is False, and q is true. What is ~p OR q?
anonymous
  • anonymous
false
KingGeorge
  • 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.
KingGeorge
  • KingGeorge
Let's go to the next step. What is ~q?
anonymous
  • anonymous
I dont get it..
anonymous
  • anonymous
true
KingGeorge
  • 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?
anonymous
  • anonymous
i guess
KingGeorge
  • KingGeorge
Alright, stop me if you start to not understand anything. Now, q is True. So what is ~q?
anonymous
  • anonymous
false
KingGeorge
  • KingGeorge
Precisely. Now, recall that p is True, and ~q is False. Can you tell me what (p AND ~q) is?
anonymous
  • anonymous
false
KingGeorge
  • KingGeorge
oops, I screwed up a bit.
KingGeorge
  • KingGeorge
Since we're assuming p is true, and q is False, ~q is True. So, (p AND ~q) is actually...?
anonymous
  • anonymous
trur
KingGeorge
  • 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?
anonymous
  • anonymous
false true
KingGeorge
  • 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.
KingGeorge
  • 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?
anonymous
  • anonymous
true
KingGeorge
  • KingGeorge
right. what about (p AND ~q)?
anonymous
  • anonymous
false since q is true
KingGeorge
  • KingGeorge
Bingo. Then, what about (~p OR q) AND (p AND ~q)?
anonymous
  • anonymous
true
UnkleRhaukus
  • UnkleRhaukus
whops i set up the table with a mistake \[\begin{array}{|c|c|c|c|c|}\hline p&\neg p&q&\neg q&\neg p\lor q&p\land \neg q&(\neg p\lor q)\land(p\land \neg q)\\\hline T&F&T&F&T&F&F\\T&F&F&T&F&T&F \\F&T&T&F\\\color{red}F&T&F&T\\\hline\end{array}\]
KingGeorge
  • 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.
KingGeorge
  • KingGeorge
Good catch^^
KingGeorge
  • 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?
anonymous
  • anonymous
false
anonymous
  • anonymous
@KingGeorge
KingGeorge
  • KingGeorge
What exactly are you saying is false?
anonymous
  • anonymous
~p or q
anonymous
  • anonymous
actually i mean that one is true since p is false
KingGeorge
  • KingGeorge
Right. what about (p AND ~q)?
anonymous
  • anonymous
true
KingGeorge
  • KingGeorge
Remember that p is False, and ~q is True, so (p AND ~q) would actually be False.
KingGeorge
  • KingGeorge
Now here's the very last step. What is (~p OR q) AND (p AND ~q)?
anonymous
  • anonymous
false and false?
KingGeorge
  • 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.
KingGeorge
  • KingGeorge
If you want to know some more about truth tables, and how to construct them, feel free to ask.
anonymous
  • anonymous
ok great so that truth table is the whole answer?
anonymous
  • anonymous
thanks so much! i have other questions but there not about truth tables
KingGeorge
  • 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.
anonymous
  • anonymous
|dw:1355892045358:dw|
anonymous
  • anonymous
The number of rows in the truth table for the compound statement
anonymous
  • anonymous
would the answer be 8?
KingGeorge
  • KingGeorge
First off,how many variables do you have?
anonymous
  • anonymous
7?
KingGeorge
  • KingGeorge
I mean like p and q are variable. So in this context, it's a variable if it can be true or false.
KingGeorge
  • KingGeorge
How many of those are there?
anonymous
  • anonymous
4?
KingGeorge
  • 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.
anonymous
  • anonymous
so it would be 16?
KingGeorge
  • KingGeorge
yup.
anonymous
  • anonymous
great do you now anything about contrapositive statments?
KingGeorge
  • KingGeorge
Sure. If you have more questions though, please put it in a new question.
anonymous
  • anonymous
ok thanks!!

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