show that the statement (~p V q) ^ (p^~q) is a logical contradiction

- anonymous

show that the statement (~p V q) ^ (p^~q) is a logical contradiction

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- schrodinger

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- KingGeorge

Have you been taught how to do truth tables?

- anonymous

no

- 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

ok

- anonymous

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

- 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

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

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

- 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

false

- KingGeorge

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

- anonymous

false

- 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

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

- anonymous

I dont get it..

- anonymous

true

- 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

i guess

- KingGeorge

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

- anonymous

false

- KingGeorge

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

- anonymous

false

- KingGeorge

oops, I screwed up a bit.

- KingGeorge

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

- anonymous

trur

- 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

false true

- 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

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

true

- KingGeorge

right. what about (p AND ~q)?

- anonymous

false since q is true

- KingGeorge

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

- anonymous

true

- 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

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

Good catch^^

- 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

false

- anonymous

@KingGeorge

- KingGeorge

What exactly are you saying is false?

- anonymous

~p or q

- anonymous

actually i mean that one is true since p is false

- KingGeorge

Right. what about (p AND ~q)?

- anonymous

true

- KingGeorge

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

- KingGeorge

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

- anonymous

false and false?

- 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

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

- anonymous

ok great so that truth table is the whole answer?

- anonymous

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

- 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

|dw:1355892045358:dw|

- anonymous

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

- anonymous

would the answer be 8?

- KingGeorge

First off,how many variables do you have?

- anonymous

7?

- 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

How many of those are there?

- anonymous

4?

- 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

so it would be 16?

- KingGeorge

yup.

- anonymous

great do you now anything about contrapositive statments?

- KingGeorge

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

- anonymous

ok thanks!!

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