Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Have you been taught how to do truth tables?
no
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

ok
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?
false
right. Now, ~p is False, and q is true. What is ~p OR q?
false
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?
I dont get it..
true
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?
i guess
Alright, stop me if you start to not understand anything. Now, q is True. So what is ~q?
false
Precisely. Now, recall that p is True, and ~q is False. Can you tell me what (p AND ~q) is?
false
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...?
trur
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?
false true
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?
true
right. what about (p AND ~q)?
false since q is true
Bingo. Then, what about (~p OR q) AND (p AND ~q)?
true
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.
Good catch^^
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?
false
What exactly are you saying is false?
~p or q
actually i mean that one is true since p is false
Right. what about (p AND ~q)?
true
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)?
false and false?
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?
7?
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?
4?
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?
yup.
great do you now anything about contrapositive statments?
Sure. If you have more questions though, please put it in a new question.
ok thanks!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question