Question

Need a little help with some propositional logic. Image coming..

Answer 1

"The negation of a contradiction is a tautology."
\[forallx(C(x) \rightarrow T((-)x)\] ... T(x) : x is a tautology. C(x): x is a contradiction

x = {propositions}

Answer 2

sorry about the poor form. I couldnt find the negation symbol..if there is one

Answer 3

I'm just having a little trouble understanding how the symbolic representation relates to the english

Answer 4

If x is true for C(x), but not true for T(x), the conditional is false

Answer 5

so I don't understand why it is true that the proposition is a tautology

Answer 6

The negation of a contradiction is a tautology, okay, this doesn't necessarily have to apply for prepositional logic, you may or may not have heard about the indirect proof? For example, Euclides proof that there are infinite prime numbers is such a proof.

Assume you have to show a statement \(q\) and you want to show that it's true, you can do that by assuming the opposite, that it's false and get it into a contradiction, that would mean you set the following implications:
\[\Large \neg q \Rightarrow F \]
Where F is false, now you know that the implication doesn't work in both ways, like the converse of an implication is wrong, but you can always use the contrapositive,
Here is the contrapositive of a statement:
\[\Large p \Rightarrow q \Longleftrightarrow \neg q \Rightarrow \neg p \]
Apply this contradiction to "The negation of a contradiciton is a tautology" this will get you:
\[\Large \neg F \Rightarrow \neg \neg q \]
or \[\Large T \Rightarrow q \]

Answer 7

by the way, you can negate statement using \neg, maybe, if my answer didn't help you yet, you can try rewriting it with \(\mathbb{LaTeX}\) it often helps

Answer 8

I'm just beginning to study this subject, so your answer is a little ahead of where i'm at unfortunately. I'm just really confused by the specific form of this question from my book. It's actually the first time I've seen x negated in that way: T(-x). But it is supposed to be true for all x, and if the proposition is true for C(x) , but is then always negated by T(-x), the consequent will always be false, right?

Answer 9

Sorry if I'm being unclear, I've only been studying this subject for about a week.

Answer 10

Sorry, not always false, but always false if the proposition is a contradiction..but then if every proposition that is a contradiction is negated, how can the conditional ever be true?

Answer 11

let me try to rewrite the statement and then see if I can dig a bit deeper into it before I confuse things more :)
\[\Large \forall x: (C(x) \rightarrow T(\neg x)) \]
where C is the contradiction and T is the tautology, neg is the negation of the preposition, like for instance "is an even number"

Answer 12

yes thats right

Answer 13

The way I read it: If its true that the prop. is a contradiction, than it will become false on the side of the consequent , so the conditional is not true. If it is false that it is a contradiction, then it will be true on the side of the consequent and the conditional will be true. But if it was false that the prop. was a contradiction, I don't understand how it can be true that its negation is a tautology. I'm so lost :P

Answer 14

sorry if I'm offending logic with my babblings :P

Answer 15

I understand where your problem is at I believe, it's always a good thing to questions such definitions, I haven't seen such a notation myself to be honest.

However, I believe that it all breaks down to the definition of the implication, so just the way you're already trying to go. A preposition is a statement that depends upon it's argument if it's true or not, like x=is a prime number, this is only true for various numbers. Such as 2, 3, 5 and so on. So a contradiction would always turn that statement into the opposite from how I understand this problem.

Like x= is a prime number
C(5) = False
C(6) = True

Is this what you're meant with the statement of the prepositons? :)

Answer 16

I believe that however, the contradiction has always give out a false output, because that is what a contradiction is.

Answer 17

Right.. I understand why it's true, I just get lost when i look at this particular notation.

Answer 18

I think it just might be a strange problem :P

Answer 19

Luckily I have it memorized now from bothering with it for so long, so I'll just move on lol

Answer 20

can you think of some other way to write it symbolically that would be easier to understand?

Answer 21

THat's really good, if you don't mind keeping that question open for a while/today, I will look later into it again and see if I can come up with a similar problem. I am pretty sure that it all breaks down on this on the definition on the implication, the implication is only wrong that if the premise is true and the consequent is false. So we would need to find a suitable application of that problem.

I just never dealt with notation such as a "Contradiction" function and a "Tautology" function, the most naive way to deal with this would be to imagine the entire left hand side as false and the entire right hand side as true.

Answer 22

Sure no problem. I appreciate the help. I've had this one written down for a few days now.

Answer 23

Is a contradiction always false?

Answer 24

Yes

Answer 25

A way to express a contradiction is to say: \[
\forall x\; (p(x)\iff \bot)
\]

Answer 26

So you could say \[
(\forall x (p(x)\iff \bot)) \implies (\forall x (\neg p(x)\iff \top))
\]

Answer 27

The best way to simplify an implication: \[
p\implies q \iff\neg p\vee q
\]