[deprosun]

"((s → c) and ((not s) → c)) → c" using Proof by Contradiction. What should be assumed?

• If s, c, and if (not s), c, and (not c): This just doesn't make sense to me because why "and c"..?
  
  
• not c: Contradiction should assume that whatever happens, c is not the case. i think. correct?
  
  
• Will it be this: c or (not c), or (c and (not c))?
  
  
• If s, c, and if (not s), c: This cant be it because, its just the same statement.

This post has been edited by deprosun: 15 September 2013 - 04:55 PM

[macosxnerd101]

If you are trying to prove that p -> q using proof by contradiction, then you assume p and ~q. Then you should that this is invalid, so it must be the case that p -> q. So for your statement, you would assume what is in the parentheses, and you would assume ~c.

[deprosun]

macosxnerd101, on 15 September 2013 - 10:08 PM, said:

If you are trying to prove that p -> q using proof by contradiction, then you assume p and ~q. Then you should that this is invalid, so it must be the case that p -> q. So for your statement, you would assume what is in the parentheses, and you would assume ~c.

Yeah, so seems like it IS number 1. But why?
If s, c, and if (not s), c, and (not c)

It says "and (not c)". shouldn't it be an implication?

[macosxnerd101]

An implication p -> q says that if p, then q. So if p is true, then q is true. The implication isn't valid if p is true and q isn't. Think of it like a contract. If p is true, then q follows contractually. So if p happens but q doesn't follow, then the implication isn't valid. That's the starting point.

[jon.kiparsky]

Bob asserts "If it is Sunday, then it must be raining". However, it is Sunday and it is not raining. Therefore Bob's assertion is false.

More related to your problem at hand:

"If Bob says it's raining in Portland, then it must be raining in Portland. On the other hand, if Bob doesn't say it's raining in Portland, then it still must be raining in Portland."

How can we prove from this that it is raining in Portland? Specifically, if we want to prove by contradiction, how do we go about it? The process is to set up a provisional world and show it to be inconsistent: what provisional world do we set up?

This post has been edited by jon.kiparsky: 15 September 2013 - 08:45 PM

[deprosun]

jon.kiparsky, on 15 September 2013 - 10:36 PM, said:

Bob asserts "If it is Sunday, then it must be raining". However, it is Sunday and it is not raining. Therefore Bob's assertion is false.

More related to your problem at hand:

"If Bob says it's raining in Portland, then it must be raining in Portland. On the other hand, if Bob doesn't say it's raining in Portland, then it still must be raining in Portland."

How can we prove from this that it is raining in Portland? Specifically, if we want to prove by contradiction, how do we go about it? The process is to set up a provisional world and show it to be inconsistent: what provisional world do we set up?

So we'd assume: If Bob says it's raining in Portland, then it must be raining in Portland, and if Bob doesn't say it's raining in Portland, then it still must be raining in Portland, then It is not raining in portland. ?

[mojo666]

The general idea is that if the statement is valid, then a false conclusion would require one or more premises to also be false. Otherwise, the statement is invalid. To perform a proof by contradiction, you assume the opposite of the conclusion (in this case assume ~c), and then using logical deduction, derive something rediculous (usually something of the form x and ~x). This rediculous statement proves that the premises are false, so the statement is valid.

We assume c is false and try to derive something rediculous from the premises (s -> c) and (~s -> c)

[deprosun]

I think I got it!