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>Take two cases. This is where you are implicitly employing proof by contradiction. Why do you take two cases and not three? Or four? What assumption allows yo

by Retra 12y ago

>Take two cases. This is where you are implicitly employing proof by contradiction. Why do you take two cases and not three? Or four? What assumption allows you to restrict your conclusions like that? Probably the Law of the Excluded Middle: (P or ~P). (This is equivalent to ~~p = p.) Once you assume or derive either of those, you have validated proof by contradiction.

ikeboy 12y ago
I proved that x has an even number of factors of 2. Why can't I then split that into scenarios when that even number is 0, and cases where it isn't? I don't see any restrictions here. Let's imagine I had shown that something was either one or two. You seem to be claiming I could not take positive proofs assuming each of them and turn it into a full proof.
- ebola1717 12y ago
  >Let's imagine I had shown that something was either one or two. You seem to be claiming I could not take positive proofs assuming each of them and turn it into a full proof. When you break it down to the logical axioms, you can't use cases like that. Try it yourself. You get something like this: Let A: (k > 0) B: (k = 0) k is a whole number, therefore A v B. A -> X !=2 B -> X !=2 Okay, from here, you need to deduce X != 2. Traditionally we do this by applying proof by cases, but as I showed in my other comment, this requires the law of the excluded middle. I don't see any other propositional logic axioms we can apply to achieve the same result.
  - ikeboy 12y ago
    >Traditionally we do this by applying proof by cases, but as I showed in my other comment, this requires the law of the excluded middle. I think that's because you don't know A v B. But in my case you can directly prove A v B. The form is (A v B, A -> C, B -> C) -> C. https://en.wikipedia.org/wiki/Proof_by_exhaustion https://en.wikipedia.org/wiki/Proof_by_exhaustion says that this is called a direct proof.

by Retra 12y ago

ikeboy 12y ago

I proved that x has an even number of factors of 2. Why can't I then split that into scenarios when that even number is 0, and cases where it isn't? I don't see any restrictions here. Let's imagine I had shown that something was either one or two. You seem to be claiming I could not take positive proofs assuming each of them and turn it into a full proof.

ebola1717 12y ago
>Let's imagine I had shown that something was either one or two. You seem to be claiming I could not take positive proofs assuming each of them and turn it into a full proof. When you break it down to the logical axioms, you can't use cases like that. Try it yourself. You get something like this: Let A: (k > 0) B: (k = 0) k is a whole number, therefore A v B. A -> X !=2 B -> X !=2 Okay, from here, you need to deduce X != 2. Traditionally we do this by applying proof by cases, but as I showed in my other comment, this requires the law of the excluded middle. I don't see any other propositional logic axioms we can apply to achieve the same result.
- ikeboy 12y ago
  >Traditionally we do this by applying proof by cases, but as I showed in my other comment, this requires the law of the excluded middle. I think that's because you don't know A v B. But in my case you can directly prove A v B. The form is (A v B, A -> C, B -> C) -> C. https://en.wikipedia.org/wiki/Proof_by_exhaustion https://en.wikipedia.org/wiki/Proof_by_exhaustion says that this is called a direct proof.