3 ms·
Yes. You’re right. I should have written it like this: Let A,B,C, and K be elements of the Natural Numbers. 1. ∃K(K≤1∧∃A,B,C(A^K+B^K=C^K )) 2. ∃K(∃A,B,C(A^K+B^K
by Kabbalist 3y ago
Yes. You’re right. I should have written it like this:
Let A,B,C, and K be elements of the Natural Numbers.
1. ∃K(K≤1∧∃A,B,C(A^K+B^K=C^K ))
2. ∃K(∃A,B,C(A^K+B^K=C^K )∧K≤1)
3. ∃K(∃A,B,C(A^K+B^K=C^K )→K≤1)
4. ∃K(K>1→∀A,B,C(A^K+B^K≠C^K ))
- lukenels 3y agoPremise 4 is trivially true for the same reason premise 3 is: choose K=1, then K>1 is false, and false implies anything.
- Kabbalist 3y agoSuppose I would pick K=2 though. What would happen then?
- psyklic 3y ago#3 and #4 say there exists a K for which the interior statement is true. K=2 doesn't make the interior true for either, so choosing it doesn't tell us whether #3 and #4 are true or false. However, if we choose K=1 then we prove #3 and #4 are true.