3 ms·
Your formalization is wrong. In fact, it's nonsensical. The predicate variable K should be bound by the existential quantifier associated with the proposition A
by p-e-w 3y ago
Your formalization is wrong. In fact, it's nonsensical. The predicate variable K should be bound by the existential quantifier associated with the proposition A^K+B^K=C^K. You cannot just separate K from the proposition with a conjunction operator and then shuffle them around, arguing from commutativity. In your "Premise 2", K is not bound at all in the first conjunction operand. You want K to mean the same thing in both operands, and with your formulas, that's not the case.
There are other problems here, but that's the most glaring one.
- Kabbalist 3y agoYes. 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.