Fermat's Last Theorem and the Principle of Duality

2 points by Kabbalist ↗ HN
Let A,B,C,K be natural numbers. Let T signify true. Let F signify false. So we have ∀K(K≤2→∃A,B,C(A^K+B^K=C^K ))=T. But this is equivalent to ∀K(K>2∨∃A,B,C(A^K+B^K=C^K ))=T. By the principle of duality we then have ∀K(K>2∧∃A,B,C(A^K+B^K=C^K ))=F . Also consider the following. Again, let A,B,C,K be natural numbers. ∃K(K≤2∧∃A,B,C(A^K+B^K=C^K ))=T. By the principle of duality we then have ∃K(K≤2∨∃A,B,C(A^K+B^K=C^K ))=F. But this is equivalent to ∃K(K>2→∃A,B,C(A^K+B^K=C^K ))=F

2 comments

[ 2.7 ms ] story [ 14.3 ms ] thread
And just with a single post you've demonstrated why Hacker news isn't the place to paste an equation. Anyway, interesting.
Are you able to read and understand it though and is it the correct use of the principle of duality?