Corrected Version of Maths Where the Pythagorean Triples Don't Exist

This is a corrected version of Maths where the Pythagorean Triples Don’t Exist. 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 )) Premise 1 should read as for all K such that K is less than or equal to 1 there exist A,B,C such that A^K+B^K=C^K. Premise 1 is true because I could find natural numbers that satisfy premise 1. Premise 2 is true because it is equivalent to Premise 1 due to the commutative property. Premise 3 is true due to Premise 2. Premise 4 is true because it is the contrapositive of Premise 3. Therefore, ∀K(K>1→∀A,B,C(A^K+B^K≠C^K )), but this seems to contradict the fact that there are Pythagorean triples. Therefore, either the reasoning is invalid, or at least one premise is wrong, or there are maths where the Pythagorean triples don’t exist.