I'm quite a noob, and this is a bit off-topic, but if it's mathematically proven that no sorting algorithm can be faster then O(n*log(n)), but we know for sure that checking if an array is sorted is O(n), then doesn't that prove that P ≠ NP?
No it doesn't, since O(n*log(n)) is in P too. P = NP doesn't imply that the complexity of finding the solution has the exact same complexity than verifying it, just that both are polynomial.
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I'm quite a noob, and this is a bit off-topic, but if it's mathematically proven that no sorting algorithm can be faster then O(n*log(n)), but we know for sure that checking if an array is sorted is O(n), then doesn't that prove that P ≠ NP?
No it doesn't, since O(n*log(n)) is in P too. P = NP doesn't imply that the complexity of finding the solution has the exact same complexity than verifying it, just that both are polynomial.