> An NP-complete problem is NP-hard, but it's also guaranteed to be an NP problem. That's the distinction. My point being that transforming tsp-opt to tsp only

> An NP-complete problem is NP-hard, but it's also guaranteed to be an NP problem. That's the distinction. My point being that transforming tsp-opt to tsp only adds a polynomial factor. So in the case of tsp-opt, it's called NP-hard rather than NP-complete because it's not a decision problem, not because it's not equivalent in terms of time complexity as a problem in NP. People simply call it NP-hard because that term is better known than NPO or NP-equivalent.