Does hardness of a special case imply hardness of a general case? I am wondering in general if it is correct to claim that if a special case of a problem is NP-Hard, the general case of that problem is NP-
Hard too?

For example: Min Set-Cover is NP-Hard does it imply without having to prove that Min Set-k-Cover where each element needs to be covered k times is NP-Hard too?
If frafl's answer doesn't give it away immediately, perhaps think in the reverse direction, if the special case is NP-hard, what would be implied by the general case being in P?