# Why do i have to show that a problem L $\in$ NP before i start a polynomial reduction?

i want to do a polynomial reduction from the Independent-Set-Problem which is NP-complete to the AUCTION-Problem, to show that AUCTION $\in$ NP-complete, but why do i always have to show first, that our problem(in this case AUCTION) $\in$ NP?

• @M.Mac No. By successfully reducing an NP-complete problem $L$ to another problem $K$, you've proved that $K$ is NP-hard. NP-complete means NP-hard and in NP. In particular, the halting problem is NP-hard but it certainly isn't in NP, so it's not NP-complete. – David Richerby Feb 6 '17 at 15:55
• @DavidRicherby I see, but is NP-hard not defined as, all problems L $\in$ NP which are succesfully reduced to another problem K? Why does it not implicate that K is in NP? – M.Mac Feb 6 '17 at 18:27