A problem $\pi\in\mathcal{NP}$ is $\mathcal{NP}$complete if for any $\pi^\prime\in\mathcal{NP}$ there is a Cook reduction of $\pi^\prime$ to $\pi$ Hence if $\pi\in\mathcal{P}$ then every $\mathcal{NP}$ problem would be in $\mathcal{P}$ A slightly stronger definition requires a Karp reduction or Karp reduction of corresponding decision problems as appropriate.
A search problem$R$ is $\mathcal{NP}$ hard if for any $R^\prime\in\mathcal{NP}$ there is a Levin reduction of $R^\prime$ to $R$
