Lemma 36.3

If L₁, L₂$\subseteq$ {0,1}*, and , then L₂$\in$ P implies L₁$\in$ P.

For any instance of L₁
$\;\;\;\;\;$map to L₂ (poly time)
$\;\;\;\;\;$solve L₂ (poly time)

Thus if we can solve L₂ in poly time we can solve L₁ in poly time.