() 1F1>!(-1) take largest gF with disjoint members g≤r-1,letY=US|Y\≤k(r-1) S∈g claim:Yintersects all SEF if otherwise::3T∈F,T∩Y=0 T is disjoint with all SEg contradiction! |G| ⇥ r ￾ 1, Y = ￾ S￾G let S take largest G ￾ F with disjoint members |Y | ⇥ k(r ￾ 1) F ￾ ￾[n] k ⇥ . |F| > k!(r ￾ 1)k claim: Y intersects all S ￾ F if otherwise: ⇥T ￾ F, T ⇧ Y = ⇤ T is disjoint with all S ￾ G contradiction!