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K =s(n) , and since this is true for all w ? L =n ,
, We are going to prove the "contrapositive inclusion
all distinct, we get that q |u|+j > q |u| . By (backward) induction, we can show that for all ? ?, we have q |u|+? > q |u| . Indeed, given ? ? ,
?1)·|u| w q1 · · · w q |u| w q 2|u|?1 w q 2|u| · · · w q 3|u|?2 w r1 · · · w r (???)·|u|+|x 2 | of ?(w). Therefore, p.1 ,
and since this is true for all w ? ? n \ L =n , we have ? n \ L =n ? ? ?1 n (A s(n) \ K =s(n) ) ,