证明：g(i) ≤ g(j) (i ≤ j)
令 d=g(i) , k<d ,
设cut = x表示 f(i) = f(x) + w[x,i] ( x < i )
构造一个式子：
( f(i) - f(i) ) - ( f(j) - f(j) )
cut=k cut=d cut=k cut=d
=( f(k) + w( k , i ) - f(d) - w( d , i ) ) - ( f(k) + w( k , j ) - f(d) - w( d , j ) )
=( w( k , i )+ w( d , j ) ) - ( w( k , j )+ w( d , i ) )
因为 k < d < i < j
所以
( w( k , i )+ w( d , j ) ) ≤ ( w( k , j )+ w( d , i ) )
( f(i) - f(i) ) - ( f(j) - f(j) ) ≤ 0
cut=k cut=d cut=k cut=d
又因为 d = g（i）
所以
( f(i) - f(i) ) ≥ 0
cut=k cut=d
( f(j) - f(j) ) ≥ 0
cut=k cut=d
f(j) ≥ f(j)
cut=k cut=d
又因为 k < d
所以 g(j)>=d=g(i)
证毕