reserve X,Y,z,s for set, L,L1,L2,A,B for List of X, x for Element of X,
  O,O1,O2,O3 for Operation of X, a,b,y for Element of X, n,m for Nat;

theorem
  n <= m & L1 c= L2 & O1 c= O2 implies L1 WHERElt(O2,n) c= L2 WHERElt(O1,m)
  proof
    assume
A1: n <= m & L1 c= L2 & O1 c= O2;
    let z be object; assume z in L1 WHERElt(O2,n); then
    consider x such that
A2: z = x & card(x.O2) in n & x in L1;
    Segm n c= Segm m & card(x.O1) c= card(x.O2) by A1,Th1,CARD_1:11,NAT_1:39;
then
    card(x.O1) in m by A2,ORDINAL1:12;
    hence thesis by A1,A2;
  end;
