reserve a,b,m,x,y,i1,i2,i3,i for Integer,
  k,p,q,n for Nat,
  c,c1,c2 for Element of NAT,
  z for set;

theorem Th6:
  i divides i1 & i1<>0 implies |.i.|<=|.i1.|
proof
  assume i divides i1 & i1<>0;
  then |.i1.|<>0 & |.i.| divides |.i1.| by ABSVALUE:2,INT_2:16;
  hence thesis by NAT_D:7;
end;
