theorem
  for B1,B2 being finite natural-membered set st B1 <N< B2 holds
  B1 misses B2
proof
  let B1,B2 be finite natural-membered set;
  assume
A1: B1 <N< B2;
  now
    set x = the Element of B1 /\ B2;
    assume a2: B1 meets B2; then
A3: x in B2 by XBOOLE_0:def 4;
    x in B1 by a2,XBOOLE_0:def 4;
    hence contradiction by A1,A3;
  end;
  hence thesis;
end;
