theorem
  C <> {} & A c= B implies exp(C,A) c= exp(C,B)
proof
A1: A c< B iff A c= B & A <> B;
  assume C <> {};
  then {} in C by ORDINAL3:8;
  then
A2: 1 c= C by Lm3,ORDINAL1:21;
  assume A c= B;
  then
A3: A in B or A = B by A1,ORDINAL1:11;
  now
    per cases;
    suppose
A4:   C = 1;
      then exp(C,A) = 1 by ORDINAL2:46;
      hence thesis by A4,ORDINAL2:46;
    end;
    suppose
      C <> 1;
      then 1 c< C by A2;
      then 1 in C by ORDINAL1:11;
      then exp(C,A) in exp(C,B) or exp(C,A) = exp(C,B) by A3,Th24;
      hence thesis by ORDINAL1:def 2;
    end;
  end;
  hence thesis;
end;
