theorem Th42:
  (f1/f2)^ = (f2|dom(f2^))/f1
proof
  thus (f1/f2)^ = (f1(#)(f2^))^ by Th38
    .= (f1^)(#)(f2^^) by Th34
    .= (f2|dom(f2^))(#)(f1^) by Th33
    .= (f2|dom(f2^))/f1 by Th38;
end;
