theorem
  g/(f1/f2) = (g(#)(f2|dom(f2^)))/f1
proof
  thus g/(f1/f2) = g (#) ((f1/f2)^) by Th31
    .= g (#) ((f2|dom(f2^))/f1) by Th35
    .= (g (#) (f2|dom(f2^)))/f1 by Th36;
end;
