theorem Th35:
  (f1/f2)^ = (f2|dom(f2^))/f1
proof
  thus (f1/f2)^ = (f1(#)(f2^))^ by Th31
    .= (f1^)(#)(f2^^) by Th27
    .= (f2|dom(f2^))(#)(f1^) by Th26
    .= (f2|dom(f2^))/f1 by Th31;
end;
