theorem
  (f1/f2)|X = f1|X / f2|X & (f1/f2)|X = f1|X / f2 &(f1/f2)|X = f1 / f2|X
proof
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th38
    .= (f1|X) (#) (f2^|X) by Th52
    .= (f1|X) (#) ((f2|X)^) by Th53
    .= (f1|X)/(f2|X) by Th38;
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th38
    .= (f1|X) (#) (f2^) by Th52
    .= (f1|X)/f2 by Th38;
  thus (f1/f2)|X = (f1(#)(f2^))|X by Th38
    .= f1 (#) (f2^)|X by Th52
    .= f1 (#) ((f2|X)^) by Th53
    .= f1/(f2|X) by Th38;
end;
