theorem
  (f/g)/(f1/g1) = (f(#)(g1|dom(g1^)))/(g(#)f1)
proof
  thus (f/g)/(f1/g1) = (f/g)(#)((f1/g1)^) by Th38
    .= (f/g)(#)(((g1|dom(g1^)))/f1) by Th42
    .= (f(#)(g1|dom(g1^)))/(g(#)f1) by Th41;
end;
