theorem Th38:
  (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies
    h*(F[;](d,f)) = H[;](h.d,h*f)
proof
  assume
A1: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  reconsider g = C --> d as Function of C,D;
A2: dom h = D & dom(h*f) = C by FUNCT_2:def 1;
  thus h*(F[;](d,f)) = h*(F.:(g,f)) by FUNCT_2:def 1
    .= H.:(h*g,h*f) by A1,Th37
    .= H[;](h.d,h*f) by A2,FUNCOP_1:17;
end;
