theorem Th50:
  (for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2)) implies
    h*(F[:](T,d)) = H[:](h*T,h.d)
proof
  assume
A1: for d1,d2 holds h.(F.(d1,d2)) = H.(h.d1,h.d2);
  per cases;
  suppose
A2: i = 0;
    then F[:](T,d) = <*>D by Lm3;
    then
A3: h*(F[:](T,d)) = <*>E;
    h*T = <*>E by A2;
    hence thesis by A3,FINSEQ_2:85;
  end;
  suppose
    i <> 0;
    then reconsider C = Seg i as non empty set;
    T is Function of C,D by Lm4;
    hence thesis by A1,Th39;
  end;
end;
