theorem
  F is commutative implies F[;](d,T) = F[:](T,d)
proof
  assume
A1: F is commutative;
  per cases;
  suppose
A2: i = 0;
    then F[;](d,T) = <*>D by Lm2;
    hence thesis by A2,Lm3;
  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,FUNCOP_1:64;
  end;
end;
