theorem
 Segm(n+1) --> e = (Segm n --> e)^<%e%>
 proof
  set p = Segm n --> e, q = Segm(n+1) --> e;
A2:  dom q = n+1
      .= len p + len <%e%> by Th31;
A3: for k st k in dom p holds q.k=p.k
    proof let k;
     assume
A4:  k in dom p;
     p c= q by FUNCT_4:4,NAT_1:63;
     hence q.k=p.k by A4,GRFUNC_1:2;
    end;
   for k st k in dom<%e%> holds q.(len p + k) = <%e%>.k
    proof let k such that
A5:    k in dom<%e%>;
A6:     k = 0 by A5,TARSKI:def 1;
      len p < n+1 by NAT_1:13;
      then len p + 0 in Segm(n+1) by NAT_1:44;
     hence q.(len p + k) = <%e%>.k by A6,FUNCOP_1:7;
    end;
  hence thesis by A2,A3,Def3;
 end;
