theorem Th27:
  rng s c= dom h implies (h/*s)^\n=h/*(s^\n)
proof
  assume
A1: rng s c= dom h;
  let m be Element of NAT;
A2: rng (s^\n) c= rng s by Th21;
   reconsider mn = m+n as Element of NAT by ORDINAL1:def 12;
  thus ((h/*s)^\n).m = (h/*s).(m+n) by NAT_1:def 3
    .= h.(s.(mn)) by A1,FUNCT_2:108
    .= h.((s^\n).m) by NAT_1:def 3
    .= (h/*(s^\n)).m by A1,A2,FUNCT_2:108,XBOOLE_1:1;
end;
