theorem
  for RNS being non empty 1-sorted, S being sequence of RNS st ex x
  being Element of RNS st rng S = {x} holds for n holds S.n = S.(n+1)
proof
  let RNS be non empty 1-sorted;
  let S be sequence of RNS;
  given z being Element of RNS such that
A1: rng S = {z};
  let n;
  n in NAT by ORDINAL1:def 12;
  then n in dom S by FUNCT_2:def 1;
  then S.n in {z} by A1,FUNCT_1:def 3;
  then
A2: S.n = z by TARSKI:def 1;
  (n+1) in NAT;
  then (n+1) in dom S by FUNCT_2:def 1;
  then S.(n+1) in {z} by A1,FUNCT_1:def 3;
  hence thesis by A2,TARSKI:def 1;
end;
