theorem
  for RNS being non empty 1-sorted, S being sequence of RNS st for n
  holds S.n = S.(n+1) holds for n, k holds S.n = S.(n+k)
proof
  let RNS be non empty 1-sorted;
  let S be sequence of RNS;
  assume
A1: for n holds S.n = S.(n+1);
  let n;
  defpred P[Nat] means S.n = S.(n+$1);
A2: now
    let k such that
A3: P[k];
    S.(n+k) = S.(n+k+1) by A1;
    hence P[k+1] by A3;
  end;
A4: P[0];
  thus for k holds P[k] from NAT_1:sch 2(A4,A2);
end;
