theorem
  for RNS being non empty 1-sorted, S being sequence of RNS st for n, k
  holds S.n = S.(n+k) holds for n, m holds S.n = S.m
proof
  let RNS be non empty 1-sorted;
  let S be sequence of RNS;
  assume
A1: for n, k holds S.n = S.(n+k);
  let n, m;
A2: now
    assume m <= n;
    then consider k being Nat such that
A3: n = m + k by NAT_1:10;
    reconsider k as Nat;
    n=m+k by A3;
    hence thesis by A1;
  end;
  now
    assume n <= m;
    then consider k being Nat such that
A4: m = n + k by NAT_1:10;
    reconsider k as Nat;
    m=n+k by A4;
    hence thesis by A1;
  end;
  hence thesis by A2;
end;
