theorem
  for RNS being non empty 1-sorted, S being sequence of RNS st for n, m
  holds S.n = S.m ex x being Element of RNS st for n being Nat holds S.n = x
proof
  let RNS be non empty 1-sorted;
  let S be sequence of RNS;
  assume that
A1: for n, m holds S.n = S.m and
A2: for x being Element of RNS ex n being Nat st S.n <> x;
  now
    let n;
    set z  = S.n;
    consider k being Nat such that
A3: S.k <> z by A2;
    thus contradiction by A1,A3;
  end;
  hence thesis;
end;
