reserve m, n for Nat;

theorem Th41:
  NatDivisors 1 = {1}
proof
  thus NatDivisors 1 c= {1}
  proof
    let x be object;
    assume x in NatDivisors 1;
    then consider k being Nat such that
A1: x = k and
    k <> 0 and
A2: k divides 1;
    k = 1 by A2,WSIERP_1:15;
    hence thesis by A1,ZFMISC_1:31;
  end;
  let x be object;
  assume
A3: x in {1};
  then reconsider m = x as Element of NAT;
  m <> 0 & m divides 1 by A3,TARSKI:def 1;
  hence thesis;
end;
