reserve n, k, r, m, i, j for Nat;

theorem Th52:
  for k being Element of NAT holds 2 * k + 1 in OddNAT & not 2 * k in OddNAT
proof
  let k be Element of NAT;
  thus 2 * k + 1 in OddNAT;
  assume 2 * k in OddNAT;
  then ex p being Element of NAT st 2 * k = 2 * p + 1;
  hence thesis;
end;
