reserve i,n,m,k,x for Nat,
  i1,i2 for Integer;

theorem Th6:
  0 in k-SD_Sub_S
proof
  defpred P[Nat] means 0 in $1-SD_Sub_S;
A1: for k being Nat st P[k] holds P[(k+1)]
  proof
    let kk be Nat;
    assume
A2: 0 in kk-SD_Sub_S;
    kk-SD_Sub_S c= (kk+1)-SD_Sub_S by Th3;
    hence thesis by A2;
  end;
A3: P[0] by Th5;
  for k being Nat holds P[k] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;
