
theorem Th6:
  for n being Nat holds 0 in dyadic(n) & 1 in dyadic(n)
proof
  defpred P[Nat] means 0 in dyadic($1) & 1 in dyadic($1);
A1: for k being Nat st P[k] holds P[(k+1)]
  proof
    let k be Nat;
    assume
A2: 0 in dyadic(k) & 1 in dyadic(k);
    dyadic(k) c= dyadic(k+1) by Th5;
    hence thesis by A2;
  end;
A3: P[0] by Th2,TARSKI:def 2;
  for k be Nat holds P[k] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;
