
theorem Th6:
  for n being Nat holds 2*n choose n >= 4|^n / (2*n)
proof
  let n be Nat;
  per cases;
  suppose
    n = 0;
    hence thesis;
  end;
  suppose
A1: n <> 0;
    set m=[\ 2*n/2 /];
A2: 2|^2 = Product <* 2,2 *> by FINSEQ_2:61
      .= 2*2 by RVSUM_1:99
      .= 4;
    reconsider m as Element of NAT by ORDINAL1:def 12;
    2*n >= 2*1 by A1,NAT_1:14,XREAL_1:64;
    then (2*n) choose m >= 2|^(2*n) / (2*n) by Th5;
    hence thesis by A2,NEWTON:9;
  end;
end;
