reserve n,m,k for Nat,
  X,Y,Z for set,
  f for Function of X,Y,
  H for Subset of X;

theorem Th11:
  m >= 1 implies n+1 <= (n+m) choose m
proof
  defpred Q[Nat] means for n being Element of NAT st $1>=1 holds n+1 <= (n+$1)
  choose $1;
A1: for k being Nat st Q[k] holds Q[k+1]
  proof
    let k be Nat;
    set k9=k+1;
    reconsider k9 as Element of NAT;
    assume
A2: Q[k];
    for n being Element of NAT st k9>=1 holds n+1 <= (n+k9) choose k9
    proof
      let n be Element of NAT;
      assume
A3:   k9>=1;
      per cases by A3,NAT_1:8;
      suppose
A4:     k+1=1;
        n+1 >= 0+1 by XREAL_1:6;
        hence thesis by A4,NEWTON:23;
      end;
      suppose
A5:     k>=1;
        (n+k9) choose k9 = (n+k+1) choose (k+1)
          .= (n+k) choose (k+1) + (n+k) choose k by NEWTON:22; then
A7:     (n+1) + (n+k) choose (k+1) <= (n+k9) choose k9 by XREAL_1:6,A2,A5;
        0+(n+1) <= (n+k) choose (k+1) + (n+1) by XREAL_1:6;
        hence thesis by A7,XXREAL_0:2;
      end;
    end;
    hence thesis;
  end;
  reconsider n9=n,m9=m as Element of NAT by ORDINAL1:def 12;
  assume
A8: m >= 1;
A9: Q[0];
  for k being Nat holds Q[k] from NAT_1:sch 2(A9,A1);
  then n9+1 <= (n9+m9) choose m9 by A8;
  hence thesis;
end;
