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

theorem Th12:
  m >= 1 & n >= 1 implies m+1 <= (n+m) choose m
proof
  defpred Q[Nat] means for n being Element of NAT st $1>=1 & n>=1 holds $1+1
  <= (n+$1) choose $1;
  assume
A1: m >= 1 & n >= 1;
  reconsider n9=n as Element of NAT by ORDINAL1:def 12;
  reconsider m9=m as Element of NAT by ORDINAL1:def 12;
A2: 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
A3: Q[k];
    for n being Element of NAT st k9>=1 & n>=1 holds k9+1 <= (n+k9) choose k9
    proof
      let n be Element of NAT;
      assume
A4:   k9>=1;
      assume
A5:   n>=1;
      per cases by A4,NAT_1:8;
      suppose
A6:     k+1=1;
        n+1 >= 0+1 & n+1 >= 1+1 by A5,XREAL_1:6;
        hence thesis by A6,NEWTON:23;
      end;
      suppose
A7:     k>=1;
        set k99=k+1;
A8:     (n+k9) choose k9 = (n+k+1) choose (k+1)
          .= (n+k) choose (k+1) + (n+k) choose k by NEWTON:22;
        k+1>=0+1 by XREAL_1:6;
        then
A9:     (n-'1) +1 <= (n-'1+k99) choose k99 by Th11;
        n-1 >= 1-1 by A5,XREAL_1:9;
        then n -' 1 = n - 1 by XREAL_0:def 2;
        then 1 <= (n+k) choose (k+1) by A5,A9,XXREAL_0:2;
        then
A10:    1+(k+1) <= (n+k) choose (k+1) + (k+1) by XREAL_1:6;
        (k+1) + (n+k) choose (k+1) <= (n+k9) choose k9
          by A3,A5,A7,A8,XREAL_1:6;
        hence thesis by A10,XXREAL_0:2;
      end;
    end;
    hence thesis;
  end;
A11: Q[0];
  for k being Nat holds Q[k] from NAT_1:sch 2(A11,A2);
  then m9+1 <= (n9+m9) choose m9 by A1;
  hence thesis;
end;
