let n, m be Nat; :: thesis: ( m >= 1 implies n + 1 <= (n + m) choose m )
defpred S₁[ Nat] means for n being Element of NAT st $1 >= 1 holds
n + 1 <= (n + $1) choose $1;
A1: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
set k9 = k + 1;
reconsider k9 = k + 1 as Element of NAT ;
assume A2: S₁[k] ; :: thesis: S₁[k + 1]
for n being Element of NAT st k9 >= 1 holds
n + 1 <= (n + k9) choose k9

proof

let n be Element of NAT ; :: thesis: ( k9 >= 1 implies n + 1 <= (n + k9) choose k9 )
assume A3: k9 >= 1 ; :: thesis: n + 1 <= (n + k9) choose k9

per cases ( k + 1 = 1 or k >= 1 ) by A3, NAT_1:8;

suppose A4: k + 1 = 1 ; :: thesis: n + 1 <= (n + k9) choose k9

n + 1 >= 0 + 1 by XREAL_1:6;
hence n + 1 <= (n + k9) choose k9 by A4, NEWTON:23; :: thesis: verum

end;

suppose A5: k >= 1 ; :: thesis: n + 1 <= (n + k9) choose k9

(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 n + 1 <= (n + k9) choose k9 by A7, XXREAL_0:2; :: thesis: verum

end;

end;

end;

hence S₁[k + 1] ; :: thesis: verum

end;

reconsider n9 = n, m9 = m as Element of NAT by ORDINAL1:def 12;
assume A8: m >= 1 ; :: thesis: n + 1 <= (n + m) choose m
A9: S₁[ 0 ] ;
for k being Nat holds S₁[k] from NAT_1:sch 2(A9, A1);
then n9 + 1 <= (n9 + m9) choose m9 by A8;
hence n + 1 <= (n + m) choose m ; :: thesis: verum