let n, m be Nat; :: thesis: ( m >= 1 & n >= 1 implies m + 1 <= (n + m) choose m )

defpred S_{1}[ 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 ) ; :: thesis: m + 1 <= (n + m) choose m

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 S_{1}[k] holds

S_{1}[k + 1]
_{1}[ 0 ]
;

for k being Nat holds S_{1}[k]
from NAT_1:sch 2(A11, A2);

then m9 + 1 <= (n9 + m9) choose m9 by A1;

hence m + 1 <= (n + m) choose m ; :: thesis: verum

defpred S

$1 + 1 <= (n + $1) choose $1;

assume A1: ( m >= 1 & n >= 1 ) ; :: thesis: m + 1 <= (n + m) choose m

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 S

S

proof

A11:
S
let k be Nat; :: thesis: ( S_{1}[k] implies S_{1}[k + 1] )

set k9 = k + 1;

reconsider k9 = k + 1 as Element of NAT ;

assume A3: S_{1}[k]
; :: thesis: S_{1}[k + 1]

for n being Element of NAT st k9 >= 1 & n >= 1 holds

k9 + 1 <= (n + k9) choose k9_{1}[k + 1]
; :: thesis: verum

end;set k9 = k + 1;

reconsider k9 = k + 1 as Element of NAT ;

assume A3: S

for n being Element of NAT st k9 >= 1 & n >= 1 holds

k9 + 1 <= (n + k9) choose k9

proof

hence
S
let n be Element of NAT ; :: thesis: ( k9 >= 1 & n >= 1 implies k9 + 1 <= (n + k9) choose k9 )

assume A4: k9 >= 1 ; :: thesis: ( not n >= 1 or k9 + 1 <= (n + k9) choose k9 )

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

end;assume A4: k9 >= 1 ; :: thesis: ( not n >= 1 or k9 + 1 <= (n + k9) choose k9 )

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

per cases
( k + 1 = 1 or k >= 1 )
by A4, NAT_1:8;

end;

suppose A6:
k + 1 = 1
; :: thesis: k9 + 1 <= (n + k9) choose k9

( n + 1 >= 0 + 1 & n + 1 >= 1 + 1 )
by A5, XREAL_1:6;

hence k9 + 1 <= (n + k9) choose k9 by A6, NEWTON:23; :: thesis: verum

end;hence k9 + 1 <= (n + k9) choose k9 by A6, NEWTON:23; :: thesis: verum

suppose A7:
k >= 1
; :: thesis: k9 + 1 <= (n + k9) choose k9

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) + (k + 1)) choose (k + 1) 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 k9 + 1 <= (n + k9) choose k9 by A10, XXREAL_0:2; :: thesis: verum

end;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) + (k + 1)) choose (k + 1) 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 k9 + 1 <= (n + k9) choose k9 by A10, XXREAL_0:2; :: thesis: verum

for k being Nat holds S

then m9 + 1 <= (n9 + m9) choose m9 by A1;

hence m + 1 <= (n + m) choose m ; :: thesis: verum