let F, G be Function of NAT ,ExtREAL ; :: thesis: for n being Nat st ( for m being Nat st m <= n holds
F . m <= G . m ) holds
(Ser F) . n <= (Ser G) . n
let n be Nat; :: thesis: ( ( for m being Nat st m <= n holds
F . m <= G . m ) implies (Ser F) . n <= (Ser G) . n )
assume A1: for m being Nat st m <= n holds
F . m <= G . m ; :: thesis: (Ser F) . n <= (Ser G) . n
defpred S₁[ Nat] means ( ( for m being Nat st m <= $1 holds
F . m <= G . m ) implies (Ser F) . $1 <= (Ser G) . $1 );

now

assume A2: for m being Nat st m <= 0 holds
F . m <= G . m ; :: thesis: (Ser F) . 0 <= (Ser G) . 0
( (Ser F) . 0 = F . 0 & (Ser G) . 0 = G . 0 ) by SUPINF_2:63;
hence (Ser F) . 0 <= (Ser G) . 0 by A2; :: thesis: verum

end;

then A3: S₁[ 0 ] ;
A4: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
A5: k is Element of NAT by ORDINAL1:def 13;
assume A6: S₁[k] ; :: thesis: S₁[k + 1]

now

assume A7: for m being Nat st m <= k + 1 holds
F . m <= G . m ; :: thesis: (Ser F) . (k + 1) <= (Ser G) . (k + 1)
then A8: F . (k + 1) <= G . (k + 1) ;

now

let m be Nat; :: thesis: ( m <= k implies F . m <= G . m )
assume m <= k ; :: thesis: F . m <= G . m
then m < k + 1 by NAT_1:13;
hence F . m <= G . m by A7; :: thesis: verum

end;

then ((Ser F) . k) + (F . (k + 1)) <= ((Ser G) . k) + (G . (k + 1)) by A6, A8, XXREAL_3:38;
then (Ser F) . (k + 1) <= ((Ser G) . k) + (G . (k + 1)) by A5, SUPINF_2:63;
hence (Ser F) . (k + 1) <= (Ser G) . (k + 1) by A5, SUPINF_2:63; :: thesis: verum

end;

hence S₁[k + 1] ; :: thesis: verum

end;

for k being Nat holds S₁[k] from NAT_1:sch 2(A3, A4);
hence (Ser F) . n <= (Ser G) . n by A1; :: thesis: verum