let X be non empty set ; :: thesis: for F being Functional_Sequence of X,REAL
for x being Element of X
for n, m being Nat st F is with_the_same_dom & x in dom (F . 0) & ( for k being Nat holds F . k is nonnegative ) & n <= m holds
(() . n) . x <= (() . m) . x

let F be Functional_Sequence of X,REAL; :: thesis: for x being Element of X
for n, m being Nat st F is with_the_same_dom & x in dom (F . 0) & ( for k being Nat holds F . k is nonnegative ) & n <= m holds
(() . n) . x <= (() . m) . x

let x be Element of X; :: thesis: for n, m being Nat st F is with_the_same_dom & x in dom (F . 0) & ( for k being Nat holds F . k is nonnegative ) & n <= m holds
(() . n) . x <= (() . m) . x

let n, m be Nat; :: thesis: ( F is with_the_same_dom & x in dom (F . 0) & ( for k being Nat holds F . k is nonnegative ) & n <= m implies (() . n) . x <= (() . m) . x )
assume A1: F is with_the_same_dom ; :: thesis: ( not x in dom (F . 0) or ex k being Nat st not F . k is nonnegative or not n <= m or (() . n) . x <= (() . m) . x )
assume A2: x in dom (F . 0) ; :: thesis: ( ex k being Nat st not F . k is nonnegative or not n <= m or (() . n) . x <= (() . m) . x )
assume A3: for m being Nat holds F . m is nonnegative ; :: thesis: ( not n <= m or (() . n) . x <= (() . m) . x )
assume A4: n <= m ; :: thesis: (() . n) . x <= (() . m) . x
set PF = Partial_Sums F;
defpred S1[ Nat] means (() . n) . x <= (() . \$1) . x;
A5: for k being Nat holds (() . k) . x <= (() . (k + 1)) . x
proof
let k be Nat; :: thesis: (() . k) . x <= (() . (k + 1)) . x
A6: (Partial_Sums F) . (k + 1) = (() . k) + (F . (k + 1)) by MESFUN9C:def 2;
A7: dom (() . (k + 1)) = dom (F . 0) by ;
( F . (k + 1) is nonnegative & () . k is nonnegative ) by ;
then ( 0 <= (F . (k + 1)) . x & 0 <= (() . k) . x ) by MESFUNC6:51;
then (((Partial_Sums F) . k) . x) + 0 <= ((() . k) . x) + ((F . (k + 1)) . x) by XREAL_1:7;
hence ((Partial_Sums F) . k) . x <= (() . (k + 1)) . x by ; :: thesis: verum
end;
A8: for k being Nat st k >= n & ( for l being Nat st l >= n & l < k holds
S1[l] ) holds
S1[k]
proof
let k be Nat; :: thesis: ( k >= n & ( for l being Nat st l >= n & l < k holds
S1[l] ) implies S1[k] )

assume A9: ( k >= n & ( for l being Nat st l >= n & l < k holds
S1[l] ) ) ; :: thesis: S1[k]
now :: thesis: ( k > n implies S1[k] )
assume k > n ; :: thesis: S1[k]
then k >= n + 1 by NAT_1:13;
then A10: ( k = n + 1 or k > n + 1 ) by XXREAL_0:1;
now :: thesis: ( k > n + 1 implies S1[k] )
assume A11: k > n + 1 ; :: thesis: S1[k]
then reconsider l = k - 1 as Nat by NAT_1:20;
k < k + 1 by NAT_1:13;
then ( k > l & l >= n ) by ;
then A12: ((Partial_Sums F) . n) . x <= (() . l) . x by A9;
k = l + 1 ;
then ((Partial_Sums F) . l) . x <= (() . k) . x by A5;
hence S1[k] by A12, XXREAL_0:2; :: thesis: verum
end;
hence S1[k] by A10, A5; :: thesis: verum
end;
hence S1[k] by A9, XXREAL_0:1; :: thesis: verum
end;
for k being Nat st k >= n holds
S1[k] from hence ((Partial_Sums F) . n) . x <= (() . m) . x by A4; :: thesis: verum