let X be non empty set ; :: thesis: for F being Functional_Sequence of X,ExtREAL
for n being Nat
for x being Element of X
for D being set st F is additive & F is with_the_same_dom & D c= dom (F . 0) & x in D holds
(Partial_Sums (F # x)) . n = ((Partial_Sums F) # x) . n
let F be Functional_Sequence of X,ExtREAL; :: thesis: for n being Nat
for x being Element of X
for D being set st F is additive & F is with_the_same_dom & D c= dom (F . 0) & x in D holds
(Partial_Sums (F # x)) . n = ((Partial_Sums F) # x) . n
let n be Nat; :: thesis: for x being Element of X
for D being set st F is additive & F is with_the_same_dom & D c= dom (F . 0) & x in D holds
(Partial_Sums (F # x)) . n = ((Partial_Sums F) # x) . n
let x be Element of X; :: thesis: for D being set st F is additive & F is with_the_same_dom & D c= dom (F . 0) & x in D holds
(Partial_Sums (F # x)) . n = ((Partial_Sums F) # x) . n
let D be set ; :: thesis: ( F is additive & F is with_the_same_dom & D c= dom (F . 0) & x in D implies (Partial_Sums (F # x)) . n = ((Partial_Sums F) # x) . n )
set PF = Partial_Sums F;
set PFx = Partial_Sums (F # x);
assume that
A1: F is additive and
A2: F is with_the_same_dom and
A3: D c= dom (F . 0) and
A4: x in D ; :: thesis: (Partial_Sums (F # x)) . n = ((Partial_Sums F) # x) . n
defpred S1[ Nat] means (Partial_Sums (F # x)) . $1 = ((Partial_Sums F) # x) . $1;
A5: for k being Nat st S1[k] holds
S1[k + 1]
proof
let k be Nat; :: thesis: ( S1[k] implies S1[k + 1] )
assume A6: S1[k] ; :: thesis: S1[k + 1]
(Partial_Sums (F # x)) . (k + 1) = ((Partial_Sums (F # x)) . k) + ((F # x) . (k + 1)) by Def1;
then (Partial_Sums (F # x)) . (k + 1) = (((Partial_Sums F) # x) . k) + ((F . (k + 1)) . x) by A6, MESFUNC5:def 13;
then A7: (Partial_Sums (F # x)) . (k + 1) = (((Partial_Sums F) . k) . x) + ((F . (k + 1)) . x) by MESFUNC5:def 13;
A8: (Partial_Sums F) . (k + 1) = ((Partial_Sums F) . k) + (F . (k + 1)) by Def4;
D c= dom ((Partial_Sums F) . (k + 1)) by A1, A2, A3, Th29;
then (Partial_Sums (F # x)) . (k + 1) = ((Partial_Sums F) . (k + 1)) . x by A4, A8, A7, MESFUNC1:def 3;
hence S1[k + 1] by MESFUNC5:def 13; :: thesis: verum
end;
(Partial_Sums (F # x)) . 0 = (F # x) . 0 by Def1;
then (Partial_Sums (F # x)) . 0 = (F . 0) . x by MESFUNC5:def 13;
then (Partial_Sums (F # x)) . 0 = ((Partial_Sums F) . 0) . x by Def4;
then A9: S1[ 0 ] by MESFUNC5:def 13;
for k being Nat holds S1[k] from NAT_1:sch 2(A9, A5);
 hence (Partial_Sums (F # x)) . n = ((Partial_Sums F) # x) . n ; :: thesis: verum