let seq3, seq1, seq2 be Real_Sequence; ( ( for k being Nat ex Fr being XFinSequence of st
( dom Fr = k + 1 & ( for n being Nat st n in k + 1 holds
Fr . n = (seq1 . n) * (seq2 . (k -' n)) ) & Sum Fr = seq3 . k ) ) implies for k being Nat ex Fr being XFinSequence of st
( dom Fr = k + 1 & ( for n being Nat st n in k + 1 holds
Fr . n = (seq2 . n) * (seq1 . (k -' n)) ) & Sum Fr = seq3 . k ) )
assume A15:
for k being Nat ex Fr being XFinSequence of st
( dom Fr = k + 1 & ( for n being Nat st n in k + 1 holds
Fr . n = (seq1 . n) * (seq2 . (k -' n)) ) & Sum Fr = seq3 . k )
; for k being Nat ex Fr being XFinSequence of st
( dom Fr = k + 1 & ( for n being Nat st n in k + 1 holds
Fr . n = (seq2 . n) * (seq1 . (k -' n)) ) & Sum Fr = seq3 . k )
let k be Nat; ex Fr being XFinSequence of st
( dom Fr = k + 1 & ( for n being Nat st n in k + 1 holds
Fr . n = (seq2 . n) * (seq1 . (k -' n)) ) & Sum Fr = seq3 . k )
consider Fr1 being XFinSequence of such that
A16:
dom Fr1 = k + 1
and
A17:
for n being Nat st n in k + 1 holds
Fr1 . n = (seq1 . n) * (seq2 . (k -' n))
and
A18:
Sum Fr1 = seq3 . k
by A15;
defpred S1[ set , set ] means for i being Nat st i = $1 holds
$2 = (seq2 . i) * (seq1 . (k -' i));
reconsider k9 = k as Element of NAT by ORDINAL1:def 12;
A19:
for i being Nat st i in k9 + 1 holds
ex z being Element of REAL st S1[i,z]
proof
let i be
Nat;
( i in k9 + 1 implies ex z being Element of REAL st S1[i,z] )
assume
i in k9 + 1
;
ex z being Element of REAL st S1[i,z]
take
(seq2 . i) * (seq1 . (k -' i))
;
S1[i,(seq2 . i) * (seq1 . (k -' i))]
thus
S1[
i,
(seq2 . i) * (seq1 . (k -' i))]
;
verum
end;
consider Fr2 being XFinSequence of such that
A20:
dom Fr2 = k9 + 1
and
A21:
for i being Nat st i in k9 + 1 holds
S1[i,Fr2 . i]
from STIRL2_1:sch 5(A19);
take
Fr2
; ( dom Fr2 = k + 1 & ( for n being Nat st n in k + 1 holds
Fr2 . n = (seq2 . n) * (seq1 . (k -' n)) ) & Sum Fr2 = seq3 . k )
thus
( dom Fr2 = k + 1 & ( for n being Nat st n in k + 1 holds
Fr2 . n = (seq2 . n) * (seq1 . (k -' n)) ) )
by A20, A21; Sum Fr2 = seq3 . k
now let n be
Nat;
( n in len Fr1 implies Fr1 . n = Fr2 . ((len Fr1) -' (1 + n)) )assume A22:
n in len Fr1
;
Fr1 . n = Fr2 . ((len Fr1) -' (1 + n))A23:
n < k + 1
by A16, A22, NAT_1:44;
then
n <= k
by NAT_1:13;
then A24:
k -' n = k - n
by XREAL_1:233;
k -' n <= (k -' n) + n
by NAT_1:11;
then A25:
k -' (k -' n) = k - (k -' n)
by A24, XREAL_1:233;
n + 1
<= len Fr2
by A20, A23, NAT_1:13;
then A26:
(len Fr2) -' (n + 1) = (k + 1) - (n + 1)
by A20, XREAL_1:233;
(
k - n <= k &
k < k + 1 )
by NAT_1:13, XREAL_1:43;
then
k - n < k + 1
by XXREAL_0:2;
then
(len Fr2) -' (n + 1) in k + 1
by A26, NAT_1:44;
then
Fr2 . ((len Fr2) -' (n + 1)) = (seq2 . (k -' n)) * (seq1 . n)
by A21, A26, A24, A25;
hence
Fr1 . n = Fr2 . ((len Fr1) -' (1 + n))
by A16, A17, A20, A22;
verum end;
hence
Sum Fr2 = seq3 . k
by A16, A18, A20, Lm5; verum