let m be non empty Element of NAT ; for x, a being Element of REAL
for i being Element of NAT st 1 <= i & i <= m holds
Replace ((0* m),i,(a * x)) = a * (Replace ((0* m),i,x))
let x, a be Element of REAL ; for i being Element of NAT st 1 <= i & i <= m holds
Replace ((0* m),i,(a * x)) = a * (Replace ((0* m),i,x))
let i be Element of NAT ; ( 1 <= i & i <= m implies Replace ((0* m),i,(a * x)) = a * (Replace ((0* m),i,x)) )
assume A1:
( 1 <= i & i <= m )
; Replace ((0* m),i,(a * x)) = a * (Replace ((0* m),i,x))
A2:
( len (Replace ((0* m),i,(a * x))) = m & len (Replace ((0* m),i,x)) = m )
by Lm6;
then A3:
len (a * (Replace ((0* m),i,x))) = len (Replace ((0* m),i,(a * x)))
by RVSUM_1:147;
for j being Nat st 1 <= j & j <= len (Replace ((0* m),i,(a * x))) holds
(Replace ((0* m),i,(a * x))) . j = (a * (Replace ((0* m),i,x))) . j
proof
let j be
Nat;
( 1 <= j & j <= len (Replace ((0* m),i,(a * x))) implies (Replace ((0* m),i,(a * x))) . j = (a * (Replace ((0* m),i,x))) . j )
assume A4:
( 1
<= j &
j <= len (Replace ((0* m),i,(a * x))) )
;
(Replace ((0* m),i,(a * x))) . j = (a * (Replace ((0* m),i,x))) . j
reconsider j =
j as
Element of
NAT by ORDINAL1:def 13;
per cases
( i = j or i <> j )
;
suppose A5:
i = j
;
(Replace ((0* m),i,(a * x))) . j = (a * (Replace ((0* m),i,x))) . jthen (Replace ((0* m),i,(a * x))) . j =
a * x
by A1, Lm7
.=
a * ((Replace ((0* m),i,x)) . j)
by A1, A5, Lm7
;
hence
(Replace ((0* m),i,(a * x))) . j = (a * (Replace ((0* m),i,x))) . j
by RVSUM_1:66;
verum end; suppose A6:
i <> j
;
(Replace ((0* m),i,(a * x))) . j = (a * (Replace ((0* m),i,x))) . jthen
(Replace ((0* m),i,x)) . j = 0
by A2, A4, Lm7;
then
(Replace ((0* m),i,(a * x))) . j = a * ((Replace ((0* m),i,x)) . j)
by A2, A4, A6, Lm7;
hence
(Replace ((0* m),i,(a * x))) . j = (a * (Replace ((0* m),i,x))) . j
by RVSUM_1:66;
verum end; end;
end;
hence
Replace ((0* m),i,(a * x)) = a * (Replace ((0* m),i,x))
by A3, FINSEQ_1:18; verum