let D1 be non empty set ; for f being FinSequence of PFuncs (D1,REAL)
for d being Element of D1
for n being Element of NAT holds (f | n) # d = (f # d) | n
let f be FinSequence of PFuncs (D1,REAL); for d being Element of D1
for n being Element of NAT holds (f | n) # d = (f # d) | n
let d1 be Element of D1; for n being Element of NAT holds (f | n) # d1 = (f # d1) | n
let n be Element of NAT ; (f | n) # d1 = (f # d1) | n
A1:
len (f # d1) = len f
by Def8;
A2:
len ((f | n) # d1) = len (f | n)
by Def8;
now per cases
( len f <= n or n < len f )
;
case A4:
n < len f
;
(f | n) # d1 = (f # d1) | nthen A5:
len (f | n) = n
by FINSEQ_1:59;
A6:
len ((f # d1) | n) = n
by A1, A4, FINSEQ_1:59;
A7:
(
dom f = Seg (len f) &
dom (f # d1) = Seg (len (f # d1)) )
by FINSEQ_1:def 3;
A8:
dom ((f | n) # d1) = Seg (len ((f | n) # d1))
by FINSEQ_1:def 3;
now per cases
( n = 0 or n <> 0 )
;
case A10:
n <> 0
;
(f | n) # d1 = (f # d1) | nA11:
dom ((f # d1) | n) = Seg (len (f | n))
by A5, A6, FINSEQ_1:def 3;
0 + 1
<= n
by A10, NAT_1:13;
then A12:
n in dom f
by A4, FINSEQ_3:25;
now let m be
Nat;
( m in dom ((f # d1) | n) implies ((f # d1) | n) . m = ((f | n) # d1) . m )assume A13:
m in dom ((f # d1) | n)
;
((f # d1) | n) . m = ((f | n) # d1) . mthen A14:
m in dom (f # d1)
by A1, A5, A7, A12, A11, RFINSEQ:6;
then reconsider G =
f . m as
Element of
PFuncs (
D1,
REAL)
by A1, A7, FINSEQ_2:11;
((f # d1) | n) . m = (f # d1) . m
by A1, A5, A7, A12, A11, A13, RFINSEQ:6;
then A15:
((f # d1) | n) . m = G . d1
by A14, Def8;
(f | n) . m = G
by A5, A12, A11, A13, RFINSEQ:6;
hence
((f # d1) | n) . m = ((f | n) # d1) . m
by A2, A8, A11, A13, A15, Def8;
verum end; hence
(f | n) # d1 = (f # d1) | n
by A2, A5, A6, FINSEQ_2:9;
verum end; end; end; hence
(f | n) # d1 = (f # d1) | n
;
verum end; end; end;
hence
(f | n) # d1 = (f # d1) | n
; verum