let n be Element of NAT ; for h, x being Real
for f1, f2 being Function of REAL ,REAL holds ((bdif (f1 - f2),h) . (n + 1)) . x = (((bdif f1,h) . (n + 1)) . x) - (((bdif f2,h) . (n + 1)) . x)
let h, x be Real; for f1, f2 being Function of REAL ,REAL holds ((bdif (f1 - f2),h) . (n + 1)) . x = (((bdif f1,h) . (n + 1)) . x) - (((bdif f2,h) . (n + 1)) . x)
let f1, f2 be Function of REAL ,REAL ; ((bdif (f1 - f2),h) . (n + 1)) . x = (((bdif f1,h) . (n + 1)) . x) - (((bdif f2,h) . (n + 1)) . x)
defpred S1[ Element of NAT ] means for x being Real holds ((bdif (f1 - f2),h) . ($1 + 1)) . x = (((bdif f1,h) . ($1 + 1)) . x) - (((bdif f2,h) . ($1 + 1)) . x);
A1:
S1[ 0 ]
proof
let x be
Real;
((bdif (f1 - f2),h) . (0 + 1)) . x = (((bdif f1,h) . (0 + 1)) . x) - (((bdif f2,h) . (0 + 1)) . x)
x in REAL
;
then
(
x in dom f1 &
x in dom f2 )
by FUNCT_2:def 1;
then
x in (dom f1) /\ (dom f2)
by XBOOLE_0:def 4;
then A2:
x in dom (f1 - f2)
by VALUED_1:12;
x - h in REAL
;
then
(
x - h in dom f1 &
x - h in dom f2 )
by FUNCT_2:def 1;
then
x - h in (dom f1) /\ (dom f2)
by XBOOLE_0:def 4;
then A3:
x - h in dom (f1 - f2)
by VALUED_1:12;
((bdif (f1 - f2),h) . (0 + 1)) . x =
(bD ((bdif (f1 - f2),h) . 0 ),h) . x
by Def7
.=
(bD (f1 - f2),h) . x
by Def7
.=
((f1 - f2) . x) - ((f1 - f2) . (x - h))
by Th4
.=
((f1 . x) - (f2 . x)) - ((f1 - f2) . (x - h))
by A2, VALUED_1:13
.=
((f1 . x) - (f2 . x)) - ((f1 . (x - h)) - (f2 . (x - h)))
by A3, VALUED_1:13
.=
((f1 . x) - (f1 . (x - h))) - ((f2 . x) - (f2 . (x - h)))
.=
((bD f1,h) . x) - ((f2 . x) - (f2 . (x - h)))
by Th4
.=
((bD f1,h) . x) - ((bD f2,h) . x)
by Th4
.=
((bD ((bdif f1,h) . 0 ),h) . x) - ((bD f2,h) . x)
by Def7
.=
((bD ((bdif f1,h) . 0 ),h) . x) - ((bD ((bdif f2,h) . 0 ),h) . x)
by Def7
.=
(((bdif f1,h) . (0 + 1)) . x) - ((bD ((bdif f2,h) . 0 ),h) . x)
by Def7
.=
(((bdif f1,h) . (0 + 1)) . x) - (((bdif f2,h) . (0 + 1)) . x)
by Def7
;
hence
((bdif (f1 - f2),h) . (0 + 1)) . x = (((bdif f1,h) . (0 + 1)) . x) - (((bdif f2,h) . (0 + 1)) . x)
;
verum
end;
A4:
for k being Element of NAT st S1[k] holds
S1[k + 1]
proof
let k be
Element of
NAT ;
( S1[k] implies S1[k + 1] )
assume A5:
for
x being
Real holds
((bdif (f1 - f2),h) . (k + 1)) . x = (((bdif f1,h) . (k + 1)) . x) - (((bdif f2,h) . (k + 1)) . x)
;
S1[k + 1]
let x be
Real;
((bdif (f1 - f2),h) . ((k + 1) + 1)) . x = (((bdif f1,h) . ((k + 1) + 1)) . x) - (((bdif f2,h) . ((k + 1) + 1)) . x)
A6:
(
((bdif (f1 - f2),h) . (k + 1)) . x = (((bdif f1,h) . (k + 1)) . x) - (((bdif f2,h) . (k + 1)) . x) &
((bdif (f1 - f2),h) . (k + 1)) . (x - h) = (((bdif f1,h) . (k + 1)) . (x - h)) - (((bdif f2,h) . (k + 1)) . (x - h)) )
by A5;
A7:
(bdif (f1 - f2),h) . (k + 1) is
Function of
REAL ,
REAL
by Th12;
A8:
(bdif f2,h) . (k + 1) is
Function of
REAL ,
REAL
by Th12;
A9:
(bdif f1,h) . (k + 1) is
Function of
REAL ,
REAL
by Th12;
((bdif (f1 - f2),h) . ((k + 1) + 1)) . x =
(bD ((bdif (f1 - f2),h) . (k + 1)),h) . x
by Def7
.=
(((bdif (f1 - f2),h) . (k + 1)) . x) - (((bdif (f1 - f2),h) . (k + 1)) . (x - h))
by A7, Th4
.=
((((bdif f1,h) . (k + 1)) . x) - (((bdif f1,h) . (k + 1)) . (x - h))) - ((((bdif f2,h) . (k + 1)) . x) - (((bdif f2,h) . (k + 1)) . (x - h)))
by A6
.=
((bD ((bdif f1,h) . (k + 1)),h) . x) - ((((bdif f2,h) . (k + 1)) . x) - (((bdif f2,h) . (k + 1)) . (x - h)))
by A9, Th4
.=
((bD ((bdif f1,h) . (k + 1)),h) . x) - ((bD ((bdif f2,h) . (k + 1)),h) . x)
by A8, Th4
.=
(((bdif f1,h) . ((k + 1) + 1)) . x) - ((bD ((bdif f2,h) . (k + 1)),h) . x)
by Def7
.=
(((bdif f1,h) . ((k + 1) + 1)) . x) - (((bdif f2,h) . ((k + 1) + 1)) . x)
by Def7
;
hence
((bdif (f1 - f2),h) . ((k + 1) + 1)) . x = (((bdif f1,h) . ((k + 1) + 1)) . x) - (((bdif f2,h) . ((k + 1) + 1)) . x)
;
verum
end;
for n being Element of NAT holds S1[n]
from NAT_1:sch 1(A1, A4);
hence
((bdif (f1 - f2),h) . (n + 1)) . x = (((bdif f1,h) . (n + 1)) . x) - (((bdif f2,h) . (n + 1)) . x)
; verum