set a = 1;
set b = 2;
set x = cos ;
set y = sin ;
set Z = ].(- 5),5.[;
consider C being PartFunc of REAL ,COMPLEX such that
A4:
C = (cos + (<i> (#) sin )) | [.1,2.]
;
take
C
; C is C1-curve-like
A5:
dom (cos + (<i> (#) sin )) = REAL
by FUNCT_2:def 1;
dom C = (dom (cos + (<i> (#) sin ))) /\ [.1,2.]
by A4, RELAT_1:90;
then A6:
[.1,2.] = dom C
by A5, XBOOLE_1:28;
A7:
].(- 5),5.[ is open
by RCOMP_1:25;
A8:
[.1,2.] c= ].(- 5),5.[
by XXREAL_1:47;
A9:
cos is_differentiable_on ].(- 5),5.[
by A7, FDIFF_1:34, SIN_COS:72;
A10:
sin is_differentiable_on ].(- 5),5.[
by A7, FDIFF_1:34, SIN_COS:73;
A11:
sin = sin | REAL
sin | ].(- 5),5.[ is continuous
by A11, FCONT_1:17;
then
(- sin ) | ].(- 5),5.[ is continuous
by SIN_COS:27, FCONT_1:22;
then A13:
- (sin | ].(- 5),5.[) is continuous
by RFUNCT_1:62;
A14: dom (- (sin | ].(- 5),5.[)) =
dom ((- sin ) | ].(- 5),5.[)
by RFUNCT_1:62
.=
(dom (- sin )) /\ ].(- 5),5.[
by RELAT_1:90
.=
REAL /\ ].(- 5),5.[
by SIN_COS:27, VALUED_1:8
.=
].(- 5),5.[
by XBOOLE_1:28
;
for x1 being Element of REAL st x1 in ].(- 5),5.[ holds
(- (sin | ].(- 5),5.[)) . x1 = diff cos ,x1
proof
let x1 be
Element of
REAL ;
( x1 in ].(- 5),5.[ implies (- (sin | ].(- 5),5.[)) . x1 = diff cos ,x1 )
assume A15:
x1 in ].(- 5),5.[
;
(- (sin | ].(- 5),5.[)) . x1 = diff cos ,x1
A16:
(- (sin | ].(- 5),5.[)) . x1 =
((- sin ) | ].(- 5),5.[) . x1
by RFUNCT_1:62
.=
(- sin ) . x1
by A15, FUNCT_1:72
.=
- (sin . x1)
by RFUNCT_1:74
.=
- ((sin | ].(- 5),5.[) . x1)
by A15, FUNCT_1:72
;
diff cos ,
x1 =
- (sin . x1)
by SIN_COS:72
.=
(- (sin | ].(- 5),5.[)) . x1
by A16, A15, FUNCT_1:72
;
hence
(- (sin | ].(- 5),5.[)) . x1 = diff cos ,
x1
;
verum
end;
then A17:
cos `| ].(- 5),5.[ is continuous
by A9, A13, A14, FDIFF_1:def 8;
A18:
cos = cos | REAL
A20:
cos | ].(- 5),5.[ is continuous
by A18, FCONT_1:17;
A21: dom (cos | ].(- 5),5.[) =
(dom cos ) /\ ].(- 5),5.[
by RELAT_1:90
.=
].(- 5),5.[
by XBOOLE_1:28, SIN_COS:27
;
for x1 being Element of REAL st x1 in ].(- 5),5.[ holds
(cos | ].(- 5),5.[) . x1 = diff sin ,x1
proof
let x1 be
Element of
REAL ;
( x1 in ].(- 5),5.[ implies (cos | ].(- 5),5.[) . x1 = diff sin ,x1 )
assume A22:
x1 in ].(- 5),5.[
;
(cos | ].(- 5),5.[) . x1 = diff sin ,x1
diff sin ,
x1 =
cos . x1
by SIN_COS:73
.=
(cos | ].(- 5),5.[) . x1
by A22, FUNCT_1:72
;
hence
(cos | ].(- 5),5.[) . x1 = diff sin ,
x1
;
verum
end;
then
sin `| ].(- 5),5.[ is continuous
by A10, A20, A21, FDIFF_1:def 8;
hence
C is C1-curve-like
by A4, A6, A7, A8, A9, A10, A17, Def3, SIN_COS:27; verum