let r be Real; :: thesis: for n being Element of NAT
for Z being open Subset of REAL
for f being PartFunc of REAL ,REAL st f is_differentiable_on n,Z holds
(diff (r (#) f),Z) . n = r (#) ((diff f,Z) . n)
let n be Element of NAT ; :: thesis: for Z being open Subset of REAL
for f being PartFunc of REAL ,REAL st f is_differentiable_on n,Z holds
(diff (r (#) f),Z) . n = r (#) ((diff f,Z) . n)
let Z be open Subset of REAL ; :: thesis: for f being PartFunc of REAL ,REAL st f is_differentiable_on n,Z holds
(diff (r (#) f),Z) . n = r (#) ((diff f,Z) . n)
let f be PartFunc of REAL ,REAL ; :: thesis: ( f is_differentiable_on n,Z implies (diff (r (#) f),Z) . n = r (#) ((diff f,Z) . n) )
defpred S₁[ Element of NAT ] means ( f is_differentiable_on $1,Z implies (diff (r (#) f),Z) . $1 = r (#) ((diff f,Z) . $1) );
A1: S₁[ 0 ]

proof

assume f is_differentiable_on 0 ,Z ; :: thesis: (diff (r (#) f),Z) . 0 = r (#) ((diff f,Z) . 0 )
(diff (r (#) f),Z) . 0 = (r (#) f) | Z by TAYLOR_1:def 5
.= r (#) (f | Z) by RFUNCT_1:65
.= r (#) ((diff f,Z) . 0 ) by TAYLOR_1:def 5 ;
hence (diff (r (#) f),Z) . 0 = r (#) ((diff f,Z) . 0 ) ; :: thesis: verum

end;

A2: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A3: S₁[k] ; :: thesis: S₁[k + 1]
assume A4: f is_differentiable_on k + 1,Z ; :: thesis: (diff (r (#) f),Z) . (k + 1) = r (#) ((diff f,Z) . (k + 1))
AA: k < k + 1 by NAT_1:19;
k <= (k + 1) - 1 ;
then A6: (diff f,Z) . k is_differentiable_on Z by A4, TAYLOR_1:def 6;
(diff (r (#) f),Z) . (k + 1) = (r (#) ((diff f,Z) . k)) `| Z by A3, AA, A4, TAYLOR_1:23, TAYLOR_1:def 5
.= r (#) (((diff f,Z) . k) `| Z) by A6, FDIFF_2:19
.= r (#) ((diff f,Z) . (k + 1)) by TAYLOR_1:def 5 ;
hence (diff (r (#) f),Z) . (k + 1) = r (#) ((diff f,Z) . (k + 1)) ; :: thesis: verum

end;

for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A1, A2);
hence ( f is_differentiable_on n,Z implies (diff (r (#) f),Z) . n = r (#) ((diff f,Z) . n) ) ; :: thesis: verum