let n be Element of NAT ; :: thesis: for Z being open Subset of REAL
for f1, f2 being PartFunc of REAL,REAL st f1 is_differentiable_on n,Z & f2 is_differentiable_on n,Z holds
(diff ((f1 - f2),Z)) . n = ((diff (f1,Z)) . n) - ((diff (f2,Z)) . n)
let Z be open Subset of REAL; :: thesis: for f1, f2 being PartFunc of REAL,REAL st f1 is_differentiable_on n,Z & f2 is_differentiable_on n,Z holds
(diff ((f1 - f2),Z)) . n = ((diff (f1,Z)) . n) - ((diff (f2,Z)) . n)
let f1, f2 be PartFunc of REAL,REAL; :: thesis: ( f1 is_differentiable_on n,Z & f2 is_differentiable_on n,Z implies (diff ((f1 - f2),Z)) . n = ((diff (f1,Z)) . n) - ((diff (f2,Z)) . n) )
defpred S₁[ Nat] means ( f1 is_differentiable_on $1,Z & f2 is_differentiable_on $1,Z implies (diff ((f1 - f2),Z)) . $1 = ((diff (f1,Z)) . $1) - ((diff (f2,Z)) . $1) );
A1: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A2: S₁[k] ; :: thesis: S₁[k + 1]
assume A3: ( f1 is_differentiable_on k + 1,Z & f2 is_differentiable_on k + 1,Z ) ; :: thesis: (diff ((f1 - f2),Z)) . (k + 1) = ((diff (f1,Z)) . (k + 1)) - ((diff (f2,Z)) . (k + 1))
A4: ( (diff (f1,Z)) . k is_differentiable_on Z & (diff (f2,Z)) . k is_differentiable_on Z ) by A3;
k < k + 1 by NAT_1:19;
then (diff ((f1 - f2),Z)) . (k + 1) = (((diff (f1,Z)) . k) - ((diff (f2,Z)) . k)) `| Z by A2, A3, TAYLOR_1:23, TAYLOR_1:def 5
.= (((diff (f1,Z)) . k) `| Z) - (((diff (f2,Z)) . k) `| Z) by A4, FDIFF_2:18
.= ((diff (f1,Z)) . (k + 1)) - (((diff (f2,Z)) . k) `| Z) by TAYLOR_1:def 5
.= ((diff (f1,Z)) . (k + 1)) - ((diff (f2,Z)) . (k + 1)) by TAYLOR_1:def 5 ;
hence (diff ((f1 - f2),Z)) . (k + 1) = ((diff (f1,Z)) . (k + 1)) - ((diff (f2,Z)) . (k + 1)) ; :: thesis: verum

end;

A5: S₁[ 0 ]

proof

assume that
f1 is_differentiable_on 0 ,Z and
f2 is_differentiable_on 0 ,Z ; :: thesis: (diff ((f1 - f2),Z)) . 0 = ((diff (f1,Z)) . 0) - ((diff (f2,Z)) . 0)
(diff ((f1 - f2),Z)) . 0 = (f1 - f2) | Z by TAYLOR_1:def 5
.= (f1 | Z) - (f2 | Z) by RFUNCT_1:47
.= ((diff (f1,Z)) . 0) - (f2 | Z) by TAYLOR_1:def 5
.= ((diff (f1,Z)) . 0) - ((diff (f2,Z)) . 0) by TAYLOR_1:def 5 ;
hence (diff ((f1 - f2),Z)) . 0 = ((diff (f1,Z)) . 0) - ((diff (f2,Z)) . 0) ; :: thesis: verum

end;

for k being Nat holds S₁[k] from NAT_1:sch 2(A5, A1);
hence ( f1 is_differentiable_on n,Z & f2 is_differentiable_on n,Z implies (diff ((f1 - f2),Z)) . n = ((diff (f1,Z)) . n) - ((diff (f2,Z)) . n) ) ; :: thesis: verum