let m be non zero Element of NAT ; :: thesis: for I being non empty FinSequence of NAT
for X being non empty Subset of (REAL m)
for f being PartFunc of (REAL m),REAL
for d being Real st X is open & f = X --> d & rng I c= Seg m holds
( (PartDiffSeq (f,X,I)) . 0 = X --> d & ( for i being Element of NAT st 1 <= i & i <= len I holds
(PartDiffSeq (f,X,I)) . i = X --> 0 ) )
let I be non empty FinSequence of NAT ; :: thesis: for X being non empty Subset of (REAL m)
for f being PartFunc of (REAL m),REAL
for d being Real st X is open & f = X --> d & rng I c= Seg m holds
( (PartDiffSeq (f,X,I)) . 0 = X --> d & ( for i being Element of NAT st 1 <= i & i <= len I holds
(PartDiffSeq (f,X,I)) . i = X --> 0 ) )
let X be non empty Subset of (REAL m); :: thesis: for f being PartFunc of (REAL m),REAL
for d being Real st X is open & f = X --> d & rng I c= Seg m holds
( (PartDiffSeq (f,X,I)) . 0 = X --> d & ( for i being Element of NAT st 1 <= i & i <= len I holds
(PartDiffSeq (f,X,I)) . i = X --> 0 ) )
let f be PartFunc of (REAL m),REAL; :: thesis: for d being Real st X is open & f = X --> d & rng I c= Seg m holds
( (PartDiffSeq (f,X,I)) . 0 = X --> d & ( for i being Element of NAT st 1 <= i & i <= len I holds
(PartDiffSeq (f,X,I)) . i = X --> 0 ) )
let d be Real; :: thesis: ( X is open & f = X --> d & rng I c= Seg m implies ( (PartDiffSeq (f,X,I)) . 0 = X --> d & ( for i being Element of NAT st 1 <= i & i <= len I holds
(PartDiffSeq (f,X,I)) . i = X --> 0 ) ) )
assume A1: ( X is open & f = X --> d & rng I c= Seg m ) ; :: thesis: ( (PartDiffSeq (f,X,I)) . 0 = X --> d & ( for i being Element of NAT st 1 <= i & i <= len I holds
(PartDiffSeq (f,X,I)) . i = X --> 0 ) )
thus A2: (PartDiffSeq (f,X,I)) . 0 = (X --> d) | X by A1, PDIFF_9:def 7
.= (X /\ X) --> d by FUNCOP_1:12
.= X --> d ; :: thesis: for i being Element of NAT st 1 <= i & i <= len I holds
(PartDiffSeq (f,X,I)) . i = X --> 0
defpred S₁[ Nat] means ( 1 <= $1 & $1 <= len I implies (PartDiffSeq (f,X,I)) . $1 = X --> 0 );
A3: S₁[ 0 ] ;
A4: for i being Nat st S₁[i] holds
S₁[i + 1]

proof

let i be Nat; :: thesis: ( S₁[i] implies S₁[i + 1] )
assume A5: S₁[i] ; :: thesis: S₁[i + 1]
assume A6: ( 1 <= i + 1 & i + 1 <= len I ) ; :: thesis: (PartDiffSeq (f,X,I)) . (i + 1) = X --> 0
A7: i <= i + 1 by NAT_1:12;

per cases ( i = 0 or i <> 0 ) ;

suppose A8: i = 0 ; :: thesis: (PartDiffSeq (f,X,I)) . (i + 1) = X --> 0

A9: ( 1 <= I /. (i + 1) & I /. (i + 1) <= m ) by Lm2, A1, A6;
thus (PartDiffSeq (f,X,I)) . (i + 1) = ((PartDiffSeq (f,X,I)) . i) `partial| (X,(I /. (i + 1))) by PDIFF_9:def 7
.= X --> 0 by A8, A9, A1, A2, Th16 ; :: thesis: verum

end;

suppose A10: i <> 0 ; :: thesis: (PartDiffSeq (f,X,I)) . (i + 1) = X --> 0

A11: ( 1 <= I /. (i + 1) & I /. (i + 1) <= m ) by Lm2, A1, A6;
thus (PartDiffSeq (f,X,I)) . (i + 1) = ((PartDiffSeq (f,X,I)) . i) `partial| (X,(I /. (i + 1))) by PDIFF_9:def 7
.= X --> 0 by A10, A11, A1, Th16, A7, A5, NAT_1:14, XXREAL_0:2, A6 ; :: thesis: verum

end;

end;

end;

for i being Nat holds S₁[i] from NAT_1:sch 2(A3, A4);
hence for i being Element of NAT st 1 <= i & i <= len I holds
(PartDiffSeq (f,X,I)) . i = X --> 0 ; :: thesis: verum