let seq1, seq2 be Functional_Sequence of REAL,REAL; :: thesis: ( seq1 . 0 = f & ( for n being Nat holds seq1 . (n + 1) = cD ((seq1 . n),h) ) & seq2 . 0 = f & ( for n being Nat holds seq2 . (n + 1) = cD ((seq2 . n),h) ) implies seq1 = seq2 )

assume that

A3: seq1 . 0 = f and

A4: for n being Nat holds seq1 . (n + 1) = cD ((seq1 . n),h) and

A5: seq2 . 0 = f and

A6: for n being Nat holds seq2 . (n + 1) = cD ((seq2 . n),h) ; :: thesis: seq1 = seq2

defpred S_{1}[ Nat] means seq1 . $1 = seq2 . $1;

A7: for k being Nat st S_{1}[k] holds

S_{1}[k + 1]
_{1}[ 0 ]
by A3, A5;

for n being Nat holds S_{1}[n]
from NAT_1:sch 2(A9, A7);

then for n being Element of NAT holds S_{1}[n]
;

hence seq1 = seq2 by FUNCT_2:63; :: thesis: verum

assume that

A3: seq1 . 0 = f and

A4: for n being Nat holds seq1 . (n + 1) = cD ((seq1 . n),h) and

A5: seq2 . 0 = f and

A6: for n being Nat holds seq2 . (n + 1) = cD ((seq2 . n),h) ; :: thesis: seq1 = seq2

defpred S

A7: for k being Nat st S

S

proof

A9:
S
let k be Nat; :: thesis: ( S_{1}[k] implies S_{1}[k + 1] )

assume A8: S_{1}[k]
; :: thesis: S_{1}[k + 1]

thus seq1 . (k + 1) = cD ((seq1 . k),h) by A4

.= seq2 . (k + 1) by A6, A8 ; :: thesis: verum

end;assume A8: S

thus seq1 . (k + 1) = cD ((seq1 . k),h) by A4

.= seq2 . (k + 1) by A6, A8 ; :: thesis: verum

for n being Nat holds S

then for n being Element of NAT holds S

hence seq1 = seq2 by FUNCT_2:63; :: thesis: verum