let seq1, seq2 be sequence of X; :: thesis: ( ( for k being Nat holds
( ( k <= n implies seq1 . k = (() . k) * ((() . n) - (() . (n -' k))) ) & ( n < k implies seq1 . k = 0. X ) ) ) & ( for k being Nat holds
( ( k <= n implies seq2 . k = (() . k) * ((() . n) - (() . (n -' k))) ) & ( n < k implies seq2 . k = 0. X ) ) ) implies seq1 = seq2 )

assume that
A1: for k being Nat holds
( ( k <= n implies seq1 . k = (() . k) * ((() . n) - (() . (n -' k))) ) & ( k > n implies seq1 . k = 0. X ) ) and
A2: for k being Nat holds
( ( k <= n implies seq2 . k = (() . k) * ((() . n) - (() . (n -' k))) ) & ( k > n implies seq2 . k = 0. X ) ) ; :: thesis: seq1 = seq2
now :: thesis: for k being Element of NAT holds seq1 . k = seq2 . k
let k be Element of NAT ; :: thesis: seq1 . b1 = seq2 . b1
per cases ( k <= n or k > n ) ;
suppose A3: k <= n ; :: thesis: seq1 . b1 = seq2 . b1
hence seq1 . k = (() . k) * ((() . n) - (() . (n -' k))) by A1
.= seq2 . k by A2, A3 ;
:: thesis: verum
end;
suppose A4: k > n ; :: thesis: seq1 . b1 = seq2 . b1
hence seq1 . k = 0. X by A1
.= seq2 . k by A2, A4 ;
:: thesis: verum
end;
end;
end;
hence seq1 = seq2 by FUNCT_2:63; :: thesis: verum