let seq1, seq2 be sequence of X; ( seq1 . 0 = 1. X & ( for n being Nat holds seq1 . (n + 1) = (seq1 . n) * z ) & seq2 . 0 = 1. X & ( for n being Nat holds seq2 . (n + 1) = (seq2 . n) * z ) implies seq1 = seq2 )
assume that
A4:
seq1 . 0 = 1. X
and
A5:
for n being Nat holds seq1 . (n + 1) = (seq1 . n) * z
and
A6:
seq2 . 0 = 1. X
and
A7:
for n being Nat holds seq2 . (n + 1) = (seq2 . n) * z
; seq1 = seq2
defpred S1[ Nat] means seq1 . $1 = seq2 . $1;
A8:
for k being Nat st S1[k] holds
S1[k + 1]
proof
let k be
Nat;
( S1[k] implies S1[k + 1] )
assume
S1[
k]
;
S1[k + 1]
hence seq1 . (k + 1) =
(seq2 . k) * z
by A5
.=
seq2 . (k + 1)
by A7
;
verum
end;
A9:
S1[ 0 ]
by A4, A6;
for n being Nat holds S1[n]
from NAT_1:sch 2(A9, A8);
then
for n being Element of NAT holds S1[n]
;
hence
seq1 = seq2
by FUNCT_2:63; verum