let seq1, seq2 be sequence of X; :: thesis: ( seq1 . 0 = 1. X & ( for n being Element of NAT holds seq1 . (n + 1) = (seq1 . n) * z ) & seq2 . 0 = 1. X & ( for n being Element of NAT holds seq2 . (n + 1) = (seq2 . n) * z ) implies seq1 = seq2 )
assume that
A4: seq1 . 0 = 1. X and
A5: for n being Element of NAT holds seq1 . (n + 1) = (seq1 . n) * z and
A6: seq2 . 0 = 1. X and
A7: for n being Element of NAT holds seq2 . (n + 1) = (seq2 . n) * z ; :: thesis: seq1 = seq2
defpred S₁[ Element of NAT ] means seq1 . $1 = seq2 . $1;
A8: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume S₁[k] ; :: thesis: S₁[k + 1]
hence seq1 . (k + 1) = (seq2 . k) * z by A5
.= seq2 . (k + 1) by A7 ;
:: thesis: verum

end;

A9: S₁[ 0 ] by A4, A6;
for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A9, A8);
hence seq1 = seq2 by FUNCT_2:113; :: thesis: verum