let seq1, seq2 be Functional_Sequence of REAL,REAL; :: thesis: ( seq1 . 0 = f | Z & ( for i being Nat holds seq1 . (i + 1) = (seq1 . i) `| Z ) & seq2 . 0 = f | Z & ( for i being Nat holds seq2 . (i + 1) = (seq2 . i) `| Z ) implies seq1 = seq2 )
assume that
A3: seq1 . 0 = f | Z and
A4: for n being Nat holds seq1 . (n + 1) = (seq1 . n) `| Z and
A5: seq2 . 0 = f | Z and
A6: for n being Nat holds seq2 . (n + 1) = (seq2 . n) `| Z ; :: thesis: seq1 = seq2
defpred S₁[ Element of NAT ] means seq1 . $1 = seq2 . $1;
A7: 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 A8: S₁[k] ; :: thesis: S₁[k + 1]
thus seq1 . (k + 1) = (seq1 . k) `| Z by A4
.= seq2 . (k + 1) by A6, A8 ; :: thesis: verum

end;

A9: S₁[ 0 ] by A3, A5;
for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A9, A7);
hence seq1 = seq2 by FUNCT_2:63; :: thesis: verum