let X be RealNormSpace; :: thesis: for seq, seq1 being sequence of X st seq is convergent & ex k being Nat st seq = seq1 ^\ k holds
lim seq1 = lim seq

let seq, seq1 be sequence of X; :: thesis: ( seq is convergent & ex k being Nat st seq = seq1 ^\ k implies lim seq1 = lim seq )
assume that
A1: seq is convergent and
A2: ex k being Nat st seq = seq1 ^\ k ; :: thesis: lim seq1 = lim seq
consider k being Nat such that
A3: seq = seq1 ^\ k by A2;
A4: now :: thesis: for p being Real st 0 < p holds
ex n being set st
for m being Nat st n <= m holds
||.((seq1 . m) - (lim seq)).|| < p
let p be Real; :: thesis: ( 0 < p implies ex n being set st
for m being Nat st n <= m holds
||.((seq1 . m) - (lim seq)).|| < p )

assume 0 < p ; :: thesis: ex n being set st
for m being Nat st n <= m holds
||.((seq1 . m) - (lim seq)).|| < p

then consider n1 being Nat such that
A5: for m being Nat st n1 <= m holds
||.((seq . m) - (lim seq)).|| < p by ;
take n = n1 + k; :: thesis: for m being Nat st n <= m holds
||.((seq1 . m) - (lim seq)).|| < p

let m be Nat; :: thesis: ( n <= m implies ||.((seq1 . m) - (lim seq)).|| < p )
assume A6: n <= m ; :: thesis: ||.((seq1 . m) - (lim seq)).|| < p
then consider l being Nat such that
A7: m = (n1 + k) + l by NAT_1:10;
reconsider l = l as Nat ;
m - k = ((n1 + l) + k) + (- k) by A7;
then reconsider m1 = m - k as Nat ;
now :: thesis: n1 <= m1
assume not n1 <= m1 ; :: thesis: contradiction
then m1 + k < n1 + k by XREAL_1:6;
hence contradiction by A6; :: thesis: verum
end;
then ( m1 + k = m & ||.((seq . m1) - (lim seq)).|| < p ) by A5;
hence ||.((seq1 . m) - (lim seq)).|| < p by ; :: thesis: verum
end;
seq1 is convergent by A1, A2, Th10;
hence lim seq1 = lim seq by ; :: thesis: verum