reserve i,j,k,n for Nat;
reserve D for non empty set,
  p for Element of D,
  f,g for FinSequence of D;

theorem
  i+1 = len f implies f = (f|i)^<*f/.len f*>
proof
  assume
A1: i+1 = len f;
  then f is non empty;
  then
A2: i+1 in dom f by A1,Th6;
  dom f = Seg(i+1) by A1,FINSEQ_1:def 3;
  hence f = f|Seg(i+1)
    .= (f|i)^<*f.(i+1)*> by A2,Th10
    .= (f|i)^<*f/.len f*> by A1,A2,PARTFUN1:def 6;
end;
