theorem
  m in dom f implies f/.m = (f|m)/.len(f|m)
proof
  assume that
A1: m in dom f;
  m<=len f by A1,FINSEQ_3:25;
  then
A2: len (f|m) = m by FINSEQ_1:59;
  1<=m by A1,FINSEQ_3:25;
  then m in Seg m;
  hence thesis by A1,A2,Th71;
end;
