theorem Th50: :: FINSEQ_1:11
  n <=len p & k in n
  implies (p|n).k = p.k & k in dom p
proof
  assume that
A1: n <=len p and
A2: k in n;
A3: Segm n c= Segm len p by A1,NAT_1:39;
  then n = dom p /\ n by XBOOLE_1:28
    .= dom(p|n) by RELAT_1:61;
  hence thesis by A2,A3,FUNCT_1:47;
end;
