theorem Th20:
  rng qe c= rng pe & vertices(pe) \ X c= V implies vertices(qe) \ X c= V
proof
  assume that
A1: rng qe c= rng pe and
A2: vertices(pe) \ X c= V;
  vertices qe c= vertices pe by A1,Th19;
  then vertices qe \ X c= vertices(pe) \ X by XBOOLE_1:35;
  hence thesis by A2;
end;
