theorem Th17:
  x in SF_have b \ SF_have a implies x is Filter of L & b in x & not a in x
proof
  assume
A1: x in SF_have b \ SF_have a;
  then
A2: not x in SF_have a by XBOOLE_0:def 5;
A3: x in SF_have b by A1,XBOOLE_0:def 5;
  then x is Filter of L by Th16;
  hence thesis by A3,A2,Th16;
end;
