reserve X for non empty TopSpace;
reserve Y for non empty TopStruct;
reserve x for Point of Y;
reserve Y for non empty TopStruct;
reserve X for non empty TopSpace;
reserve x,y for Point of X;

theorem
  for x, y being Point of X holds MaxADSet(x) misses MaxADSet(y) iff (ex
  E being Subset of X st E is closed & MaxADSet(x) c= E & E misses MaxADSet(y))
  or ex F being Subset of X st F is closed & F misses MaxADSet(x) & MaxADSet(y)
  c= F
proof
  let x, y be Point of X;
  thus MaxADSet(x) misses MaxADSet(y) implies (ex E being Subset of X st E is
closed & MaxADSet(x) c= E & E misses MaxADSet(y)) or ex F being Subset of X st
  F is closed & F misses MaxADSet(x) & MaxADSet(y) c= F
  proof
    assume
A1: MaxADSet(x) misses MaxADSet(y);
    hereby
      per cases by A1,Th53;
      suppose
        ex V being Subset of X st V is open & MaxADSet(x) c= V & V
        misses MaxADSet(y);
        then consider V being Subset of X such that
A2:     V is open and
A3:     MaxADSet(x) c= V and
A4:     V misses MaxADSet(y);
        now
          take F = V`;
          thus F is closed by A2;
          thus F misses MaxADSet(x) by A3,SUBSET_1:24;
          thus MaxADSet(y) c= F by A4,SUBSET_1:23;
        end;
        hence thesis;
      end;
      suppose
        ex W being Subset of X st W is open & W misses MaxADSet(x) &
        MaxADSet(y) c= W;
        then consider W being Subset of X such that
A5:     W is open and
A6:     W misses MaxADSet(x) and
A7:     MaxADSet(y) c= W;
        now
          take E = W`;
          thus E is closed by A5;
          thus MaxADSet(x) c= E by A6,SUBSET_1:23;
          thus E misses MaxADSet(y) by A7,SUBSET_1:24;
        end;
        hence thesis;
      end;
    end;
  end;
  assume
A8: (ex E being Subset of X st E is closed & MaxADSet(x) c= E & E
  misses MaxADSet(y)) or ex F being Subset of X st F is closed & F misses
  MaxADSet(x) & MaxADSet(y) c= F;
  assume MaxADSet(x) meets MaxADSet(y);
  then
A9: MaxADSet(x) = MaxADSet(y) by Th22;
  now
    per cases by A8;
    suppose
      ex E being Subset of X st E is closed & MaxADSet(x) c= E & E
      misses MaxADSet(y);
      then consider E being Subset of X such that
      E is closed and
A10:  MaxADSet(x) c= E and
A11:  E misses MaxADSet(y);
      E /\ MaxADSet(y) = {} by A11;
      hence contradiction by A9,A10,XBOOLE_1:28;
    end;
    suppose
      ex F being Subset of X st F is closed & F misses MaxADSet(x) &
      MaxADSet(y) c= F;
      then consider F being Subset of X such that
      F is closed and
A12:  F misses MaxADSet(x) and
A13:  MaxADSet(y) c= F;
      F /\ MaxADSet(x) = {} by A12;
      hence contradiction by A9,A13,XBOOLE_1:28;
    end;
  end;
  hence contradiction;
end;
