reserve X for non empty TopSpace;
reserve x for Point of X;
reserve U1 for Subset of X;

theorem Th20:
  X is normal iff for A,C being Subset of X st A <> {} & C <> [#]
X & A c= C & A is closed & C is open ex G being Subset of X st G is open & A c=
  G & Cl G c= C
proof
  thus X is normal implies for A,C being Subset of X st A <> {} & C <> [#] X &
A c= C & A is closed & C is open ex G being Subset of X st G is open & A c= G &
  Cl G c= C
  proof
    assume
A1: for A,B being Subset of X st A <> {} & B <> {} & A is closed & B
is closed & A misses B ex G,H being Subset of X st G is open & H is open & A c=
    G & B c= H & G misses H;
    let A,C be Subset of X such that
A2: A <> {} and
A3: C <> [#] X and
A4: A c= C and
A5: A is closed and
A6: C is open;
    set B=[#](X) \ C;
    B c= A` by A4,XBOOLE_1:34;
    then
A7: A misses B by SUBSET_1:23;
    B <> {} & C` is closed by A3,A6,PRE_TOPC:4;
    then consider G,H being Subset of X such that
A8: G is open and
A9: H is open and
A10: A c= G and
A11: B c= H and
A12: G misses H by A1,A2,A5,A7;
    take G;
    for p being object holds p in Cl G implies p in C
    proof
      let p be object;
      assume
A13:  p in Cl G;
      then reconsider y=p as Point of X;
      H` is closed & G c= H` by A9,A12,SUBSET_1:23;
      then
A14:  y in H` by A13,PRE_TOPC:15;
      H` c= B` by A11,SUBSET_1:12;
      then y in B` by A14;
      hence thesis by PRE_TOPC:3;
    end;
    hence thesis by A8,A10;
  end;
  assume
A15: for A,C being Subset of X st A <> {} & C <> [#] X & A c= C & A is
  closed & C is open ex G being Subset of X st G is open & A c= G & Cl G c= C;
  for A,B being Subset of X st A <> {} & B <> {} & A is closed & B is
closed & A misses B ex G,H being Subset of X st G is open & H is open & A c= G
  & B c= H & G misses H
  proof
    let A,B be Subset of X such that
A16: A <> {} and
A17: B <> {} and
A18: A is closed and
A19: B is closed & A misses B;
    set C = [#] X \ B;
    [#] X \ C = B by PRE_TOPC:3;
    then
A20: C <> [#] X by A17,PRE_TOPC:4;
    A c= B` & C is open by A19,PRE_TOPC:def 3,SUBSET_1:23;
    then consider G being Subset of X such that
A21: G is open and
A22: A c= G and
A23: Cl G c= C by A15,A16,A18,A20;
    take G;
    take H = [#] X \ Cl G;
    thus G is open & H is open by A21,PRE_TOPC:def 3;
    C` c= (Cl G)` by A23,SUBSET_1:12;
    hence A c= G & B c= H by A22,PRE_TOPC:3;
    H c= G` by PRE_TOPC:18,XBOOLE_1:34;
    hence thesis by SUBSET_1:23;
  end;
  hence thesis;
end;
