
theorem Th53:
  for T being _Tree for a, b, c being Vertex of T for S being non
  empty set st for s being set st s in S holds (ex t being _Subtree of T st s =
  the_Vertices_of t) & (a in s & b in s or a in s & c in s or b in s & c in s)
  holds meet S <> {}
proof
  let T be _Tree;
  let a, b, c be Vertex of T;
  let S be non empty set;
  assume
A1: for s being set st s in S holds (ex t being _Subtree of T st s =
  the_Vertices_of t) & (a in s & b in s or a in s & c in s or b in s & c in s);
  set m = MiddleVertex(a,b,c);
  set Pca = T.pathBetween(c,a);
  set Pbc = T.pathBetween(b,c);
  set Pac = T.pathBetween(a,c);
  set Pab = T.pathBetween(a,b);
  set VPab = Pab.vertices();
  set VPac = Pac.vertices();
  set VPbc = Pbc.vertices();
  set VPca = Pca.vertices();
  VPab /\ VPbc /\ VPca = {m} by Def3;
  then
A2: m in VPab /\ VPbc /\ VPca by TARSKI:def 1;
  then
A3: m in VPab /\ VPbc by XBOOLE_0:def 4;
  then
A4: m in VPbc by XBOOLE_0:def 4;
  VPca = VPac by Th32;
  then
A5: m in VPac by A2,XBOOLE_0:def 4;
A6: m in VPab by A3,XBOOLE_0:def 4;
  now
    let s be set;
    assume
A7: s in S;
    then
A8: ex t being _Subtree of T st s = the_Vertices_of t by A1;
    per cases by A1,A7;
    suppose
      a in s & b in s;
      then VPab c= s by A8,Th34;
      hence m in s by A6;
    end;
    suppose
      a in s & c in s;
      then VPac c= s by A8,Th34;
      hence m in s by A5;
    end;
    suppose
      b in s & c in s;
      then VPbc c= s by A8,Th34;
      hence m in s by A4;
    end;
  end;
  hence thesis by SETFAM_1:def 1;
end;
