reserve A for non empty set,
  a for Element of A;
reserve A for set;
reserve B,C for Element of Fin DISJOINT_PAIRS A,
  x for Element of [:Fin A, Fin A:],
  a,b,c,d,s,t,s9,t9,t1,t2,s1,s2 for Element of DISJOINT_PAIRS A,
  u,v,w for Element of NormForm A;
reserve K,L for Element of Normal_forms_on A;
reserve f,f9 for (Element of Funcs(DISJOINT_PAIRS A, [:Fin A,Fin A:])),
  g,h for Element of Funcs(DISJOINT_PAIRS A, [A]);

theorem Th21:
  K ^ -K = {}
proof
  set x = the Element of K ^ -K;
  assume
A1: K ^ -K <> {};
  then reconsider a = x as Element of DISJOINT_PAIRS A by SETWISEO:9;
  consider b,c such that
A2: b in K and
A3: c in -K and
A4: a = b \/ c by A1,NORMFORM:34;
A5: a`1 = b`1 \/ c`1 by A4;
A6: a`2 = b`2 \/ c`2 by A4;
  consider g such that
A7: s in K implies g.s in s`1 \/ s`2 and
A8: c = [{ g.t1 : g.t1 in t1`2 & t1 in K }, { g.t2 : g.t2 in t2`1 & t2
  in K }] by A3,Th11;
A9: g.b in b`1 \/ b`2 by A2,A7;
  now
    per cases by A9,XBOOLE_0:def 3;
    case
A10:  g.b in b`1;
      hence g.b in a`1 by A5,XBOOLE_0:def 3;
      g.b in { g.t2 : g.t2 in t2`1 & t2 in K } by A2,A10;
      then g.b in c`2 by A8;
      hence g.b in a`2 by A6,XBOOLE_0:def 3;
    end;
    case
A11:  g.b in b`2;
      hence g.b in a`2 by A6,XBOOLE_0:def 3;
      g.b in { g.t1 : g.t1 in t1`2 & t1 in K } by A2,A11;
      then g.b in c`1 by A8;
      hence g.b in a`1 by A5,XBOOLE_0:def 3;
    end;
  end;
  then a`1 /\ a`2 <> {} by XBOOLE_0:def 4;
  then a`1 meets a`2;
  hence contradiction by NORMFORM:25;
end;
