theorem Th13:
  for K st K = {} holds -K = {[{},{}]}
proof
  let K;
  assume
A1: K = {};
A2: { [{ g.t1 : g.t1 in t1`2 & t1 in K }, { g.t2 : g.t2 in t2`1 & t2 in K }]
  : s in K implies g.s in s`1 \/ s`2 } = {[{},{}]}
  proof
    thus { [{ g.t1 : g.t1 in t1`2 & t1 in K }, { g.t2 : g.t2 in t2`1 & t2 in K
    } ] : s in K implies g.s in s`1 \/ s`2 } c= {[{},{}]}
    proof
      let x be object;
      assume x in { [{ g.t1 : g.t1 in t1`2 & t1 in K }, { g.t2 : g.t2 in t2`1
      & t2 in K }] : s in K implies g.s in s`1 \/ s`2 };
      then consider g such that
A3:   x = [{ g.t1 : g.t1 in t1`2 & t1 in K }, { g.t2 : g.t2 in t2`1 &
      t2 in K }] and
      s in K implies g.s in s`1 \/ s`2;
A4:   x`2 = { g.t2 : g.t2 in t2`1 & t2 in K } by A3;
A5:   now
        set y = the Element of x`2;
        assume x`2 <> {};
        then y in x`2;
        then ex t1 st y = g.t1 & g.t1 in t1`1 & t1 in K by A4;
        hence contradiction by A1;
      end;
A6:   x`1 = { g.t1 : g.t1 in t1`2 & t1 in K } by A3;
      now
        set y = the Element of x`1;
        assume x`1 <> {};
        then y in x`1;
        then ex t1 st y = g.t1 & g.t1 in t1`2 & t1 in K by A6;
        hence contradiction by A1;
      end;
      then x = [{},{}] by A3,A5;
      hence thesis by TARSKI:def 1;
    end;
    thus {[{},{}]} c= { [{ g.t1 : g.t1 in t1`2 & t1 in K }, { g.t2 : g.t2 in
    t2`1 & t2 in K }] : s in K implies g.s in s`1 \/ s`2 }
    proof
      set g = the Element of Funcs(DISJOINT_PAIRS A, [A]);
      let x be object;
      assume x in {[{},{}]};
      then
A7:   x = [{},{}] by TARSKI:def 1;
A8:   now
        set y = the Element of { g.t1 : g.t1 in t1`2 & t1 in K };
        assume { g.t1 : g.t1 in t1`2 & t1 in K } <> {};
        then y in { g.t1 : g.t1 in t1`2 & t1 in K };
        then ex t1 st y = g.t1 & g.t1 in t1`2 & t1 in K;
        hence contradiction by A1;
      end;
A9:   now
        set y = the Element of { g.t2 : g.t2 in t2`1 & t2 in K };
        assume { g.t2 : g.t2 in t2`1 & t2 in K } <> {};
        then y in { g.t2 : g.t2 in t2`1 & t2 in K };
        then ex t1 st y = g.t1 & g.t1 in t1`1 & t1 in K;
        hence contradiction by A1;
      end;
      s in K implies g.s in s`1 \/ s`2 by A1;
      hence thesis by A7,A8,A9;
    end;
  end;
  [{},{}] is Element of DISJOINT_PAIRS A by Th12;
  hence thesis by A2,ZFMISC_1:46;
end;
