theorem
  not a \/ b in DISJOINT_PAIRS X implies ex p being Element of X st p in
  a`1 & p in b`2 or p in b`1 & p in a`2
proof
  set p = the Element of (a \/ b)`1 /\ (a \/ b)`2;
  assume not a \/ b in DISJOINT_PAIRS X;
  then (a \/ b)`1 meets (a \/ b)`2;
  then
A1: (a \/ b)`1 /\ (a \/ b)`2 <> {};
  (a \/ b)`1 /\ (a \/ b)`2 is Subset of X by FINSUB_1:16;
  then reconsider p as Element of X by A1,TARSKI:def 3;
  p in (a \/ b)`2 by A1,XBOOLE_0:def 4;
  then p in a`2 \/ b`2;
  then
A2: p in b`2 or p in a`2 by XBOOLE_0:def 3;
  take p;
  p in (a \/ b)`1 by A1,XBOOLE_0:def 4;
  then p in a`1 \/ b`1;
  then p in a`1 or p in b`1 by XBOOLE_0:def 3;
  hence thesis by A2,Th27;
end;
