reserve A, B for non empty preBoolean set,
  x, y for Element of [:A,B:];
reserve X for set,
  a,b,c for Element of [:A,B:];
reserve a for Element of [:Fin X, Fin X:];
reserve A for set;
reserve x,y for Element of [:Fin X, Fin X:],
  a,b for Element of DISJOINT_PAIRS X;

theorem
  y in DISJOINT_PAIRS X & x in DISJOINT_PAIRS X implies (y \/ x in
  DISJOINT_PAIRS X iff y`1 /\ x`2 \/ x`1 /\ y`2 = {})
proof
  assume that
A1: y in DISJOINT_PAIRS X and
A2: x in DISJOINT_PAIRS X;
A3: (y`1 \/ x`1) /\ (y`2 \/ x`2) = (y`1 \/ x`1) /\ y`2 \/ (y`1 \/ x`1) /\ x
  `2 & (y`1 \/ x`1) /\ y`2 = y`1 /\ y`2 \/ x`1 /\ y`2 by XBOOLE_1:23;
  x`1 misses x`2 by A2,Th23;
  then
A4: x`1 /\ x`2 = {};
  y`1 misses y`2 by A1,Th23;
  then
A5: y`1 /\ y`2 = {};
A6: (y`1 \/ x`1) /\ x`2 = y`1 /\ x`2 \/ x`1 /\ x`2 by XBOOLE_1:23;
A7: (y \/ x)`1 = y`1 \/ x`1 & (y \/ x)`2 = y`2 \/ x`2;
  thus y \/ x in DISJOINT_PAIRS X implies y`1 /\ x`2 \/ x`1 /\y`2 = {}
     by A5,A4,A7,A3,A6,XBOOLE_0:def 7,Th23;
  assume y`1 /\ x`2 \/ x`1 /\ y`2 = {};
  then (y \/ x)`1 misses (y \/ x)`2 by A5,A4,A3,A6;
  hence thesis;
end;
