theorem
  x in dom f & y in dom f & x <> y implies (disjoin f).x misses (disjoin f).y
proof
  assume that
A1: x in dom f and
A2: y in dom f and
A3: x <> y;
  set z = the Element of ((disjoin f).x) /\ ((disjoin f).y);
  assume
A4: ((disjoin f).x) /\ ((disjoin f).y) <> {};
A5: (disjoin f).x = [:f.x,{x}:] by A1,Def3;
A6: (disjoin f).y = [:f.y,{y}:] by A2,Def3;
A7: z in (disjoin f).x by A4,XBOOLE_0:def 4;
A8: z in (disjoin f).y by A4,XBOOLE_0:def 4;
A9: z`2 in {x} by A5,A7,MCART_1:10;
A10: z`2 in {y} by A6,A8,MCART_1:10;
  z`2 = x by A9,TARSKI:def 1;
  hence contradiction by A3,A10,TARSKI:def 1;
end;
