reserve n,m,k for Nat,
  x,X for set,
  A for Subset of X,
  A1,A2 for SetSequence of X;

theorem Th69:
  lim_sup A1 \ lim_sup A2 c= lim_sup (A1 (\) A2)
proof
  let x be object;
  assume
A1: x in lim_sup A1 \ lim_sup A2;
  then not x in lim_sup A2 by XBOOLE_0:def 5;
  then consider n1 being Nat such that
A2: for k holds not x in A2.(n1+k) by KURATO_0:5;
  assume not x in lim_sup (A1 (\) A2);
  then consider n2 being Nat such that
A3: for k holds not x in (A1 (\) A2).(n2+k) by KURATO_0:5;
A4: now
    let k;
    not x in (A1 (\) A2).(n2+k) by A3;
    then not x in A1.(n2+k) \ A2.(n2+k) by Def3;
    hence not x in A1.(n2+k) or x in A2.(n2+k) by XBOOLE_0:def 5;
  end;
  x in lim_sup A1 by A1,XBOOLE_0:def 5;
  then consider k1 being Nat such that
A5: x in A1.(n1+n2+k1) by KURATO_0:5;
A6: x in A1.(n2+(k1+n1)) by A5;
  not x in A2.(n1+(n2+k1)) by A2;
  hence contradiction by A4,A6;
end;
