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

theorem Th72:
  A2 is convergent implies lim_sup (A1 (\) A2) = lim_sup A1 \ lim_sup A2
proof
  assume
A1: A2 is convergent;
  thus lim_sup (A1 (\) A2) c= lim_sup A1 \ lim_sup A2
  proof
    let x be object;
    assume
A2: x in lim_sup (A1 (\) A2);
A3: now
      let n;
      consider k1 being Nat such that
A4:   x in (A1 (\) A2).(n+k1) by A2,KURATO_0:5;
      x in A1.(n+k1) \ A2.(n+k1) by A4,Def3;
      then x in A1.(n+k1) & not x in A2.(n+k1) by XBOOLE_0:def 5;
      hence ex k st x in A1.(n+k) & not x in A2.(n+k);
    end;
    assume not x in lim_sup A1 \ lim_sup A2;
    then
A5: not (x in lim_sup A1 & not x in lim_sup A2) by XBOOLE_0:def 5;
    per cases by A1,A5,KURATO_0:def 5;
    suppose
      not x in lim_sup A1;
      then consider n1 being Nat such that
A6:   for k holds not x in A1.(n1+k) by KURATO_0:5;
      ex k2 being Nat st x in A1.(n1+k2) & not x in A2.(n1+k2)
      by A3;
      hence contradiction by A6;
    end;
    suppose
      x in lim_inf A2;
      then consider n2 being Nat such that
A7:   for k holds x in A2.(n2+k) by KURATO_0:4;
      ex k3 being Nat st x in A1.(n2+k3) & not x in A2.(n2+k3)
      by A3;
      hence contradiction by A7;
    end;
  end;
  thus thesis by Th69;
end;
