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

theorem Th87:
  A1 is convergent implies lim_sup (A (\+\) A1) = A \+\ lim_sup A1
proof
  assume
A1: A1 is convergent;
  thus lim_sup (A (\+\) A1) c= A \+\ lim_sup A1
  proof
    let x be object;
    assume
A2: x in lim_sup (A (\+\) A1);
A3: now
      let n;
      consider k1 being Nat such that
A4:   x in (A (\+\) A1).(n+k1) by A2,KURATO_0:5;
      x in A \+\ A1.(n+k1) by A4,Def9;
      then x in A & not x in A1.(n+k1) or not x in A & x in A1.(n+k1) by
XBOOLE_0:1;
      hence ex k st x in A & not x in A1.(n+k) or not x in A & x in A1.(n+k);
    end;
    assume
A5: not x in A \+\ lim_sup A1;
    per cases by A5,XBOOLE_0:1;
    suppose
A6:   not x in A & not x in lim_sup A1;
      then consider n1 being Nat such that
A7:   for k holds not x in A1.(n1+k) by KURATO_0:5;
      ex k1 being Nat st ( x in A & not x in A1.(n1+k1 ) or not
      x in A & x in A1.(n1+k1)) by A3;
      hence contradiction by A6,A7;
    end;
    suppose
A8:   x in A & x in lim_sup A1;
      then x in lim_inf A1 by A1,KURATO_0:def 5;
      then consider n2 being Nat such that
A9:   for k holds x in A1.(n2+k) by KURATO_0:4;
      ex k2 being Nat st ( x in A & not x in A1.(n2+k2 ) or not
      x in A & x in A1.(n2+k2)) by A3;
      hence contradiction by A8,A9;
    end;
  end;
  thus thesis by Th85;
end;
