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

theorem Th42:
  A \+\ Union A1 c= Union (A (\+\) A1)
proof
  let x be object;
  assume
A1: x in A \+\ Union A1;
  per cases by A1,XBOOLE_0:1;
  suppose
A2: x in A & not x in Union A1;
    then not x in A1.0 by PROB_1:12;
    then x in A \+\ A1.0 by A2,XBOOLE_0:1;
    then x in (A (\+\) A1).0 by Def9;
    hence thesis by PROB_1:12;
  end;
  suppose
A3: not x in A & x in Union A1;
    then consider n1 being Nat such that
A4: x in A1.n1 by PROB_1:12;
    x in A \+\ A1.n1 by A3,A4,XBOOLE_0:1;
    then x in (A (\+\) A1).n1 by Def9;
    hence thesis by PROB_1:12;
  end;
end;
