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

theorem Th56:
  (superior_setsequence(A (\/) A1)).n = A \/ (superior_setsequence A1).n
proof
  (superior_setsequence(A (\/) A1)).n = Union ((A (\/) A1) ^\n) by Th2
    .= Union (A (\/) (A1 ^\n)) by Th17
    .= A \/ Union (A1 ^\n) by Th39
    .= A \/ (superior_setsequence A1).n by Th2;
  hence thesis;
end;
