theorem
  for X being non empty set,A being Subset of X, S being a_partition of
  X holds A in S implies union(S \ {A}) = X \ A
proof
  let X be non empty set;
  let A be Subset of X;
  let S be a_partition of X;
  assume
A1: A in S;
  {A} c= S by A1,TARSKI:def 1;
  then reconsider AA = {A} as Subset of S;
  thus union (S \ {A}) = (union S) \ (union AA) by Th43
    .= X \ (union {A}) by Def4
    .= X \ A;
end;
