reserve x,A,B,X,X9,Y,Y9,Z,V for set;

theorem
  Y misses Z implies (X \ Y) \/ Z = (X \/ Z) \ Y
proof
  assume
A1: Y misses Z;
  thus (X \/ Z) \ Y = (X \ Y) \/ (Z \ Y) by Th42
    .= (X \ Y) \/ Z by A1,Th83;
end;
