reserve E,X,Y,x for set;
reserve A,B,C for Subset of E;

theorem
  (for x being Element of E holds x in A iff x in B & not x in C)
  implies A = B \ C
proof
  assume
A1: for x being Element of E holds x in A iff x in B & not x in C;
  now
    let x be Element of E;
    assume x in A;
    then x in B & not x in C by A1;
    hence x in B \ C by XBOOLE_0:def 5;
  end;
  hence A c= B \ C by Th2;
  now
    let x be Element of E;
    assume x in B \ C;
    then x in B & not x in C by XBOOLE_0:def 5;
    hence x in A by A1;
  end;
  hence thesis by Th2;
end;
