Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck in this proof, I am given:

$$A\setminus(B\setminus(C\setminus D)) = (A\cup C)\setminus (B\cup D)$$. I did this, but cannot come to solution where i can say, this is true or not.

$A\setminus(B\setminus(C\setminus D)) \Leftrightarrow x\in(A\setminus(B\setminus(C\setminus D))) \Leftrightarrow x\in A \land x\notin(B\setminus(C\setminus D)) \Leftrightarrow x\in A \land x\notin B \land \neg(x\notin(C\setminus D)) \Leftrightarrow x\in A \land x\notin B \land \neg(x\notin C \land \neg(x\notin D) ) \Leftrightarrow x\in A \land x\notin B \land x\in C \lor x\notin D \Leftrightarrow ...???$

please help. :(

share|cite|improve this question
Something is odd. If you put A,B,D = $\emptyset$, $C = \Omega$, LHS is empty but RHS is not. – Gautam Shenoy Apr 28 '13 at 17:06
up vote 2 down vote accepted

You've made some errors in your logic. Be careful writing things like $X \not \in Y$, because it's really shorthand for $\neg ( X \in Y )$. When you expanded your brackets, because you worked outside-in, what you ended up doing essentially is putting the closing bracket too soon. For instance, you wrote $$x \in A \wedge x \not \in (B \setminus (C \setminus D)) \Leftrightarrow x \in A \wedge x \not \in B \wedge \neg (x \not \in C \setminus D)$$ This is false. In fact you should have written $$x \in A \wedge x \not \in (B \setminus (C \setminus D)) \Leftrightarrow x \in A \wedge \neg(x \in B \wedge (x \not \in C \setminus D))$$ These are not equivalent. Expanded further, the RHS of the latter is $$x \in A \wedge (x \not \in B \vee (x \in C \setminus D))$$ by de Morgan's law.

share|cite|improve this answer
great, thanks Clive – doniyor Apr 28 '13 at 17:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.