Set Theory notation simplification Let A and B be sets.
i) Express { $x \mid x ∈ A ∨ x \notin  B$ } as simply as possible in the notation of set theory,
without using set-builder notation.
ii) Express $∀x (x ∈ A ∨ x \notin  B)$ as simply as possible in the notation of set theory,
without any quantiﬁers.
Well for the first part, I get the following: $A-B$
but, I'm confused as to how the other in part two differs? How does the universal quantifier change the meaning?
 A: 1) I will assume that all elements are restricted to some domain $V$; otherwise, the question doesn't make much sense. We have,
$$\{x| x\in A \vee x\notin B\} = \{x| x\in A \} \cup \{x| x\notin B\} = A \cup (V\setminus B) \equiv A \cup \bar B.$$
2) The formula says that every $x$ either belongs to $A$ or doesn't belong to $B$ (or both). That is, equivalently, there is no $x$ that is not in $A$ and in $B$. We can write that as
$$B\setminus A = \varnothing.$$
Another way to write this is
$$B \subset A.$$ 
A: The first one is not an exclusion operation. An exclusion operator is:
$$\{x:x\in{A}\land{x}\not\in{B}\}$$
The set in your problem subsumes $A-B$, but it also includes the cases $x\in{A}\land{x}\in{B}$ and $x\not\in{A}\land{x}\not\in{B}$.
The difference between the first and the second problem is that the first one is a description of a set, the second is a statement about two sets.
$$\forall{x}(x\in{A}\lor{x}\not\in{B})$$
means two things: that $B$ is a proper subset of $A$, and that $A$ is a proper subset of the domain of discourse for $x$.
