\neg (x < n < y) is not equivalent to x\ge n \ge y. The correct solution uses De Morgan's law, on the expanded version x<n \land n < y.

Generally, do study

**DeMorgan**'s law. Can't hurt. Also know how to rewrite the implication using \lor and \neg.

Always be mindful of what arguments predicates might take, and do not omit them: Say P(x) when you mean a predicate P evaluated for x. The x is not optional, and points will get taken off if you skip it.

- Domains of quantifiers don't change if you pull a negation sign through them.
- When I said "try to use equivalences", I meant "you should be able to do this using equivalences." Truth tables will not help you much going forward in the class, as in, they won't get you any points.
D=\mathbb Z and D=\{ \mathbb Z \} are not the same thing. In one case, the set D is the set of all integers. In the other, it's the set consisting of the set of all integers. Suppose you have an apple and an orange and a pear (each representing an integer). Then the first is a bag containing all three. The second is a bag containing a bag containing all three.

Also, the grader was lenient on questions of form. That won't be the case on future homework (or quizzes). Messy solutions will get points taken off.