The book solution is missing a quantifier on $Y$. I can't see any reason for this except that it must be a typo.
All three solutions involve use of commas, which is not standard punctuation in logic, and may lead to confusion and mistakes if you use them without being very careful about which logical content you mean by them. It is much better to write exactly what you want with actual logical connectives.
Written out completely, your first formula would be
$$\tag{1} \forall X.\forall Y.(\operatorname{mother}(X) \land \operatorname{daughter}(Y,X)\to\operatorname{love}(X,Y)) $$
and I would say this is a correct solution, under the assumption that you're working with a uniary "mother" predicate rather than, say, a "female" predicate to distinguish mothers from fathers. It's a bit asymmetric to have "daughter" relation that tells about the gender of the offspring but not of the parent, so arguably a nicer representation would be
$$\forall X.\forall Y.(\operatorname{female}(X) \land \operatorname{female}(Y) \land \operatorname{offspring}(Y,X)\to\operatorname{love}(X,Y)) $$
but the choice of which predicates to use in the model is something different from how to use the chosen predicates to write formulas, and it is probably the latter that is the point of this exercise.
The formula (1) above is equivalent to
$$\tag{2} \forall X.\big(\operatorname{mother}(X) \to \forall Y.(\operatorname{daughter}(Y,X)\to\operatorname{love}(X,Y))\big) $$
which I thought was what you mean by your second formula before editing the question.
But notice that writing everything out, one of your commas has become a $\land$ and the other became a $\to$. Commas as really don't work well as do-what-I-mean logical operators. (And they won't be accepted in a classroom setting where the point of the exercise is to show that you actually know what you mean).
With a $\land$ instead of $\to$ you get
$$\tag{3} \forall X.\big(\operatorname{mother}(X) \land \forall Y.(\operatorname{daughter}(Y,X)\to\operatorname{love}(X,Y))\big) $$
which is not the same as (1) and (2). It says, instead that everybody is a mother and loves whichever daughters she has.
And your formula in the edited question
$$\tag{4} \big(\forall X.\operatorname{mother}(X) \land \forall Y.\operatorname{daughter}(Y,X)\big)\to\operatorname{love}(X,Y) $$
doesn't work at all -- it puts the quantifiers inside the parentheses, so they don't range over $\operatorname{love}(X,Y)$ at all. This is the same as saying
$$\tag{4'} \big(\forall Z.\operatorname{mother}(Z) \land \forall W.\operatorname{daughter}(W,Z)\big)\to\operatorname{love}(X,Y) $$
which says "If every two random persons are mother and daughter then $X$ loves $Y$" and depends on the context for telling us who $X$ and $Y$ are.
Edit: The question was updated; this answer refers to the original posting.
You haven't specified the meanings of Howl, Have, etc. I have made the most charitable assumptions, but if I'm wrong you may need to make some changes (for instance, if "Howl" doesn't have "at night" baked in).
- This is probably fine.
- Your two options are logically equivalent, and they're logically equivalent to how I would translate it, but my personal preference would be for something like $\forall x\left((\exists y(\ldots))\to\ldots\right)$. Even though it's equivalent, the phrasing "has any cats" makes me think of an embedded $\exists y$.
- Your answers are odd because the sentence says "have" but you don't use the $\mathrm{Have}(x,y)$ construction.
- This is close, but your FOL form is false in a world with no cats at all, even though "Peter has either a cat or a dog" could still be true if he has a dog.
- This is probably fine.
Best Answer
Most introductory books on axiomatic set theory directly define set theoretical concepts in the language of first order logic. Once you know these definitions, all you need to do is translate, translate and translate until you are satisfied. One such book that I use can be found here: http://www.amazon.com/Theory-Continuum-Problem-Dover-Mathematics/dp/0486474844. If you buy this book, I warn you that it has errors. Fortunately, you can find a lot of the corrections here: http://comet.lehman.cuny.edu/fitting/errata/book_errors/RevisedSetBookErrors/split.html.
I also noticed that in the answer to your earlier question, set builder notation $\{x:\Phi(x)\}$ was used a lot. While this is not necessarily wrong, it can lead you dangerously close to Bertrand Russell's paradox. Briefly, just let $\Phi(x)$ be $x \not\in x$. Then, the entity $X = \{x:x \not\in x\}$ cannot itself be a set in the sense that it can never sit on the left hand side of the symbol $\in$ (i.e. $X \in \mathcal{S}$ is invalid syntax for any symbol $\mathcal{S}$). If it could, then either $X \in X$ or $X \not\in X$. Either case leads to a contradiction.
Thus, I will show you an alternate way of encoding set theory in first order logic. To start with, the book I referred to above studies a popular axiomatization of set theory: Von Neumann–Bernays–Gödel (NBG) set theory. This theory talks not only about sets but also about classes. The intuition behind a class is that it is an arbitrary collection of things. A set is a class that is contained in some other class. Thus, sets are just "smaller" collections of things. Now, to show you how NBG encodes set theory in first order logic, we need to be a bit more precise with our language.
I reserve the CAPITAL letters $A, B, C, ..., X, Y, Z$ with or without sub(super)scripts to denote classes. I reserve the small letters $a, b, c, ..., x, y, z$ with or without sub(super)scripts to denote sets. Appearance aside, the capital and small letters also have different usage. How so? Well, we first introduce a binary symbol $\in$. Then we impose the following condition on it: only small letters can sit on the left of this symbol while both small and capital letters can sit on its right. Hence $x \in A$ is valid while $B \in A$ is not. Thus, we have implicitly said that only sets (small letters) can be elements.
With these preliminaries, we can start defining things. I skip many technicalities to keep the exposition brief. For instance, I do not mention any axioms of set theory itself. I can come back to them later on if you want me to. For now:
Now you have enough material to translate expressions with Cartesian products and power sets into expressions in first-order logic in a straightforward but cumbersome manner:
As you can see, a full translation like that is definitely possible but it is very very tedious! For example, try translating "$A \times B = B \times A$ if and only if $A = B$". It is not fun!