Writing and proving a predicate logic word problem

Question

I was assigned some homework for an intro class that covers predicate logic. I got through it, but I'm unsure if I have set it up and proven it correctly. The book were using is rather vague on the set-up process and just provides some examples, but not thought process/explanation. The explanation on some of the inference rules is also not clear enough, so I wanted to make sure that I'm using these correctly. Furthermore, our professor told us that when using (∃x) it should be by an ∧, and when using ∀x it should be followed by an →. I tried to adhere to that as I best I could to formulate the logic.

The Problem:

Using predicate logic and the symbols {F(x),C(x),O(x,y),D(x)}, prove the following argument: Every farmer owns a cow. No dentist owns a cow. Therefore no dentist is a farmer.

My set-up and proof:

(∃y)[C(y)∧(∀x)(F(x)→O(x,y))]∧(∀x)(∀y)[D(x)∧C(y)→O(x,y)']→(∀x)[D(x)→F(x)']

1.  (∃y)[C(y)∧(∀x)(F(x)→O(x,y))]                    hypothesis
2.  (∀x)(∀y)[(D(x)∧C(y))→O(x,y)']                   hypothesis
3.  C(a)∧(∀x)(F(x)→O(x,a))                          1, existential instantiation
4.  C(a)∧F(x)→O(x,a)                                3, universal instantiation
5.  (∀y)[(D(x)∧C(y))→O(x,y)']                       2, universal instantiation
6.  D(x)∧C(a)→O(x,a)'                               5, universal instantiation
7.  D(x)→C(a)→O(x,a)'                               6, exportation
8.  C(a)                                            4, simplification
9.  F(x)→O(x,a)                                     4, simplification
10. O(x,a)'→F(x)'                                   9, contraposition
11. D(x)→F(x)'                                      7, 10 hypothetical syllogism
12. (∀x)[D(x)→F(x)']                                11, universal generalization

EDIT:

After reworking the problem based on the suggestions I received, I have come up with the following:

(∀x)[F(x)→(∃y)(C(y)∧O(x,y))]∧(∀x)(∀y)[(C(y)∧O(x,y))→D(x)']→(∀x)[D(x)→F(x)']

1.  (∀x)[F(x)→(∃y)(C(y)∧O(x,y))]                hyp.
2.  (∀x)(∀y)[(C(y)∧O(x,y))→D(x)']               hyp.
3.  (∀x)[F(x)→(C(a)∧O(x,a))]                    1, existential instantiation
4.  F(x)→(C(a)∧O(x,a))                          1, universal instantiation
5.  (∀y)[(C(y)∧O(x,y))→D(x)']                   2, universal instantiation
6.  (C(a)∧O(x,a))→D(x)'                         2, universal instantiation
7.  (D(x)')'→(C(a)∧O(x,a))'                     6, contraposition
8.  D(x)→(C(a)∧O(x,a))'                         7, double negation
9.  D(x)→C(a)'∨O(x,a)'                          8, De-Morgan
10. (C(a)∧O(x,a))'→F(x)'                        4, contraposition
11. C(a)'∨O(x,a)'→F(x)'                         10, De-Morgan
12. D(x)→F(x)'                                  9, 11 modus ponens
13. (∀x)[D(x)→F(x)']                            12, universal generalization

I've decided to stick with this answer. The problem was extra, but through struggling with it and re-writing I have a better feel for some of the rules. At this point I feel like I just need to clear up some things about this topic in general. Thank you for the help/suggestions.

Answer 1

Using predicate logic and the symbols $\{F(x),C(x),O(x,y),D(x)\}$, prove the following argument: Every farmer owns a cow. No dentist owns a cow. Therefore no dentist is a farmer.

$(∃y)[C(y)∧(∀x)(F(x)→O(x,y))]∧(∀x)(∀y)[D(x)∧C(y)→O(x,y)']→(∀x)[D(x)→F(x)']$

You have written: "If there is some cow that every farmer owns, and all dentists don't own all cows, then all dentists are not farmers." Which, while true, is not what you were asked to say.

Because, $(∃y)[C(y)∧(∀x)(F(x)→O(x,y))$ means there is at least one cow that is co-owned by all farmers. The same cow.

You wanted $(\forall x)(\exists y)[F(x)\to (C(y)\wedge O(x,y))]$ "everyone who is a farmer owns something which is a cow."