Cat(x) for “x is a cat,” Mouse(x) for “x is a mouse” Eats(x, y) for “x eats y”
Sentences 1: ∀x∃y Cat(x)∧ Mouse(y) ⇨ Eats(x,y)
Sentences 2: ∃x∃y∃z Cat(x)∧ (Mouse(y) ∧ Mouse(z) ∧ ¬(y=z)) ⇨ Eats(x,y)
Sentences 3: ∀x Cat(x)∧ Mouse(y) ⇨ Eats(x,y)
Could you please help me if I have made an error somewhere. I am unsure if these are correct
Let's break it down step-by-step. Do note that there are many ways of writing a logic statement equivalently.
Every cat eats exactly one mouse.
I'm going to work in reverse; show you the answer, then work backwards so you can see how I arrived there. $$\forall x(\text{Cat}(x)\implies\exists!y(\text{Mouse}(y)\wedge\text{Eats}(x, y)))$$
We start with $\forall x(\text{Cat}(x)\implies\dots)$. What this says is that for all things $x$, if $x$ is a cat, then that implies something. Specifically, that something is the statement that every cat eats exactly one mouse. Therefore, if $x$ is a cat, it must imply that there exists exactly one ($\exists !$) thing $y$ that is a mouse, $\text{Mouse(y)}$, and was also eaten by the cat, $\text{Eats}(x,y)$. Hence, we can deduce from this FOL statement that for any cat, there exists exactly one mouse which it eats.
Some cats eat at least two mice
$$\exists x(\text{Cat}(x)\wedge\exists y\exists z(y\neq z\wedge\text{Mouse}(y)\wedge\text{Mouse}(z))\wedge\text{Eat}(x,y)\wedge\text{Eat}(x,z))$$
We start with $\exists x(\text{Cat}(x)\wedge\dots)$, the declaration of the existence of a cat, and something ($\wedge\dots$) is also true, probably having to do with said cat. This something in particular, is that it has eaten not one, but two distinct mice. Thus, we lead into $\exists y\exists z$, asserting the existence of two more things, and that they are distinct ($y\neq z$, or equivalently, $\neg(y=z)$). Also, both of which are mice, $\wedge\text{Mouse}(y)\wedge\text{Mouse}(z)$, and they were both eaten by this cat $\wedge\text{Eat}(x,y)\wedge\text{Eat}(x,z)$. Because we started this statement with $\exists$ instead of $\exists!$, we know of the existence of at least one cat that ate two mice, but there could be more.
Every cat eats at most one mouse
$$\forall x(\text{Cat}(x)\implies\neg \exists y(\text{Mouse}(y)\wedge\text{Eats}(x,y))\vee\exists! z(\text{Mouse}(z)\wedge\text{Eats}(x,z)))$$
Again, we start with $\forall x(\text{Cat}(x)\implies\dots)$, asserting that for all things $x$, if they are a cat, $\text{Cat}(x)$, then this implies something. This time, being a cat implies you eat at most one mouse. This leaves two options:
There were no things, $\neg \exists y$, which were mice, $\text{Mouse}(y)$, that the cat ate: $\wedge \text{Eats}(x,y)$.
or, $\vee$, there was one thing, $\exists ! z$, which was a mouse, $\text{Mouse}(z)$, and was eaten by the cat, $\wedge \text{Eats}(x,y)$.
By the way, this is assuming that $\forall x \forall y(\text{Cat}(x)\iff\neg\text{Mouse}(x))$. Although it is obvious to you and I that being a cat and being a mouse are exclusive opportunities, you aren't always able to assume these things, especially once the logic becomes less related to the real world, so you'd have to assert such at the beginning.
Also, it is personal preference where to put the quantifiers (i.e; $\forall x$, $\exists y$, and $\exists! z$). I put them immediately before the first instance of whatever you're referring to, to make them easier to read and understand, but the statements work the same if you group them together at the beginning.
