# Understanding predicate logic wording

I'm confused with translating english to predicate logic. An example that I couldn't really solve is this. Let $x$ be the domain of animals, $P(x)$ be the claim that $x$ is a lion, $Q(x)$ be the claim that $x$ is fierce and $R(x)$ be the claim that $x$ drinks coffee.

My guess is that "All lions are fierce" translates to $\forall x(P(x)\rightarrow Q(x))$. Is this correct? Wouldn't this imply that all lions are fierce is true as long as I don't have a lion?

How about "some lions do not drink coffee"? Would $\exists x(P(x)\rightarrow\neg R(x))$ be correct? I'm having trouble understanding what this means and I'm still debating the difference between my answers and $\exists x(P(x)\land Q(x))$.

Your second example isn't correct, because any non-lion would be a value for $x$ making $P(x)\to\neg R(x)$ true. (Remember that an implication with a false antecedent is true.) Also, any non-coffee-drinker (like me) would be a value for $x$ making $P(x)\to\neg R(x)$ true. You need $\exists x\,(P(x)\land\neg R(x))$.
The second statement should be $\exists x (P(x) \land \neg R(x))$. We can make the statement $\exists x (P(x) \rightarrow \neg R(x))$ again vacuously true by having something that is not a lion, and indeed by not having any lions at all, but clearly 'some lions do not drink any coffee' will require at least one lion in order for it to be true.
As a rule of thumb; $\forall$ claims very often go hand in hand with a $\rightarrow$, and $\exists$ almost always go hand in hand with a $\land$. I have seen and worked with many logic statements, but I have never encountered a case where we combine an $\exists$ with a $\rightarrow$ for the very reason as indicated: we would like to say that there are objects of a certain type, but if you combine this with a $\rightarrow$, then you can make the statement vacuously true by pointing to something that is not of the desired type, and thus by not having any objects of the desired type at all!