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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 $\exists x(P(x)\land Q(x))$.

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Your first example is correct. It does, as you said, imply that all lions are fierce as long as there are no lions. That conclusion is (vacuously) correct; in the real world, where there are no unicorns, all unicorns are fierce (and they're all pink, and they're all green). If you don't see why, just imagine trying to exhibit a counterexample: a unicorn that isn't fierce(or isn't pink, etc.)

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 $\exists x\,(P(x)\land\neg R(x))$.

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Your first translation is correct. And yes, if there are not any lions, then the statement is what is called vacuously true: all zero lions are fierce! But if there are any lions, then all of them will have to be fierce as well, and that is what you are really aiming for.

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!

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