First-Order Logic Converse True?

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I really cannot find a good title for this question. In NLTK Chapter 10 Section 3.1 (last part), there is this paragraph:

1. An occurrence of a variable $x$ in a formula $\phi$ is free in $\phi$ if that occurrence doesn't fall within the scope of all x or some x in $\phi$.

2. Conversely, if $x$ is free in formula $\phi$, then it is bound in all x.$\phi$ and exists x.$\phi$.

3. If all variable occurrences in a formula are bound, the formula is said to be closed.

I understand sentence 1. That means in some formula ((all x. cat(x)) -> meow(x)), the second occurrence of x is unbounded. However, how does 2 make sense? Let's say there is some formula ((exists y. milk(y)) -> white(y)). How do you prove 2? y in white(y) is considered free in $\phi$ but is not bound in exists y.milk(y).