Do these FOL formula both represent "You can fool some of the people all of the time"? You can fool some of the people all of the time.

This can be represented in FOL as follows
$$\exists x \; \forall t  \; (\text{person}(x) \land \text{time}(t)) \Rightarrow \text{can-fool}(x,t) \tag{1}\label{1}$$
Is $\exists x \; \forall t \; \text{can-fool}(\text{person}(x), \text{time}(t))$ equivalent to (\ref{1}) ?
(1) can be paraphrased as &quot;There exists an x, and for any t if x is a person and t is a time, then x can be fooled at time t&quot; (I would use fool-able instead of can-fool, as it is closer to the intended meaning).
(2) would be &quot;There exists an x, and for any t, you can fool x is a person and t is a time.&quot;
They are not equivalent: person(x)  and time(t) are boolean predicates, which return a truth value: they are true if x is a person, and t is a time, respectively. So in (1) they act as a constraint on the values that x and t can take. If x was a saucepan, then person(x) would be false, and thus you wouldn't be able to claim that you can fool a saucepan.
So can-fool takes two arguments: one for which person is true, and one for which time is true. But in (2), the arguments are actually the boolean truth values: if x was &quot;Falstaff&quot; and t was &quot;yesterday&quot;, then in (1) the premise would be true, as person(&quot;Falstaff&quot;) and time(&quot;yesterday&quot;) are true, and so you conclude can-fool(&quot;Falstaff&quot;, &quot;yesterday&quot;).
In (2) that becomes can-fool(person(&quot;Falstaff&quot;), time(&quot;yesterday&quot;)), which evaluates to can-fool(true, true), and that won't work.