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 Answer 1


(1) can be paraphrased as "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" (I would use fool-able instead of can-fool, as it is closer to the intended meaning).

(2) would be "There exists an x, and for any t, you can fool x is a person and t is a time."

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 "Falstaff" and t was "yesterday", then in (1) the premise would be true, as person("Falstaff") and time("yesterday") are true, and so you conclude can-fool("Falstaff", "yesterday").

In (2) that becomes can-fool(person("Falstaff"), time("yesterday")), which evaluates to can-fool(true, true), and that won't work.


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