There is someone who can fool exactly one person besides himself or herself
Predicate given: F(x, y): x can fool y.
My answer attempt:
ヨxヨy (F(x, y) ⋀ ((x = y) V ( (x ≠ y) ⋀ ∀k(F(x, k) -> (k = y))))

Way of thinking:
There exists a person x and there exists a person y for which x can fool y and if so, then either x IS y OR if x is not y, and if for all k, x can fool k, then k is indeed y.
Please correct me if I am wrong.

  • $\begingroup$ If $F(x,x)$ the first clause is true, as is the $x=y$ part in the second $\lor$-statement. So the whole statement is then true automatically $\endgroup$ – Henno Brandsma Feb 19 '17 at 9:56

You're almost there. I would say:

$$\exists x (\exists y: F(x,y) \land (\forall z: (F(x,z) \land (z \neq x)\rightarrow z=y)$$

In words: there is some $x$ that can fool some $y$ and if we ever find someone that is fooled by $x$ and is not $x$ itself, it must have been $y$.

  • $\begingroup$ Thank you, your answer makes perfect sense. But, I can't find out where my answer went wrong. I have a doubt in the portion: ( (x ≠ y) ⋀ ∀k(F(x, k) -> (k = y))) $\endgroup$ – Ciado Feb 19 '17 at 6:12
  • $\begingroup$ If $F(x,x)$ holds your statement is true, regardless whether there are other $y$ that $x$ fools. $\endgroup$ – Henno Brandsma Feb 19 '17 at 9:54

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