# Express the following sentence using logical connectives, predicates and quantifiers:

There is someone who can fool exactly one person besides himself or herself
Predicate given: F(x, y): x can fool y.
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.

• 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 – Henno Brandsma Feb 19 '17 at 9:56

$$\exists x (\exists y: F(x,y) \land (\forall z: (F(x,z) \land (z \neq x)\rightarrow z=y)$$
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 $y$.
• If $F(x,x)$ holds your statement is true, regardless whether there are other $y$ that $x$ fools. – Henno Brandsma Feb 19 '17 at 9:54