# Predicate Logic: Everybody can fool Fred

Let F(x,y) denote the statement "x can fool y" and D denote the domain of all people. So far I have the quantified proposition of "Everybody can fool Fred" as, ∀x∈D ∃y∈D: F(x,y) But I know that I have to be specific when it comes to Fred. So my question is how to I specify Fred in the domain of all people. Do I have to create another variable or do I have to specify y some how? Any help would be appreciated. Thank you.

• Your quantified proposition says something entirely different, namely that everyone can fool someone. This does not convey that the same person (Fred) can be fooled by everyone. – hardmath Sep 8 '17 at 18:36
• You're quite close. There's a quantifier denoted by $\exists !$ which means "exists and is unique", so you can write the proposition as $$\exists ! y\in D~\forall~x\in D~:~F(x,y)$$ – Prasun Biswas Sep 8 '17 at 18:40
• @PrasunBiswas, this says that everyone has one unique person that they can fool. It doesn't say that all those unique individuals are the same. To get that, you have to switch the order of the quantifiers. – G Tony Jacobs Sep 8 '17 at 18:42
• You're right. I've edited the comment. – Prasun Biswas Sep 8 '17 at 18:44

$\forall x\in D$ $\exists y\in D: F(x,y)$ is the statement, "Everyone can fool someone."
$\exists y\in D$ $\forall x\in D: F(x,y)$ is the statement, "There is a person everyone can fool."
As for your statement, I would write it as $\forall x\in D: F(x,Fred)$.
You can use an individual constant for Fred. So that's not a variable, but a constant that denotes Fred, for example $f$. Also, if the domain is all people, then there is no need to specify that in the symbolization, and you can just use:
$$\forall x \ F(x,f)$$