\problemdata
{S00}
{logic}
{fool}
{predicate}

\begin{problem} Let $F(x,y)$ be the predicate ``$x$ can fool $y$'' and the
universe of discourse be the set of all people.  Use quantifiers to
express the following statements.
\begin{problemparts}
\problempart Everybody can fool Fred. \\
\solution{
$\forall x F(x,\textrm{Fred})$
}
\problempart Evelyn can fool everybody. \\
\solution{
$\forall y F(\textrm{Evelyn},y)$
}
\problempart Everybody can fool somebody. \\
\solution{
$\forall x \exists y F(x,y)$
}
\problempart No one can fool everybody. \\
\solution{
$\neg \exists x \forall y F(x,y)$
}
\problempart Everybody can be fooled by somebody. \\
\solution{
$\forall y \exists x F(x,y)$
}
\problempart Nancy can fool exactly two people. \\
\solution{
$\exists x \exists y 
    (x \neq y  
        \land F(\textrm{Nancy}, x)
        \land F(\textrm{Nancy}, y)
        \land \forall z 
                (z \neq x \land z \neq y \implies \neg F(\textrm{Nancy},z)))$
Alternatively: 
$\exists x \exists y 
    (x \neq y  
        \land F(\textrm{Nancy}, x)
        \land F(\textrm{Nancy}, y)
        \land \forall z 
                (z = x \lor z = y \lor \neg F(\textrm{Nancy},z)))$

Note that by convention, different logical operators bind more or less tightly.
Quantifiers and negation bind most tightly; 
$\land$ and $\lor$ bind more tightly than $\implies$.
}
\problempart There is exactly one person whom everybody can fool. \\
\solution{
$\exists! y \forall x F(x,y)$. 

Note that $\exists! y P(y)$ is logically equivalent to 
$\exists y (P(y) \land \forall z (z \neq y \implies \neg P(z)))$. 
Thus, it is possible to rewrite the statement as 
$\exists y (\forall x F(x,y) 
                \land \forall z (z \neq y \implies \neg \forall x F(x,z)))$. 
}
\problempart There is someone who can fool exactly one person besides him- or
herself. \\
\solution{
$\exists x \exists! y (x \neq y \land F(x,y))$.  
(This can be rewritten without $\exists!$ as we did above.)
%or $\exists y (x \neq y \land F(x,y) 
%               \land \forall z (z \neq y \implies \neg (x \neq z \land F(x,z))))$. 
}
\end{problemparts}
\end{problem}

