# Exercise 7.66#

In the topos $$\textbf{Set}$$, where $$Ξ© = πΉ$$, consider the predicate $$p: βΓβ€\to πΉ$$ given by

\begin{split} \begin{align*} p(n, z) &= \begin{cases} \text{true} & \text{if } n \leq |z| \\ \text{false} & \text{if } n > |z| \end{cases} \end{align*} \end{split}
1. What is the set of $$n \in β$$ for which the predicate $$\forall(z: β€). p(n, z)$$ holds?

2. What is the set of $$n \in β$$ for which the predicate $$\exists(z: β€). p(n, z)$$ holds?

3. What is the set of $$z \in β€$$ for which the predicate $$\forall(n: β). p(n, z)$$ holds?

4. What is the set of $$z \in β€$$ for which the predicate $$\exists(n: β). p(n, z)$$ holds?

## Authorβs solution#

We have the predicate $$π : β Γ β€ β πΉ$$ given by $$π(π, π§)$$ iff $$π β€ |π§|$$.

1. The predicate $$β(π§ : β€). π(π, π§)$$ holds for $$\{0\} β β$$.

2. The predicate $$β(π§ : β€). π(π, π§)$$ holds for $$β β β$$.

3. The predicate $$β(π : β). π(π, π§)$$ holds for $$β β β€$$.

4. The predicate $$β(π : β). π(π, π§)$$ holds for $$β€ β β€$$.