> To prove that \\(r(x,y) = x \wedge y\\) it's therefore enough to show
>
> $$ a \le_A x \wedge y \textrm{ if and only if } a \le_A x \textrm{ and } a \le_A y. $$
>
> **MD Puzzle 2'.** Can someone show this?

Let's look directly at the definition of \\(\wedge\\) from Fong and Spivak, pg. 17:

> Definition 1.60. Let \\((P, \leq)\\) be a preorder, and let \\(A \subseteq P\\) be a subset. We say that an element
> \\(p \in P\\) is the meet of \\(A\\) if
>
> 1. for all \\(a \in A\\), we have \\(p \leq a\\), and
> 2. for all \\(q\\) such that \\(q \leq a\\) for all \\(a \in A\\), we have that \\(q \leq p\\).
>
> We write \\(p = \bigwedge A\\), or \\(p = \bigwedge_{a \in A} a\\). If \\(A\\) just consists of two elements, say \\(A = \\{a, b\\}\\), we
can denote \\(\bigwedge A\\) simply by \\(a \wedge b\\).

So let's assume \\( a \le_A x\\) and \\(a \le_A y\\). We want to show \\(a \le_A x \wedge y \\). By assumption we have \\(\forall z \in \\{x,y\\}. a \leq z\\). Then by (2) in Definition 1.60 we have \\(a \leq \bigwedge \\{x,y\\}\\), which can be rewritten as \\(a \leq x \wedge y\\) according to Spivak and Fong's short hand.


Next let's assume \\(a \le_A x \wedge y \\). We want to show \\( a \le_A x\\) and \\(a \le_A y\\). Our assumption \\(a \le_A x \wedge y \\) is shorthand for \\(a \leq \bigwedge \\{x,y\\}\\). By (1) we have \\(\forall z \in \\{x,y\\}. a \leq z\\). But that's just the same as \\( a \le_A x\\) and \\(a \le_A y\\) as desired.