> 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.