Associativity of a lattice Suppose that $P=(X,\leq)$ is a lattice. Prove that $\forall x,y,z\in X$: 
$$(x\wedge y)\wedge z =x\wedge(y\wedge z) \hspace{6mm}\mbox{(associativity)},$$ 
where $\wedge$ indicates the meet of the two sets, i.e., the greatest lower bound.
I have proved commutativity  but I am not sure how to prove associativity. 
 A: Let $$s = (x \wedge y)\wedge z$$
It follows by definition that $s \leq (x \wedge y)$ and  $s \leq z$ (that is, $s$ is an lower bound for $(x \wedge y)$ and $z$) and that if $\xi \leq (x \wedge y)$ and $\xi \leq z$ is another lower bound, then $\xi \leq s$ 
But if $s \leq x \wedge y $ then it follows that $s \leq x$ and $s \leq y$ for order is transitive and $x\wedge y \leq x$ and $x \wedge y \leq y$ . Similarly $\xi \leq x \wedge y$ implies $\xi \leq x$ and $\xi \leq y$
Therefore s satisfies the property that $$s \leq x, y, z $$ and $$\xi \leq x, y, z \implies \xi \leq s$$
Now notice that this notion is "symmetric" in $x, y$ and $ z$. Expand the other side similarly and you'll get to the same definition
A: Let $a=(x\wedge y)\wedge z$ and $b = x \wedge (y \wedge z)$. Then $a \le x\wedge y$ and $a \le z$, so we must also have $a \le x$ and $a \le y$. Similarly for $b$, $b \le x$, $b \le y $, and  $b \le z$.
The above relations, and the definition of $\wedge$ tell us that $a \le b$ and $b \le a$, so we must have that $a=b$.
To see the use of the definition of $\wedge$, notice that since $b \le x$ and $b \le y$, it must be that $b \le x\wedge y$, so that $b \le a$
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