I read somewhere that in a lattice, the definition of $a\land b$ being the greatest lower bound of $a$ and $b$ means that for all $c$, one has $$c\le a\land b \iff c\le a \text{ and } c\le b.$$
I don't find this definition to be intuitive. Does it agree with the usual definition from analysis, which says that $a\land b$ is the greatest lower bound of $\{a,b\}$ if
- $a\land b$ is a lower bound of $\{a,b\}$, i.e., $a\land b \le a$ and $a\land b\le b$,
- If $d$ is a lower bound of $\{a,b\}$, then $d\le a\land b $?
Here's my try to prove that they are equivalent. Suppose the first definition holds. Then since $\le$ is a partial order, $a\land b\le a\land b$. So $a\land b\le a,\ a\land b\le b$ by taking $c=a\land b$ and using the forward implication of the first definition. Suppose $d$ is a second lower bound of $\{a,b\}$. Then $d\le a,d\le b$. By the other implication in the first definition, $d\le a\land b$. So the first definition implies the second. Is this correct?
I'm not sure how to prove the other direction. Here we need to prove that something holds for all $c$, but we are only given $a$ and $b$.
Let's prove the other direction, starting with "$\Rightarrow$":
If $c \le a \land b$ holds, we know from 1. that $a \land b \le a$, so by transitivity of partial orders we get $c \le a$, and $c \le b$ the same way.
Now we prove the "$\Leftarrow$" direction of that equivalence:
If $c \le a$ and $c \le b$, then obviously $c$ is a lower bound of $\{a,b\}$, which means it can take the place of $d$ in property 2.: $c \le a \land b$
So both definitions of a greatest lower bound are equivalent.