1. **Stating the problem:** We are given a lattice $(L, \leq)$ and elements $a, b \in L$. We want to understand and prove the equivalence: $$a \leq b \iff a \wedge b = a \iff a \vee b = b$$ 2. **Recall definitions:** - In a lattice, $a \wedge b$ is the greatest lower bound (meet) of $a$ and $b$. - $a \vee b$ is the least upper bound (join) of $a$ and $b$. - The partial order $\leq$ is defined such that $a \leq b$ means $a$ is less than or equal to $b$ in the lattice order. 3. **Show $a \leq b \implies a \wedge b = a$:** - Since $a \leq b$, $a$ is a lower bound of $a$ and $b$. - The meet $a \wedge b$ is the greatest lower bound, so it must be at least $a$. - But $a$ is already a lower bound, so $a \wedge b = a$. 4. **Show $a \wedge b = a \implies a \leq b$:** - Since $a \wedge b$ is the greatest lower bound, and $a \wedge b = a$, it means $a$ is a lower bound of $a$ and $b$. - In particular, $a \leq b$. 5. **Show $a \leq b \implies a \vee b = b$:** - Since $a \leq b$, $b$ is an upper bound of $a$ and $b$. - The join $a \vee b$ is the least upper bound, so it must be at most $b$. - But $b$ is already an upper bound, so $a \vee b = b$. 6. **Show $a \vee b = b \implies a \leq b$:** - Since $a \vee b$ is the least upper bound, and $a \vee b = b$, it means $b$ is an upper bound of $a$ and $b$. - In particular, $a \leq b$. **Final conclusion:** We have shown the equivalences: $$a \leq b \iff a \wedge b = a \iff a \vee b = b$$ This characterizes the partial order in terms of meet and join operations in a lattice.