A) We need to relate $\wedge$ and $\vee$ in this question, and the only rule that contains both operations is rule (6). And in fact, that works: $\begin{align*} x &= x\wedge y \\ \Rightarrow x\vee y &= (x\wedge y)\vee y \\ \Rightarrow x\vee y &= y\text{ by rule (6)} \end{align*}$ The reverse implication is similar.
B) Reflexivity is $x\leq x$; using the definition this becomes $x\wedge x = x$, which is rule (3). Transitivity says that if $x\leq y$ and $y\leq z$, then $x\leq z$, i.e., if $x\wedge y= x$ and $y\wedge z = y$, then $x\wedge z$ must equal $x$. Starting with $x\wedge z$ and looking for something to substitute, we get $\begin{align*} x\wedge z &= (x\wedge y)\wedge z\\ &= x\wedge (y\wedge z)\text{ by rule (5)}\\ &= x\wedge y = x. \end{align*}$ As for antisymmetry, suppose $x\leq y$ and $y\leq x$; we want to show that $x=y$. The definition of $\leq$ tells us that $x\wedge y = x$ and $y\wedge x = y$, but by rule (4) these are equal.
C) First we show that $x\wedge y\leq x$, that is, $(x\wedge y) \wedge x= x$. This is true, because $(x\wedge y)\wedge x = x\wedge (x\wedge y) = (x\wedge x)\wedge y = x\wedge y$ by rules (4), (5), and (3), respectively. Similarly, $x\wedge y\leq y$, so $x\wedge y$ is smaller than both $x$ and $y$ in this partial order.
Now suppose $z$ is smaller than $x$ and $y$; we show that $z\leq x\wedge y$, making $x\wedge y$ necessarily the largest element smaller than $x$ and $y$. We have $\begin{align*} z\wedge(x\wedge y) &= (z\wedge x)\wedge y\\ &= z\wedge y\text{ since }z\leq x\\ &= z\text{ since }z\leq y. \end{align*}$ Therefore $z\leq (x\wedge y)$ as desired.
D) We need to show that $x$ and $y$ are smaller than $x\vee y$, and that if $x$ and $y$ are smaller than $z$, then $x\vee y$ is smaller than $z$. By (A), $x\leq y$ is the same as $x\wedge y = x$ is the same as $x\vee y = y$. So we're trying to prove that
- $x \vee (x\vee y) = x\vee y$
- $y \vee (x\vee y) = x\vee y$
- If $x \vee z = z$ and $y \vee z = z$, then $(x\vee y)\vee z = z$.
The proofs from (C) work exactly the same way here, only with $\wedge$ replaced by $\vee$.
E) Proving $x\leq T$ is the same as proving $x\wedge T = x$, which is true by rule (1). Proving $F\leq x$ is the same as proving $F\wedge x = F$, which is true by rule (2).
Good luck on your test!