How to frame the dual statement in a lattice ordered set or an algebraic lattice in general

Question

I am learning the theory of posets and lattices which will eventually lead to Boolean Algebra. I am stuck with the proper understanding of the concept of duality. Followings are what I have gathered so far:

1. If $ (P,\preccurlyeq ) $ is a poset, then $ (P,\succcurlyeq) $ is also a poset where $ \succcurlyeq $ is defined in terms of $ \preccurlyeq $ as $$ y\succcurlyeq x\iff x\preccurlyeq y. $$ Then $ (P,\succcurlyeq) $ is called the dual of the poset $ (P,\preccurlyeq ) $.
2. Every statement in $ P $ involving the relation $ \preccurlyeq $ has a corresponding dual statement in $ P^{\partial} $ obtained by replacing the relation $ \preccurlyeq $ with $ \succcurlyeq $. A statement in $ (P,\preccurlyeq ) $ is true if and only if the dual statement in $ (P,\succcurlyeq ) $ is true.
3. A poset $ (L,\preccurlyeq ) $ becomes a lattice ordered set (LOS) if for any two elements $ x,y\in L $, both $ \sup\{x,y\} $ and $ \inf\{x,y\} $ exist in $ L $.
4. If $ L $ is a LOS and $ L^{\partial} $ is the dual of $ L $ as a poset, then $ L^{\partial} $ is also a lattice. Then for any two elements $ x,y\in L $, $ \sup\{x,y\} $ in $ L^{\partial} $ is same as $ \inf\{x,y\} $ in $ L $ and $ \inf\{x,y\} $ in $ L^{\partial} $ is same as $ \sup\{x,y\} $ in $ L $.
5. An algebraic lattice (AL) $ (L,\vee,\wedge) $ is a non-empty set $ L $ equipped with two binary operations $ \vee $ and $ \wedge $ so that the system satisfies the commutative, associative, absorptive and by extension, the idempotent laws.
6. Every AL is a LOS and vice versa with the following appropriations:
   • AL to LOS: $ x\preccurlyeq y\iff x\wedge y=x $.
   • LOS to AL: $ x\vee y=\sup\{x,y\} $ and $ x\wedge y=\inf\{x,y\} $.
7. The dual of the AL $ (L,\vee,\wedge) $ is the AL $ (L,\wedge,\vee) $.

At this point, my major point of concern is how to frame the dual of a statement in a LOS or an AL in general? My first guess was to replace ($ \preccurlyeq $ and $ \succcurlyeq $), ($ \vee $ and $ \wedge $) and ($ \sup $ and $ \inf $) simultaneously. However, that didn't go well with the following contradiction:

In the isotone property, the statement "$ y\preccurlyeq z $ implies $ x\vee y\preccurlyeq x\vee z $" dually proves $ x\wedge y\preccurlyeq x\wedge z $ as well. Here you replace $ \vee $ by $ \wedge $ but the relation $ \preccurlyeq $ remains unchanged in the dual statement. However, in the distributive inequality, the dual statement of $$ x\vee(y\wedge z)\preccurlyeq (x\vee y)\wedge (x\vee z) $$ reverses the $ \vee $ and $ \wedge $ and the relation simultaneously. (These arguments are used in the book "Applied Abstract Algebra" by Rudolf Lidl and Günter Pilz.)

Thus, I am stuck on the question

How to frame a dual statement in a LOS or an AL in general?

In the process, I came across with this question, but the answers did not satisfy me.

Answer 1

Your first guess (in the paragraph just after the seventh enumeration point) is correct and what you present as a counter-example for it, it isn't.

The dual statement of $$y \leq z \implies x \vee y \leq x \vee z$$ is $$y \geq z \implies x \wedge y \geq x \wedge z,$$ as your guess would dictate.
However, we can re-write this statement as $$z\leq y \implies x \wedge z \leq x \wedge y,$$ which, after swapping the roles of $y$ and $z$, becomes $$y \leq z \implies x \wedge y \leq x \wedge z.$$ It perhaps not immediate that this last statement is dual of the first, but it is equivalent to the second displayed one, which has exactly the right shape, and so it is indeed dual of the first.