Existence and uniqueness of particular binary operations on a lattice

Let $(A, \le)$ be a lattice. Consider the following properties for a commutative operation $\cdot$ on $A$: $$c \cdot (a \wedge b) = (c \cdot a) \wedge (c \cdot b)$$ $$c \cdot (a \vee b) = (c \cdot a) \vee (c \cdot b)$$ $$a \cdot b = (a \wedge b) \cdot (a \vee b)$$ for all $a,b,c \in A$. (Actually, the third property comes from the first two, as pointed out by Xodarap in his answer below.)

For example, both $\wedge$ and $\vee$ satisfy the properties if and only if the lattice is distributive. Furthermore, if the lattice has a greatest (resp. least) element, then it is an identity for $\wedge$ (resp. $\vee$).

My question is the following:

What are some natural conditions on the lattice that guarantee the existence of an operation satisfying the properties above other than $\wedge$ and $\vee$? What other assumptions can we make (possibly on $\cdot$) to guarantee uniqueness?

I am mostly interested in associative operations with an identity.

The question arose from the problem of writing the sum and product of natural numbers in terms of the lattice operations respectively of $(\mathbb{N}, \le)$ and $(\mathbb{N}, \mid)$. In those cases I already know the existence of such operations, which are hidden in the definitions of the orders. Indeed, both orders are of the form $a \preceq b$ if and only if there exists $c$ such that $b = a*c$.

Actually, the construction of an order $\preceq$ like that can be carried out in any commutative monoid where the identity is irreducible and the cancellation law holds, but $\preceq$ is not necessarily a lattice order, unless $*$ enjoys a unique factorization property. Furthermore, uniqueness does not seem to follow from these premises only.

• Consider the operation that takes any two points to a single fixed point 0. This would satisfy all the equalities above, right? Maybe you want the operation to also have an inverse? – Henrique Oct 13 '13 at 17:15
• Yes, it would. It wouldn't have an identity though, and your example has just convinced me that this should be a requirement. I am not sure about inverses, because neither the sum nor the product of natural numbers have inverse operations. – Luca Bressan Oct 13 '13 at 17:25
• Ah, good point. Maybe you should edit the question so people see it as a requirement. – Henrique Oct 13 '13 at 19:17

Note that $(a\vee b)(a\wedge b)$ is equal to $(a^2\vee ab)\wedge(ab\vee b^2)$ - clearly $ab\leq (a^2\vee ab),(ab\vee b^2)$ so $ab$ is a lower bound of this product. An analogous one shows that $ab\geq (a\vee b)(a\wedge b)$ and therefore we conclude that $ab=(a\vee b)(a\wedge b)$.
• Thank you, I didn't notice that the third property was a consequence of the first two. But I still can't see why the first two would hold in all lattice-ordered commutative monoids: it's clear that $c(a\wedge b) \le ca \wedge cb$, but from $d \le ca,\, cb$ how do you deduce that $d \le c(a\wedge b)$? – Luca Bressan Oct 15 '13 at 19:40