"Pseudo-Boolean" lattice (almost every element has several complements) If $(L,\leq)$ is a lattice with bottom element $0$ and top element $1$ and $x\in L$ we say that $y$ is a complement of $x$ if $x\vee y = 1$ and $x\wedge y = 0$.
Is there a lattice $(L,\leq)$ with more than $2$ elements and the following properties?

*

*Every element of $L\setminus\{0,1\}$ has at least $2$ complements, and

*for every $a\in L\setminus\{0,1\}$ there is $b\in L$ such that $a \vee b \notin \{b,1\}$ and $a\wedge b \notin \{b,0\}$ (implying in particular that $a,b$ are incomparable).

 A: Among finite lattices, no. A finite lattice (with more than 2 elements) has atoms. Let $a$ be an atom. Condition 2 says there must be an incomparable $b$ that meets $a$ above bottom, which is clearly impossible.
Among infinite lattices, yes. Consider the following lattice, consisting of two slanted infinite ladders, and augmented bottom and top elements. If $a$ is in the left ladder (say), it has infinitely many complements in the right ladder, satisfying condition 1. Also, there is an element $b$ in the same ladder (pictured), which is incomparable with $a$, joins below $1$, and meets above $0$, satisfying condition 2.

A: One can ensure property $2$ simply by taking reduced products.
Suppose that $I$ is an index set and $L_{i}$ is a lattice for each $i\in I$ such that each $x\in L_{i}$ has at least 2 complements.
Then suppose that $\mathcal{F}$ is a filter on $I$ such that $P(I)/\mathcal{F}$ is atomless (i.e. where $\mathcal{F}$ is nowhere an ultrafilter). Then I claim that the reduced product $\prod_{i\in I}L_{i}/\mathcal{F}$ satisfies properties 1 and 2. Let $[(x_{i})_{i\in I}]_{\mathcal{F}}\in\prod_{i\in I}L_{i}/\mathcal{F}$.
Observe that every element of $\prod_{i\in I}L_{i}/\mathcal{F}$ has at least two complements. In fact, the formula $$\forall x\exists y,z,x\wedge y=0\,\text{ and }\,x\vee y=1\,\text{ and }\,x\wedge z=0\,\text{ and }\,x\vee z=1$$ is a Horn sentence, and Horn formulae are always preserved by taking reduced products.
One can show that for each $\mathbf{x}\in\prod_{i\in I}L_{i}/\mathcal{F}$ with $\mathbf{x}\not\in\{0,1\}$, there is some $\mathbf{y}\in\prod_{i\in I}\{0,1\}/\mathcal{F}$ with $\mathbf{x}\vee\mathbf{y}\neq 1,\mathbf{x}\wedge\mathbf{y}\neq 1$ and where $\mathbf{x},\mathbf{y}$ are incomparable.
Automated counterexamples
This counterexample can be produced algorithmically.
The Feferman-Vaught theorem is a result that allows one to compute the truth value of a sentence $\phi$ in a reduced power $\mathcal{A}^{I}/\mathcal{F}$ as long as one is able to compute the truth value of sentences in $P(I)/\mathcal{F}$ and $\mathcal{A}$. In particular, since the theory of atomless Boolean algebras is $\omega$-categorical and hence complete, if $P(I)/\mathcal{F},P(J)/\mathcal{F}$ are atomless, then $\mathcal{A}^{I}/\mathcal{F}$ and $\mathcal{A}^{J}/\mathcal{G}$ are elementarily equivalent. Furthermore, if $\mathcal{A}$ is finite and $P(I)/\mathcal{F}$ is atomless and infinite and $\phi$ is a first order sentence, then the question of whether $\mathcal{A}^{I}/\mathcal{F}\models\phi$ is decidable and independent of $I,\mathcal{F}$.
