It is easy to think up interesting, natural models of the algebraic theory presented as follows, such that in these models, $x^\dagger$ is always the multiplicative inverse of $x$.



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> **Question.** What are some interesting, natural models in which $x^\dagger$ is not necessarily the multiplicative inverse?
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**Sorts.** $U$


**Functions.** $$\wedge : U \times U \rightarrow U, \qquad \vee : U \times U \rightarrow U$$


$$1 : U, \qquad \times : U \times U \rightarrow U, \qquad (x \mapsto x^{\dagger}) : U \rightarrow U$$


**Axioms.** (I write $\times$ concatenatively).


1. $(U,\wedge,\vee)$ is a distributive lattice
2. $(U,1,\times)$ is a commutative monoid
3. Multiplication distributes over both $\wedge$ and $\vee$
4. $1^\dagger = 1$
5. $(xy)^\dagger = x^\dagger y^\dagger$
6. $x^{\dagger\dagger} = x$
7. $(x \wedge y)^\dagger = x^\dagger \vee y^\dagger$
8. $(x \vee y)^\dagger = x^\dagger \wedge y^\dagger$


**Examples of such things in which $x^\dagger$ is always the multiplicative inverse.**


* $\mathbb{Q}\_{>0}$ or $\mathbb{R}\_{>0}$ with $\wedge$ and $\vee$ interpreted as $\mathrm{min}$ and $\mathrm{max}$ respectively.
* $\mathbb{Z}$ with $(1,\times,x \mapsto x^\dagger)$ interpreted as $(0,+,x \mapsto -x)$, and $\wedge$ and $\vee$ interpreted as $\mathrm{min}$ and $\mathrm{max}$ respectively.
* Cartesian products of these



Looking for interesting, natural models of this algebraic theory in which $x^\dagger$ is not always the multiplicative inverse of $x$
**not a solution**  

Anything that satisfies 1,2,3; then define $x^\dagger = 1$ for all $x$. 


**added December 27**  

Try this example...  

$U := \mathbb Z \times \mathbb Z$. 


* multiplication is performed by componentwise addition, $(a,b)(c,d) = (a+c,b+d)$
* the lattice operations are also performed componentwise, $(a,b) \vee (c,d) = (a\vee c,b\vee d)$, $(a,b) \wedge (c,d) = (a\wedge c,b\wedge d)$
* but reflect the symmetry, $(a,b)^\dagger = (-b,-a)$


Now there are lots of things to check, to see if it really works.


