# Can numbers (not sets of numbers) in Linear Integer Arithmetic form a Boolean algebra?

As far as I understood, Boolean algebra is just one of the many first-order theories (1). It has the signature $$\{\sqcap, \sqcup, \neg, \bot, \top\}$$ and the axioms: associativity, commutativity, absorption, identity, distributivity, and complements (2). (I used $$\sqcup$$ and $$\sqcap$$ to distinguish them from the Boolean operators $$\vee$$ and $$\wedge$$, they seem to be different things.)

1. Can we say that the theory of linear integer arithmetic (with the signature $$\{+, -, n\cdot_{n\in\{..., -2, -1, 1, 2, ...\}}, <, ..., -2, -1, 0, 1, 2, ...\}$$) forms a Boolean algebra? I.e., we can map elements of the Boolean-algebra theory to the elements of the LIA theory? How would you map $$\sqcap$$, $$\sqcup$$, $$\neg$$, $$\bot$$, $$\top$$? I guess, $$\top \mapsto 1$$, $$\bot \mapsto 0$$, $$\neg \mapsto -1\cdot$$ (multiplication by constant -1), but what about $$\sqcap$$ and $$\sqcup$$?

2. What about other FOTs, e.g. Presburger arithmetic (which does not have multiplication by a constant $$const\cdot$$)? Is it true that many FOTs form a Boolean algebra?

(Related references would be appreciated.)

Update: Essentially, the question is whether the numbers in the theory of linear integer arithmetic form a Boolean algebra. I don't think that is true. But it is possible to form Boolean algebra using sets of numbers, i.e., predicates of the linear integer arithmetic form a Boolean algebra. For example, $$a<10 \sqcup b<10$$ becomes $$a<10 \vee b<10$$, $$a<10 \sqcap b<10$$ becomes $$a<10 \wedge b<10$$, the negation operator $$\neg$$ of the Boolean algebra is simply the logical operator $$\neg$$, and so on. Hm, I wonder if the numbers of the non-linear arithmetic can form a Boolean algebra.