What's a structure where $a-a = 1$ and $a\cdot a^{-1} = 0$? I'm trying to specify a structure that has the basic features of a Boolean algebra but not necessarily restricted to binary sets. However, I observed that I almost have a field except that adding/multiplying the inverse ($-a = a^{-1} =: \bar a$) yields the multiplicative/additive inverse ($a \cdot \bar a = 0$, $a + \bar a = 1$ where I interpret "and" as $\cdot$ and "or" as $+$).
To be more precise: My structure $S$ must satisfy the following:
$(S, +, 0)\text{ is an abelian monoid}$
$(S, \cdot, 1)\text{ is an abelian monoid}$
$\cdot\text{ distributes over }+$
$+\text{ distributes over }\cdot$
$\forall a \in S \exists \bar a \in S:\quad a + \bar a = 1 \;\wedge\; a \cdot \bar a = 0 \;\wedge\; \bar{\bar a} = a$
$\forall a,b \in S:\quad a\cdot(a+b) = a \;\wedge\; a+(a\cdot b) = a$
Is there a name for this structure or am I doing something wrong?
 A: A set $A$ together with operations $+$ and $\cdot$ that satisfy your conditions will be a boolean algebra under the operations $\oplus$ and $\otimes$ given by
$$a\oplus b = (a+b)\cdot (\overline{ab}),\quad a\otimes b = a\cdot b.$$
The only sticking points were the idempotency laws, and you've explained how to get them from your conditions.
Conversely, a Boolean algebra $(A,\oplus,\otimes)$ will satisfy the conditions you ask by defining 
$$a+b = (a\oplus b) - (a\otimes b),\quad a\cdot b = a\otimes b,\quad \overline{a} = 1-a.$$
So your structure is essentially equivalent to a boolean algebra, which in turn is essentially equivalent to a complemented and  distributive lattice.
You state in the comments that you believe it will be a complete lattice. I do not think this is the case. Binary operations give you a way to talk about finitary derived operations, but it makes no sense to talk about "series" or "infinite products"
 (which you would probably need to define joins and meets of arbitrary sets) in the abstract. Certainly, there are complemented distributive lattices that are not complete; they will provide examples where your conditions hold but that are not complete lattices. 
(It seems the comment was rather the result of misuse of terminology, though.)
Added. An example of a complemented distributive lattice that is not complete is given by the collection of all Borel subsets of the real line; that is, let $S$ be the $\sigma$-algebra of Borel subsets of $\mathbb{R}$, let $+$ denote union, let $\cdot$ denote intersection, and let $\overline{a}$ denote complementation in $\mathbb{R}$. Then $(S,+,\emptyset)$ is a monoid, since the union of two Borel sets is Borel; $(S,\cdot,\mathbb{R})$ is a monoid, since the intersection of two Borel subsets is Borel. Since union distributes over intersection and vice versa, $+$ distributes over $\cdot$ and $\cdot$ distributes over $+$. For every Borel subset $a$, $a\cup \overline{a} = \mathbb{R}$ and $a\cap \overline{a} = \emptyset$. And for all Borel subsets $a$ and $b$, $a\cap(a\cup b) = a$, $a\cup(a\cap b) = b$. So the Borel subsets satisfy your conditions; they form a distributive complemented lattice under the operations of union and intersection. But they are not a complete lattice, because they are not closed under arbitrary unions. If we let $A$ be a subset of $\mathbb{R}$ that is not a Borel subset (e.g., a Vitali set), then 
the family of singletons taken from $A$ has no least upper bound in the collection of Borel subsets: given any Borel subset that contains $A$, the set must properly contain $A$, so you can always "throw away" a single point, showing the set is not a least upper bound. 
