I know that M. H. Stone proved that there is a bijection between boolean algebras and boolean rings. The bijection I know is the following: to any given Boolen algebra $(L,\, \vee, \wedge)$ we associate the ring with addition and mutiplication given by the following relations: $$x+y:=(x\wedge y')\vee (x'\wedge y)$$ where the $x'$ is the complement of $x$; $$xy=x\wedge y.$$
Anyway trying to prove that relation I find that starting with $(L, \vee, \wedge)$ it is possible to define the respective ring without the absorption laws. Is it true or have I necesserily made some mistake in the proof?
(Question from the comments which is in some sense the point of the original question): Is it possible to prove absorption from "$a \wedge a = a$" and the others? Yes:
Lemma 1: for all $a$, $a \wedge 0 = 0$.
$a \wedge \neg a = 0$ (complement)
$a \wedge a \wedge \neg a = 0$ ($a \wedge a = a$)
$a \wedge 0 = 0$ (complement)
Lemma 2: $a \wedge (a \vee b) = a$ (i.e absorption one way round)
$(a \vee 0) \wedge (a \vee b) = a \vee (0 \wedge b)$ (distributivity)
$a \wedge (a \vee b) = a \vee 0$ (identity, Lemma 1, and commutativity)
$a \wedge (a \vee b) = a$ (identity)
The other absorption axiom is $a \vee (a \wedge b) = a$. We prove this:
$a \wedge (a \vee b) = (a \wedge a) \vee (a \wedge b)$ (distributivity)
$a = a \vee (a \wedge b)$ (Lemma 2 and $a \wedge a = a$)