Proving that a "Boolean" algebra containing an element $i$ such that $i\land0=1$ is not distributive. The following is a substantial rewriting of the question by someone other than the OP. Earlier versions can be found in the history. - Noah

Throughout, we're looking at structures in the language $\{\vee,\wedge,',0,1\}$, where (per usual) $\vee,\wedge$ are binary function symbols, $'$ is a unary function symbol, and $0,1$ are constants.
Huntington showed (theorems X and XIII especially) that the following axioms suffice to characterize Boolean algebras:


*

*$0$ and $1$ are identities for $\vee$ and $\wedge$, respectively.

*Commutativity of $\wedge$ and $\vee$.

*The complementation rules $x\wedge x'=0$ and $x\vee x'=1$.

*Distributivity of $\vee$ over $\wedge$ and of $\wedge$ over $\vee$.
(That is, Huntington showed that associativity and absorption are derivable from the others.)
Say that an algebra is strange iff it satisfies all of Huntington's axioms except the distributivity laws and has some element $i$ with $i\wedge 0=1$. Clearly the only strange algebra which is distributive is the trivial (= one-element) algebra.

Question: Is there a nontrivial strange algebra? If so, how does distributivity fail in nontrivial strange algebras (e.g. is there a canonical counterexample)?

In some sense we already know how distributivity must fail in a strange algebra: some instance of distributivity used in the proof of the remaining Boolean algebra axioms from Huntington's four must fail. However, there are several such uses, and there could be no nontrivial strange algebras in the first place.
 A: Without distributivity, these axioms are extremely weak and so there are lots of ways to get strange algebras.  For instance, you can start with any structure $A$ over the language $\{0,1,\wedge,\vee\}$ satisfying the identity and commutative axioms (so just any set together with two commutative operations which have identity elements!) and then consider $B=A\cup A'$ where $A'$ some set disjoint from $A$ with a bijection $f:A\setminus\{0,1\}\to A'$ which we write as $f(a)=a'$ (in other words, we just formally adjoint new elements to be the complements of all the elements of $A$ except $0$ and $1$).  We can then make $B$ into a strange algebra by defining $0'=1$, $1=0'$, $(a')'=a$ for $a\in A$, $a\vee a'=1$, $a\wedge a'=0$, and defining $a\wedge b$ and $a\vee b$ to be arbitrary commutative operations in all other cases that have not yet been defined (i.e. when at least one of $a$ and $b$ is in $A'$ and $a$ and $b$ are not complements of each other).  In particular, we can choose to have $i\wedge 0=1$ for some $i\in A'$ that is different from $0'$.
For instance, starting from a three-element set $A=\{0,1,x\}$, we will get a strange algebra $B=\{0,1,x,y\}$ where $y=x'$ and $x=y'$ and we can have $y\wedge 0=1$.  (Note that there are lots of degrees of freedom here: we can define all of $0\wedge 0$, $1\vee 1$, $x\wedge x$, $x\vee x$, $x\wedge 0$, $x\vee 1$, $y\wedge y$, $y\vee y$, and $y\vee 1$ to be whatever we want!)
There are many variations on this construction; for instance, there's no reason we must have $(a')'=a$, so we could instead take $B=A\cup A'\cup A''\cup\dots$ where we formally adjoin $n$-fold complements to every element of $A\setminus\{0,1\}$.
As for where exactly distributivity must fail, you can just look to the proof of $i\wedge 0=0$ using distributivity, which goes like this: $$0=i\wedge i'=i\wedge(i'\vee 0)=(i\wedge i')\vee (i\wedge 0)=0\vee(i\wedge 0)=i\wedge 0.$$  So if we have $i\wedge 0=1$ and $0\neq 1$, the specific case of distributivity in this proof must fail, i.e. $i\wedge (i'\vee 0)$ must fail to be equal to $(i\vee i')\wedge (i\wedge 0)$.  (Note though that we could have $0=1$ and still have a nontrivial strange algebra; for instance, $\wedge$ and $\vee$ could be the same operation which is commutative and has an identity and an inverse for every element.  So in full generality, I wouldn't think that there's any "canonical way" that distributivity must fail.)
