I have to study systems of equations in a Boolean algebra, the matrix is $m\times n$ with $m\neq n$. The Boolean algebra is actually the simplest one, it contains only $0$ and $1$, let us denote it by $\mathbb{B}$. What I need to know is a necessary and sufficient condition for an application from $\mathbb{B}^n$ to $\mathbb{B}^m$ to be one-to-one. I read in the paper ["Linear Boolean Equations and Generalized Minterms" by S. Rudeanu (Discrete Math 43 (1983) 241-248)](http://www.sciencedirect.com/science/article/pii/0012365X83901619) that Löwenheim proved some theorems in a paper written in 1919. Hence my question : 



> 
> Is there any more recent reference about this subject (systems of
>  equations in Boolean algebra) ? And where can I find a proof of
>  Löwenheim's theorem (that could help to understand) ?
> 
> 
> 


All references I can found in some papers I can find (with difficulty) on the Web are unavailable online, and unavailable in my library. 


EDIT: Here is Löwenheim's theorem I mention above: In a Boolean algebra $(\mathbb{B},\cup,.,',0,1)$ (I guess that $.$ is the intersection and $'$ the negation), to each $(b\_1,\dots,b\_m)\in \mathbb{B}^m$, we can associate the system of equations $$\bigcup\_{j=1}^na\_{ij}x\_j=b\_i \ \ \ (i=1,\dots,m).$$
Löwenheim proved in a 1919 paper that the system is consistent (I guess that that means that there is a solution) for a given $(b\_1,\dots,b\_m)\in \mathbb{B}^m$ if and only if $$b\_i\leq \bigcup\_{j=1}^n a\_{ij} \prod\_{h=1,h\neq i}^m (a'\_{hj}\cup b\_h)\ \ \ (i=1,\dots,m).$$
I guess that $\leq$ means the inclusion (it is not explained in the paper). In the same paper, the author calls $x+y=xy'\cup x'y$ the ring sum which is "xor".



Systems of equations in Boolean Algebra
Let $A$ be an $m\times n$ Boolean matrix. Then the mapping $v\mapsto Av$ is 1:1 iff $n\leq m$ and some subset of $n$ rows of gives a permutation matrix. 


The reason is duality of modules over the Boolean semiring shows that $A$ is 1:1 iff the transpose is onto. Since the standard basis vectors of a free $\mathbb B$-module are join irreducible a Boolean matrix gives an onto map iff each standard basis vector appears as a column. 


**Added per request for more detail.** 


A $\mathbb B$-module is the same thing as a join-semilattice $M$. The dual $M^\*$ is the set of all join-semilattice homomorphisms $M\to \mathbb B$ with pointwise join. If $M$ is finite, then the dual consists of the mappings $f\_m$ with $m\in M$ where $f\_m(x)=0$ if $x\leq m$ and $1$ else. Indeed, choose $m$ to be the join of all elements mapping to $0$. 


It follows if $N$ is a submodule of $M$ then the restriction $M^\*\to N^\*$ is surjective. Also the canonical map to the double dual is an ISO for $M$ finite. It now follows a map of finite $\mathbb B$-modules is injective iff the dual map is surjective. For a matrix map the dual is the transpose. 


If we view $\mathbb B^n$ as a join semilattice, the basis consists of atoms. It follows a matrix transformation is onto iff each basis vector appears in some column. Dualizing gives the result. 


