Belnap’s logic contains the the truth values 'true' ($t$), 'false' ($f$), 'unknown' ($\bot$) and 'paradox' ($\top$). Each of these is represented by a pair of bits:
\begin{align} t &\rightarrow (1,0) \\ f &\rightarrow (0,1) \\ \bot &\rightarrow (0,0) \\ \top &\rightarrow (1,1) \end{align}
The operations are defined as follows:
\begin{align} \land &: \bigl((x_1,y_1),(x_2,y_2)\bigr) &&\rightarrow \bigl(\min(x_1,x_2), \max(y_1,y_2)\bigr) \\ \lor &: \bigl((x_1,y_1),(x_2,y_2)\bigr) &&\rightarrow \bigl(\max(x_1,x_2), \min(y_1,y_2)\bigr) \\ \lnot &: (x,y) &&\rightarrow (y,x) \end{align}
I am wondering, whether Belnap’s four valued-valued logic, with the set of truth values $\{t,f,\bot,\top\}$ and the operations $\land,\lor,\lnot$ is a boolean algebra, and if so why?
EDIT: The complements-rule ($a ∨ ¬a = 1$ and $a ∧ ¬a = 0$) doesn’t work, does it?