# FOUR-algebra - boolean algebra?

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?

As you say, the complements axioms $a\vee \neg a=1$ and $a\wedge \neg a=0$ do not work. A really quick way to see this is that those axioms imply $a\neq \neg a$ (unless $0=1$ in which case the Boolean algebra can only have one element), but $\bot$ and $\top$ do satisfy $a=\neg a$.
Note that if you ignore the given definition of $\neg$, then you actually do have a Boolean algebra (just with a different definition of negation). Indeed, if you flip the values of the second bits (so $t$ is written as $(1,1)$, $f$ as $(0,0)$, $\bot$ as $(0,1)$, and $\top$ as $(1,0)$), then the operations $\wedge$ and $\vee$ just become the usual Boolean operations on $\{0,1\}^2$ (namely, coordinatewise min and max, respectively). The correct negation operation for this Boolean algebra is coordinatewise negation, in other words $\neg(x,y)=(1-x,1-y)$ (a formula which does not change if you flip all values of the second bit).