Vtyjkyuk

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:

beginalign
t &rightarrow (1,0) \
f &rightarrow (0,1) \
bot &rightarrow (0,0) \
top &rightarrow (1,1)
endalign

The operations are defined as follows:

beginalign
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)
endalign

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 $avee neg a=1$ and $awedge neg a=0$ do not work. A really quick way to see this is that those axioms imply $aneq neg a$ (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).