Is Belnap's four valued-logic a boolean algebra?

Question

Belnap's logic contains the the truth values 'true' (t), 'false' (f), 'unknown' ($\bot$) and 'paradox' (T). Each of these is represented by pair a of bits:

t $\rightarrow$ (1,0)
f $\rightarrow$ (0,1)
$\bot$ $\rightarrow$ (0,0)
T $\rightarrow$ (1,1)

The operations are defined as follows:
$\wedge$ : $((x_1, y_1),(x_2, y_2)) \rightarrow$ $($min$\{x_1, x_2\}$, max$\{y_1,y_2\}$)
$\vee$ : $((x_1, y_1),(x_2, y_2)) \rightarrow$ $($max$\{x_1, x_2\}$, min$\{y_1,y_2\}$)
$\neg$ : ($x, y$) $\rightarrow$ ($y, x$)

I am wondering, whether Belnap's four valued-valued logic, with the set of truth values $\{t, f, \bot,$ T$\}$ and the operations $\wedge, \vee, \neg$ is a boolean algebra, and if so why?

Answer 1

Of course just checking the axioms one by one would be tedious. No doubt someone will give an elegant explanation - just for fun here's a little Python that verifies one of DeMorgan's laws:

t=(1,0)
f=(0,1)
u=(0,0)
p=(1,1)

values = (t,f,u,p)

def And(x,y):
  return (min(x[0],y[0]),max(x[1],y[1]))

def Or(x,y):
  return (max(x[0],y[0]),min(x[1],y[1]))

def Not(x):
  return (x[1], x[0])

for x in values:
  for y in values:
    if not Not(And(x,y)) == Or(Not(x),Not(y)):
      print 'oops'

print 'done'

If all else fails you could easily do that for the other axioms as well...

Please don't tell anyone I posted this. The thing about adding four spaces to the start of each line actually works, heh.