4
$\begingroup$

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?

$\endgroup$
1
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.