Are true and false each considered self-dual?

Normally a Boolean function is self dual if:

$$f(x_1, x_2, ..., x_n) = \lnot f(\lnot x_1, \lnot x_2, ..., \lnot x_n)$$

For example the "not" function is self-dual:

$$\lnot x_1 = \lnot (\lnot (\lnot x_1)) = \lnot x_1$$

But the "and" function is not:

$$x_1 \land x_2 \neq \lnot(\lnot x_1 \land \lnot x_2) = x_1 \lor x_2$$

But how do we model something like true $1$ or false $0$? Are they self-dual? I don't know how to model them as functions since they don't take arguments.

For example if $f = \text{True}$ then I don't even know how we'd write a self-dual when there are no function arguments to negate.

They are actually each other's dual, because you can see the True as a conjunction of $0$ conjuncts (it is trivially true that 'all' $0$ conjuncts are true) and the False as a disjunction of $0$ disjuncts (if there are no disjuncts, then there is not at least one true disjunct) and the conjunction and disjunction are each other's dual.
You can view $0$ and $1$ as nullary functions. Just consider the nullary case of the scheme you showed. $f()=\neg f()$. From this you can see that $0$ and $1$ are not self-dual. $0\neq\neg 0=1$.
• Given a function $f$, its dual $f^D$, and a variable $x$ that does not appear in $f$, then $(x \wedge f) \vee (\neg x \wedge f^D)$ is self-dual. – Fabio Somenzi Mar 19 '18 at 1:40