# 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.

## 2 Answers

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. $0\neq\neg 0=1$.

• Are the "not" functions basically the only self-dual ones? Most of the ones I check seem to not be self-dual (implies, implied-by, iff, and, or, etc) – user525966 Mar 19 '18 at 1:10
• @user525966 The identity functions are self-dual too – Bram28 Mar 19 '18 at 1:15
• @Bram28 It looks like the identity function is everything (T_0, T_1, L, M, S) – user525966 Mar 19 '18 at 1:30
• 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
• @user525966 Of course ... an identity function is pretty useless, since you can just use the variable it picks out instead, i.e. the identity function does not help you create new truth-functions. – Bram28 Mar 21 '18 at 1:15