On the functional-completeness of the sheffer stroke I have seen functional-completeness (in regards to boolean functions) defined as: > A set X of truth-functions (of 2-valued logic) is functionally complete if and only if for each of the five defined classes, there is a member of X which does not belong to that class With those 5 classes being: > Truth-Preserving > > False-Preserving > > Linear > > Monotone > > Self-Dual So since | is functionally-complete, this means it is not any one of the 5 type of functions listed above, right? But I am not quite seeing how it is not self-dual, it seems like it IS self-dual. Could someone go over the self-dual functions and show the sheffer stroke is not self-dual?

A Boolean operator $f(x_1, \dotsc, x_n)$ is _self-dual_ iff $ f(\
eg x_1, \dotsc, \
eg x_n) = \
eg f(x_1, \dotsc, x_n)$.

In the case of the Sheffer stroke (NAND), $(p|q) \equiv \
eg (p \land q) \equiv (\
eg p \lor \
eg q)$, so $(\
eg p | \
eg q) \equiv (p\lor q)$, whereas $\
eg (p|q) \equiv (p \land q)$. As $\lor$ and $\land$ are distinct Boolean functions, $|$ isn't self-dual.