Are Boolean functions Turing complete
Informally, a (programming) language is Turing complete if every computable function has a representation. A general computable function accepts an input of arbitrary size. Boolean functions, on the other hand, accept an input of a fixed size. Hence Boolean functions don't even qualify as potentially Turing-complete.

The relevant notion of completeness here is a complete basis of connectives. A set of connectives ($k$-ary functions on Boolean values for arbitrary $k$) is complete if every Boolean function on $x_1,\ldots,x_n$ (for arbitrary $n \geq 1$) can be represented using the connectives. The following sets are complete: the de Morgan basis $\{\lnot,\lor,\land\}$ and the basis $\{\lnot,\Rightarrow\}$. In contrast, $\{\lnot,\oplus\}$ is not complete: it can only express linear functions.