I am trying to show that if $f\notin T_0\cup T_1\cup S,$ where $T_0$ and $T_1$ are the sets of zero and one-preserving Boolean functions respectfully and $S$ is the set of all self-dual boolean functions, then $f$ is a Sheffer function (Sheffer function is a function for which $\{f\}$ is a complete set).
We can easily show that $f$ is not monotonic, as $f(0,\ldots,0)=1$ and $f(1,\ldots,1)=0$, but $f(0, 0, \dots,0)>f(1,1,\ldots,1)$, so $f$ is not monotonic.
I don't know how to show that $f$ is not linear, though.
Let's suppose $f$ is linear, then $$f=a_0\oplus a_1x_1\oplus\ldots\oplus a_nx_n$$ and as $f(0, \ldots, 0)=1,$ then $a_0=1$. We also have $f(1,\ldots,1)=0$, so $$1\oplus a_1\oplus\ldots \oplus a_n=0$$ I don't know what to do from here.