I have two 3-ary propositional functions given by the table
$$ \begin{array}{|c|c|c|c|c|} v(a) & v(b) & v(c) & v(f(a, b, c)) & v(g(a, b, c)) \\ \hline 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 \\ \hline \end{array} $$
And I need to decide which of these functions are complete. I made some googling but I didn't get the way how to solve such type of tasks. Books say that $L(\vee,\wedge,\neg)$, $L(\wedge,\neg)$, ... are complete and $L(\wedge)$, $L(\vee)$... are not. But how to apply this on the functions given by table?
I gues I need to make propositional formula from $f$ and $g$. Am I right? If so what's next?
We can try exploiting the fact that we know that : $\lnot, \land$ is complete.
The ternary connective $g(a,b,c)$ is the minority connective : $v(g(a,b,c))$ always disagrees with the majority of $v(a),v(b),v(c)$.
We can formally set :
We can define $\lnot$, because $v(g(a,a,a))=1$ iff $v(a)=0$.
But we cannot define $a \land b$, because $v(a \land b)=1$ iff $v(a)+v(b) = 2$.
Thus, it is not complete.
It seems to me that $f(a,b,c)$ is complete.
We can formally set :
In this way, we can define $\lnot$, because $v(f(a,a,a))=1$ iff $v(a)=0$.
We can check also that $f(a,b,c)$ is equivalent to : $\lnot c \lor (\lnot a \land \lnot b)$.
Obviously : $\lnot a \land \lnot a \equiv \lnot a$; thus $f(a,a,c)$ is $\lnot a \lor \lnot c$ i.e. $\lnot (a \land c)$ that is NAND, or Sheffer stroke and we know that it is complete.