TITLE: Show functionally completeness property for propositional logic
QUESTION [6 upvotes]: Let $n>0, n\in \mathbb{Z}$ and let t,f denote true and false.
For every function
$$g:\{t,f \}^n \to \{t,f\} $$ There is a propositional forumala $B$,
  using only the connectives $\neg, \wedge$ and built up from  from the
  atomic formulas $P_1,\ldots, P_n$ such that for every truth assignment
  $$\mathcal{A}:\{P_1,\ldots, P_n\}\to \{t,f \}$$
$$A\models B \text{ if and only if } g(\mathcal{A}(P_1),\ldots ,
> \mathcal{A}(P_1)) = t$$

I don't really know how to prove this. I understand that I somehow must code the above as a formula and then "reducing" it to a equivalent formula consisting only of $\neg, \wedge$ using well known equivalences and then prove the above with induction. I don't really where to start. Any help or references to where this is proved in detail would be greatly appreciated.

REPLY [5 votes]: The proof naturally goes as an induction on $n$:
(Basis Case) Let $n=1$. It suffices to show that there is a formula consisting of $\neg, \wedge$ with the all possible unary connectives defined by the following 4 truth tables:
$$\begin{array} {|c|}
\hline
P & \neg(P\wedge\neg P) \\ \hline
1 & 1 \\ \hline
0 & 1 \\ \hline
\end{array}
\begin{array} {|c|}
\hline
P & P \\ \hline
1 & 1 \\ \hline
0 & 0 \\ \hline
\end{array}
\begin{array} {|c|}
\hline
P & \neg P \\ \hline
1 & 0 \\ \hline
0 & 1 \\ \hline
\end{array}
\begin{array} {|c|}
\hline
P & P\wedge\neg P \\ \hline
1 & 0 \\ \hline
0 & 0 \\ \hline
\end{array}$$
(Inductive Case) Inductive Hypothesis: Suppose that for every $n$-ary connective defined by its valuation function, there are formulae consisting of $\neg, \wedge$. We need to show that for every $n+1$-ary connectives, propositions have been found. (Can you continue from here?)