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.

  • $\begingroup$ See Karnaugh map. $\endgroup$ – Git Gud Oct 29 '14 at 12:56
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    $\begingroup$ I believe you mean $n \in \mathbb{N}$.. $\endgroup$ – Bruno Bentzen Oct 29 '14 at 14:48

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?)

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