Truth table and induction It is true that every truth table can be represented by some wff built using only the connectives $\neg, \implies$ and $\iff$ - let's call it "negation-arrow-wff" for convenience. I want to be able to prove this, using induction (perhaps). Is the following approach correct? 
Induction on number of propositional variables. In the inductive step I take a truth table with $n-1$ propositional variables $P_1,P_2,\dots,P_{n-1}$ represented by some negation-arrow-wff $W_{n-1}=W_{n-1}(P_1,P_2,\dots,P_{n-1})$. Then "adjoining" an $n$th propositional variable $P_n$ to the truth table, I have to consider 16 cases for the 16 possibilities that may arise as I try to form a negation-arrow-wff $W_n=W_n(P_1,P_2,\dots,P_n)$ from the propositional variable $P_n$ and a negation-arrow-wff $W_{n-1}$. If I can show that in all 16 cases $W_n$ can be built from $P_n$ and $W_{n-1}$ using only the allowed connectives then this completes the inductive step.
 A: No, I don't think induction is a fruitful approach — at least, not that induction.
Where do you get $16$ from, anyway? That value is hard-coded for $n = 4$ variables, right? With $n-1$ variables, you previously had $2^{n-1}$ rows in the truth table; now you'll have twice as many on introducing $P_n$, i.e. $2^n$ rows. 
A better approach: Note that you can express both $\land$ and $\lor$ using $\neg, \to$:
$$\begin{align}
p \land q \equiv \neg(p\to\neg q) \\
p \lor q \equiv (\neg p\to q).
\end{align}$$
Now, given a truth table with $n$ variables, you can write an equivalent formula in disjunctive normal form (DNF): a disjunction ($\lor$) of conjunctions ($\land$) of the $n$ variables or negations of them. You'll have one conjunction for every row in the truth table whose final column is $\top$ (true), ignoring rows whose final column has $\bot$ (false). For each $\top$ row, use $P_i$ if $P_i$ has value $\top$ in that row, and use $\neg P_i$ if $P_i$ has value $\bot$ in that row. 
Number the rows from $1$ to $2^n$. For $k = 1, \dotsc 2^n$ and $i = 1, \dotsc n$, let $v_{k,i}$ be the value of $P_i$ in row $k$. Let 
$$
P^k_i = \begin{cases} P_i      &\quad\text{if $v_{k,i} = \top$,} \\
                      \neg P_i &\quad\text{if $v_{k,i} = \bot$.}
\end{cases}
$$.
Let $R$ be the set of true rows in the truth table. The the equivalent formula is:
$$
\bigvee_{k\in R} \bigwedge_{i=1}^n P^k_i. 
$$
