Q) For $x \in \{0,1\}$, let $\lnot x$ denote the negation of $x$, that is $$\lnot \, x = \begin{cases}1 & \mbox{iff } x = 0\\ 0 & \mbox{iff } x = 1\end{cases}$$. If $x \in \{0,1\}^n$, then $\lnot \, x$ denotes the component wise negation of $x$; that is: $$(\lnot \, x)_i=\left (\lnot \, x_i \mid i \in [1..n] \right )$$
Consider a circuit $C$, computing a function $f: \{0,1\}^{n} \to \{0,1\}$ using AND $(\land)$, OR $(\lor)$,and NOT $(\lnot)$ gates. Let $D$ be the circuit obtained from $C$ by replacing each AND gate by an OR gate and replacing each OR gate by an AND. Suppose $D$ computes the function $g$. Then it is given as $g(x) = \lnot f(\lnot x).$ It is the definition of self-dual boolean function and I was trying to prove it using induction on '$n$'.
Base Step : For $n=2$, assuming boolean variables $x_1,x_2 \in \{0,1\}$ and circuit $C$ has one or more than $1$ AND gate then $f(x) = f(x_1,x_2) = x_1.x_2$. Now, $\lnot f(\lnot x) = \lnot f(\lnot x_1,\lnot x_2) = \lnot(\lnot x_1 . \lnot x_2) = x_1 + x_2$ which is same as output of OR gate with input $x_1,x_2$ and it is circuit $D$ with one or more OR gate.
Similarly, I can show it for Circuit $C$ with one or more OR gate.
Inductive Hypothesis : For $x \in \{0,1\}^n$ and $x_1,x_2,.....,x_n \in \{0,1\},$ $g(x) = \lnot f(\lnot x).$
Inductive Step : If circuit cotains only AND gates then if we add one more boolean variable $x_n$ and one AND gate then output of circuit C is $f(x) = x_1.x_2.....x_n.x_{n+1}$. So, here, according to De-Morgan's Law for $n+1$ boolaen variables $g(x) = \lnot f(\lnot x).$ is true. Similarly, it is true for OR gates. Now, when circuit $C$ has combination of AND and OR gates then how to proceed it ? Any help would be appreciated.
HINT: Please note that your circuit can be arbitrarily complex, even with just $2$ variables and just a bunch of AND gates, e.g. you can have a circuit corresponding to $(x_1 \cdot x_2) \cdot (x_2 \cdot (x_1 \cdot x_1))$
So, don't try to prove this by induction on $n$, but rather try to prove this based on how the circuit is structured. Thus, the base case is a single variable, while the inductive step is taking any two circuits and applying either an AND or an OR to their outputs.