Could anyone check if my proof is ok/ suggest any improvement please? I couldn't find a way to utilise the induction hypothesis so I am not sure if this is ok.
Let $φ$ be a formula built up using the connectives $¬, ∧, ∨$. The dual $φ$' of $φ$ is the formula obtained from $φ$ by replacing all occurrences of $∧$ by $∨$, of $∨$ by $∧$, and all propositional variables by their negations. Show that $φ$' is logically equivalent to $¬φ$.
Induction on occurrence of operator.
Base case: operator occurence of $φ$'=$0$ - this is only possible if double negation elimination has been applied, since if $φ$=$P$ $\to$ $φ$'=$\lnot P$, or $φ$=$\lnot P$ $\to$ $φ$'=$\lnot \lnot P$. $φ$' has at least one operator no matter what. Therefore $φ$ must already have a $\lnot$ and get another $\lnot$ added on when transformed into $φ$', then have them cancelled by double negation. Thus $φ$' $\equiv$ $\lnot φ$.
Hypothesis: Assume that for all $φ$' whose length $\le n$, $φ$' $\equiv φ$.
Since $φ$ is only built up with 3 types of operators, there can only be 3 possible scenarios:
$φ$'=$\psi$ $\land$ $\phi$: Therefore $φ=\lnot \psi \lor \lnot \phi$. Applying DeMorgan then $φ$'$=\lnotφ=\lnot(\lnot\psi\lor\lnot\phi)$.
$φ$'=$\psi$ $\lor$ $\phi$: Similar approach to 1.
$φ$'=$\lnot \psi$:Therefore $φ=\lnot \psi$, and $φ$'$=φ=\lnot \psi$.