Language: consider propositional logic over the connectives $\land$, $\lor$, and $\lnot$.
Consider a propositional formula $\phi$ which contains at least one negation symbol and is neither a contradiction nor a tautology. We are also given that no negation symbols can be cancelled in $\phi$ by any valid biconditional such as $$\alpha \land (\lnot \alpha \lor \beta) \iff \alpha \land \beta$$ $$\lnot \lnot \alpha \iff \alpha$$
Now let's push negation inward to reduce $\phi$ to negation normal form and call that normal form $\phi'$. Just to be clear, $\phi \iff \phi'$.
Can we conclude that $\phi'$ also contains at least one negation symbol? If not, please provide a counterexample.
EDIT: The allowable biconditionals were clarified to include any valid biconditional that would cancel a negation. The intention is to capture the notion that $\phi$ contains no cancellable negations.