Let $S$ be the set of all sentence symbols. Let $B\subseteq\overline{S}$ be the set of wffs $\beta$ such that $\beta^{*}$ is tautologically equivalent to $(\neg\beta)$ . Now, since a wff $\beta$ consisting of a single sentence symbol $A_{n}$ does not include any connective symbols, $\beta^{*}=(\neg A_{n})=(\neg\beta)$ , i.e. all sentence symbols are in $B$ . Suppose that $\gamma,\beta\in B$ . Then, $(\neg\gamma)^{*}=(\neg\gamma^{*})$ , as we do the substitutions inside the $\gamma$ only, and the latter is tautologically equivalent to $(\neg(\neg\gamma))$ . Further, $(\gamma\square\beta)^{*}=(\gamma^{*}\overline{\square}\beta^{*})$ , where $\overline{\wedge}=\vee$ and $\overline{\vee}=\wedge$ , and the latter is tautologically equivalent to $((\neg\gamma)\overline{\square}(\neg\beta))$ , which is equivalent to $(\neg(\gamma\square\beta))$ (see De Morgan’s laws at p.27). In summary, we conclude that $(\neg\gamma),(\gamma\square\beta)\in B$ , and, by induction, $B$ consists of at least those wffs that use the connective symbols $\wedge$ , $\vee$ and $\neg$ , in particular, $\alpha\in B$ .