(a) Let $S$ be the set of all sentence symbols. Let $B\subseteq\overline{S}$ be the set of all wffs $\alpha$ such that $\overline{u}(\alpha)=\overline{v}(\alpha^{*})$ . Then, by assumption, all sentence symbols are in $B$ (here we assume that if there is a sentence symbol $A$ different from all $A_{n}$ , then $\overline{u}(A)=u(A)=v(A)=\overline{v}(A)=\overline{v}(A^{*})$ , though, I think, what is meant here is that $A_{n}$ ’s form the set of all sentence symbols, as was defined earlier in the text). Now, if $\alpha,\beta\in B$ , then $\overline{u}((\neg\alpha))=\neg\overline{u}(\alpha)=\neg\overline{v}(\alpha^{*})=\overline{v}((\neg\alpha^{*}))=\overline{v}((\neg\alpha)^{*})$ , where, by definition, $\neg T=F$ and $\neg F=T$ , and, similarly, $\overline{u}(\alpha\square\beta)=\overline{u}(\alpha)\square\overline{u}(\beta)=\overline{v}(\alpha^{*})\square\overline{v}(\beta^{*})=\overline{v}((\alpha^{*}\square\beta^{*}))=\overline{v}((\alpha\square\beta)^{*})$ , where, by definition, $T\square F$ is valuated according to the truth tables. One thing to mention here is that we used equalities $(\neg\alpha^{*})=(\neg\alpha)^{*}$ and $(\alpha^{*}\square\beta^{*})=(\alpha\square\beta)^{*}$ which simply mean that the substitution in a formula is equivalent to the substitution in its components, which is easy to accept or show using induction. Finally, by the induction principle, we conclude that $B=\overline{S}$ , and $\phi\in B$ .

(b) ($\phi$ is a tautology) iff (every truth assignment $u$ on $S$ implies $\overline{u}(\phi)=T$ ) THEN (here is where iff does not work) (every truth assignment $v$ on $S$ implies $\overline{v}(\phi^{*})=\overline{u}(\phi)=T$ , where $u$ is defined as in (a)) iff ($\phi^{*}$ is a tautology). To see that the other direction does not work, consider, for example, the wff $\phi=(A_{1}\vee(\neg A_{2}))$ , which is not a tautology, together with the substitution $\alpha_{n}=A_{1}$ . Then, $\phi^{*}=(A_{1}\vee(\neg A_{1}))$ , which is a tautology.