Remember that $+$ is the “exclusive or” connector. Again, similar to the two previous exercises, we look for a property such that a) it holds for sentence symbols, b) it holds for $\alpha\square\beta$ , $\square\in\{\top,\bot,\neg,\leftrightarrow,+\}$ , whenever it holds for $\alpha$ and $\beta$ , and c) it does not hold for any connective, in particular, for $\downarrow$ and $\vert$ . We consider the property as suggested, and call any wff using sentence symbols $A$ and $B$ only and satisfying the property, “even”. We are going to use induction. Let $S=\{A,B\}$ . Let $\mathcal{B}\subseteq\overline{S}$ be the set of all even wffs. We have four possible truth assignments on $S$ . Among those $T$ is assigned to $A$ twice, and $T$ is assigned to $B$ twice as well. Therefore, both sentence symbols are even wffs, i.e. $S\subseteq\mathcal{B}$ . Suppose, $\alpha$ and $\beta$ are even having $a$ and $b$ $T$ -values, correspondingly, and $c$ truth assignments such that both are true ($a$ and $b$ are even, $c$ does not have to be). Then, $\top$ is even ($4$ $T$ ’s), $\bot$ is even ($0$ $T$ ’s), $\neg\alpha$ is even ($4-a$ $T$ ’s), $\alpha\leftrightarrow\beta$ is even ($4-a-b+2c$ $T$ ’s), and $\alpha+\beta$ is even ($a+b-2c$ $T$ ’s). By induction, $\mathcal{B}$ contains all the wffs that can be built using $\{\top,\bot,\neg,\leftrightarrow,+\}$ connective symbols only. Note, that $\overline{S}-\mathcal{B}$ is not empty, as neither $\downarrow$ nor $\vert$ is even. Hence, those cannot be tautologically equivalent to any wffs built using connectives from $\{\top,\bot,\neg,\leftrightarrow,+\}$ only.