We are going to use the following fact (*): for any truth assignment $v$ that includes all sentence symbols of $\phi$ , if $v(A)=T$ then $\overline{v}(\phi)=\overline{v}(\phi_{\top}^{A})$ , otherwise $\overline{v}(\phi)=\overline{v}(\phi_{\bot}^{A})$ . Also, we note that (**) for any truth assignment $v$ , $\overline{v}(\phi_{*}^{A})=\overline{v}(\phi_{\top}^{A})\vee\overline{v}(\phi_{\bot}^{A})$ .

(a) If a truth assignment $v$ satisfies $\phi$ then, according to (*), depending on the value of $v(A)$ , $\overline{v}(\phi_{\top}^{A})$ or $\overline{v}(\phi_{\bot}^{A})$ is $T$ , and, hence, according to (**), $\overline{v}(\phi_{*}^{A})=T$ .

(b) Exercise 4 of Section 1.2 tells us that $\phi\vDash\psi$ iff $\phi\rightarrow\psi$ is a tautology. Now, according to Exercise 8 of Section 1.2, we can substitute $\top$ or $\bot$ for $A$ in this expression (or, since that exercise was proved for the language without $\top$ and $\bot$ , substitute $(B\vee\neg B)$ or $(B\wedge\neg B)$ for $A$ ), and the resulting wff will still be a tautology. Now, note that in the first case we obtain $\phi_{\top}^{A}\rightarrow\psi$ , while in the other case we obtain $\phi_{\bot}^{A}\rightarrow\psi$ (we actually need to prove that substituting $(B\vee\neg B)$ or $(B\wedge\neg B)$ is equivalent to substituting $\top$ or $\bot$ , but we will skip this part). So, both are tautologies (we could use an argument using truth assignments to avoid referring to previous exercises). Now, for any truth assignment $v$ such that $v$ satisfies $\phi_{*}^{A}$ , we can consider the truth assignment $v'$ restricted to all sentence symbols of $v$ except for $A$ such that $v'$ still satisfies $\phi_{*}^{A}$ (see Exercise 6 of Section 1.2 for more details), and we can extend the truth assignment $v'$ to $u$ or $w$ by letting $u(A)=F$ and $w(A)=T$ . Then, according to Exercise 6 of Section 1.2, $\overline{u}(\phi_{*}^{A})=\overline{w}(\phi_{*}^{A})=\overline{v'}(\phi_{*}^{A})=\overline{v}(\phi_{*}^{A})=T$ , and according to (**), at least one of $\overline{v}(\phi_{\top}^{A})$ or $\overline{v}(\phi_{\bot}^{A})$ is $T$ . Hence, $v$ satisfies $\psi$ .

(c) One direction follows from (a). Now, assume that $\phi_{*}^{A}$ is satisfiable. If $\phi$ is not satisfiable then every wff is tautologically implied by $\phi$ , including $\bot$ . But then, by (b), $\bot$ is tautologically implied by $\phi_{*}^{A}$ , which is not true as $\phi_{*}^{A}$ is satisfiable. Therefore, $\phi$ is satisfiable as well.

To conclude this exercise, we illustrate the concept by some examples. First, consider $\phi=A\wedge B$ . $(A\wedge B)_{*}^{A}=(\top\wedge B)\vee(\bot\wedge B)$ which is tautologically equivalent to $B$ . We can think of it as the best we can say if $A\wedge B$ is known to be true but we cannot express $A$ , or, alternatively, without using $A$ , is that $B$ is true. Similarly, $(A\vee B)_{*}^{A}$ is tautologically equivalent to $\top$ , as in this case, without mentioning $A$ all we can say is that $\top$ is true, in other words, we can say nothing specific.