Consider a sequence \(\alpha_1,\alpha_2,\ldots\) of wffs. For each wff \(\varphi\) let \(\varphi^*\) be the result of replacing the sentence symbol \(A_n\) by \(\alpha_n\) for each \(n\).

Part (a)

Problem

Let \(v\) be a truth assignment for the set of all sentence symbols; define \(u\) to be the truth assignment for which \(u(A_n) = \overline{v}(\alpha_n)\). Show that \(\overline{u}(\varphi)=\overline{v}(\varphi^*)\)

Solution

Let \(S\) be the set of wffs for which \(\overline{u}(\varphi)=\overline{v}(\varphi^*)\). If \(\varphi = A_n\), then \(\varphi^* = \alpha_n\). Thus, since \(u(A_n) = \overline{v}(\alpha_n)\) and \(\overline{u}\) is an extension of \(u\), we have that \(\overline{u}(A_n) = \overline{v}(\alpha_n)\), i.e. \(\overline{u}(\varphi) = \overline{v}(\varphi^*)\). Therefore, \(S\) contains all sentence symbols.

Now, I just need to show that \(S\) is closed under the five sentence building operations. Suppose \(\varphi\) and \(\gamma\) are members of \(S\). Since \(\overline{u}(\varphi) = \overline{v}(\varphi^*)\), we will have that \(\overline{u}((\lnot\varphi)) = \overline{v}((\lnot\varphi^*))\) since \(\overline{u}\) and \(\overline{v}\) both adhere to the preestablished axioms for truth assignment extensions. Now since \(\overline{v}((\lnot\varphi^*)) = \overline{v}((\lnot\varphi)^*)\), we have that \(\overline{u}((\lnot\varphi)) = \overline{v}((\lnot\varphi)^*)\), showing that \((\lnot\varphi)\in S\).

Let \(\square\) denote one of \(\land,\lor,\to,\leftrightarrow\). Then \(\overline{u}(\varphi) = \overline{v}(\varphi^*)\) and \(\overline{u}(\gamma) = \overline{v}(\gamma^*)\) will imply that \(\overline{u}((\varphi\square\gamma)) = \overline{v}((\varphi^*\square\gamma^*))\) by similar logic as above. But since \((\varphi^*\square\gamma^*) = (\varphi\square\gamma)^*\), we have that \(\overline{u}((\varphi\square\gamma)) = \overline{v}((\varphi\square\gamma)^*)\), showing that \((\varphi\square\gamma)\in S\).

Since \(S\) contains all the sentence symbols and is closed under all sentence building operations, we conclude that \(S\) is the set of all wffs by the induction principle. Therefore, for any wff \(\varphi\), it will be that \(\overline{u}(\varphi) = \overline{v}(\varphi^*)\).

Part (b)

Problem

Show that if \(\varphi\) is a tautology, then so is \(\varphi^*\).

Solution

Let \(v\) be an arbitrary truth assignment, and let \(u\) be the truth assignment defined by \(u(A_n) = \overline{v}(\alpha_n)\). Then by part (a) we know that \(\overline{u}(\varphi) = \overline{v}(\varphi^*)\). Since \(\varphi\) is a tautology, \(\overline{u}(\varphi) = T\). Thus, \(\overline{v}(\varphi^*) = T\). Since \(v\) was arbitrary, it follows that \(\varphi^*\) is a tautology.