Problem

Let \(\mathcal{S}\) be the set of all sentence symbols, and assume that \(v:\mathcal{S}\to\{F,T\}\) is a truth assignment. Show there is at most one extension \(\overline{v}\) meeting conditions 0-5 listed at the beginning of this section.

Solution

Suppose that \(\overline{v}_1\) and \(\overline{v}_2\) are both such extensions, and let \(X\) be the set of wffs for which \(x\in X\) iff \(\overline{v}_1(x) = \overline{v}_2(x)\). Since \(\overline{v}_1\) and \(\overline{v}_2\) are extensions of the truth assignment \(v\), we must have that \(\overline{v}_1(A)=\overline{v}_2(A)\) for every sentence symbol \(A\in\mathcal{S}\). Thus, \(X\) contains all of the sentence symbols.

Now, suppose that \(\alpha,\beta\in\Sigma\), so that \(\overline{v}_1(\alpha)=\overline{v}_2(\alpha)\) and \(\overline{v}_1(\beta)=\overline{v}_2(\beta)\). Since both \(\overline{v}_1\) and \(\overline{v}_2\) meet condition 0-5 listed at the beginning of the section, we must have that \(\overline{v}_1((\lnot\alpha)) = \overline{v}_2((\lnot\alpha))\). Thus, \((\lnot\alpha)\in S\).

By conditions 0-5 we also have that \begin{gather}\overline{v}_1(\alpha\land\beta) = \begin{cases}T\quad\text{if } \overline{v}_1(\alpha)=T\text{ and }\overline{v}_1(\beta) = T\ F\quad \text{otherwise}\end{cases}\end{gather} \begin{gather}\overline{v}_2(\alpha\land\beta) = \begin{cases}T\quad\text{if } \overline{v}_2(\alpha)=T\text{ and }\overline{v}_2(\beta) = T\ F\quad \text{otherwise}\end{cases}\end{gather} Since \(\overline{v}_1(\alpha) = \overline{v}_2(\alpha)\) and \(\overline{v}_1(\alpha) = \overline{v}_2(\alpha)\). it must also be that \(\overline{v}_1(\alpha\land\beta) = \overline{v}_2(\alpha\land\beta)\). Thus, \((\alpha\land\beta)\in S\), meaning \(S\) is closed under \(\land\). By similar reasoning, it is not difficult to see that \begin{gather}\overline{v}_1(\alpha\lor\beta) = \overline{v}_2(\alpha\lor\beta)\ \overline{v}_1(\alpha\to\beta) = \overline{v}_2(\alpha\to\beta)\ \overline{v}_1(\alpha\leftrightarrow\beta) = \overline{v}_2(\alpha\leftrightarrow\beta)\end{gather} so that \((\alpha\lor\beta),(\alpha\to\beta),(\alpha\leftrightarrow\beta)\in S\). Since \(S\) contains all the sentence symbols and is closed under \(\lnot,\land,\lor,\to,\) and \(\leftrightarrow,\) by the induction principle it must be that \(S\) is the set of all wffs. Thus, for any wff \(\alpha\), \(\overline{v}_1(\alpha) = \overline{v}_2(\alpha)\). Therefore, \(\overline{v}_1=\overline{v}_2\).