The book is Mendelson's Introduction to Logic, 6th edition.
Here is the statement of Lemma 2.16:
Let $J$ be a consistent, complete scapegoat theory. Then $J$ has a model $M$ whose domain is the set $D$ of closed terms of $J$.
Definition of scapegoat theory in his book is:
A theory $K$ is a scapegoat theory if, for any wf $\mathscr B(x)$ that has $x$ as its only free variable, there is a closed term $t$ such that $\vdash_K (\exists x) \neg \mathscr B(x) \implies \neg \mathscr B(t)$.
Here is how Mendelson starts proving it:
For any individual constant $a_i$ of $J$, let $(a_i)^M = a_i$. For any function letter $f^k_n$ of $J$ and for any closed terms $t_1,\dots,t_n$ of $J$, let $(f^k_n)^M (t_1,...,t_n) = f^k_n(t_1,...,t_n)$. (Notice that $f^k_n(t_1,...,t_n)$ is a closed term. Hence, $(f^k_n)^M$ is an $n$-ary operation on $D$.) For any predicate letter $A^k_n$ of $J$, let $(A^k_n)^M$ consist of all $n$-tuples $(t_1,...,t_n)$ of closed terms $t_1,\dots,t_n$ of $J$ such that $\vdash_J A^k_n(t_1,...,t_n)$ . It now suffices to show that, for any closed wf $\mathscr C$ of $J$: \begin{align*} (*) \vDash_M \mathscr C \text{ if and only if } \vdash_J \mathscr C. \end{align*}
He proves it with induction on the number of connectives and quantifiers. I don't type steps that are not needed for my questions.
Case 4. $\mathscr C$ is $(\forall x_m ) \mathscr D$.
Case 4b. $\mathscr D$ is not a closed wf. Since $\mathscr C$ is closed, $\mathscr D$ has $x_m$ as its only free variable, say $\mathscr D$ is $F(x_m)$. Then $\mathscr C$ is $(\forall x_m)F(x_m)$.
(I don't type one of the directions here.)
Assume $\vdash_J \mathscr C$ and not $\vDash_M \mathscr C$. Thus, \begin{align} &\text{(1)} \vdash_J (\forall x_m)F(x_m),\\ &\text{(2) not} \vDash_M (\forall x_m)F(x_m). \end{align} By (2), some sequence of elements of the domain $D$ does not satisfy $(\forall x_m)F(x_m)$. Hence, some sequence $s$ does not satisfy $F(x_m)$. (Following sentence has a possible typo.) Let $t$ be the $i$'th component of $s$. Notice that $s^*(u) = u$ (Note: this is a function mapping terms of the language to elements of the domain) for all closed terms $u$ of $J$ (by the definition of $(a_i)^M$ and $(f^k_n)^M$). Observe also that $F(t)$ has fewer connectives and quantifiers than $\mathscr C$ and, therefore, the inductive hypothesis applies to $F(t)$, that is, $(*)$ holds for $F(t)$. Hence, by Lemma 2(a) on page 60, $s$ does not satisfy $F(t)$. So, $F(t)$ is false for $M$. But, by (1) and particularization rule, $\vdash_J F(t)$, and so, by $(*)$ for $F(t)$, $\vDash_M F(t)$. This contradiction shows that, if $\vdash_J \mathscr C$, then $\vDash_M \mathscr C$.
Here is Lemma 2 (a):
Let $t$ be free for $x_i$ in $\mathscr B(x_i)$. Then $s$ satisfies $\mathscr B(t)$ if and only if $s'$, obtained by replacing the $i$th element of $s$ with $s^*(t)$ satisfies $\mathscr B(x_i)$.
A possible typo
I baffled my head around the picked $i$'th element. It doesn't make sense to me and it seems that Mendelson meant $m$'th element. Here is why: pick $t$ as $m$'th element, then $s$ is a sequence that suits the hypothesis of Lemma 2 (a), since $s^*(t)=t$, and so it does not satisfy $F(t)$, but $t$ is closed, and since $x_m$ was the only free variable, the formula is closed, hence $F(t)$ is false. Everything from his proof follows.
Questions
- Is it a typo?
- If it is not, could you provide some clues, why $i$'th element makes sense here and why the proof follows from it?
