| a) For every instruction $\delta(q_i, \sigma) = \langle q_j, \sigma', R \rangle$, the sentence: | |
| $$ \forall x \forall y ((Q_{q_i}(x, y) \land S_\sigma(x, y)) \rightarrow \\ (Q_{q_j}(x', y') \land S_{\sigma'}(x, y') \land \varphi(x, y) \land \psi(y')) ) $$ | |
| In other words, the same as the corresponding **sentence** in $\tau(M, w)$, | |
| except we add $\psi(y')$ at the end. ($\psi(y')$ ensures that the number $y'$ | |
| of the “next” configuration is different from all previous numbers 0, | |
| 0', ...) | |
| b) For every instruction $\delta(q_i, \sigma) = \langle q_j, \sigma', L \rangle$, the sentence: | |
| $$ \forall x \forall y ((Q_{q_i}(x, y) \land S_\sigma(x, y)) \rightarrow \\ (Q_{q_j}(x, y') \land S_{\sigma'}(x', y') \land \varphi(x, y))) \land \\ \forall y ((Q_{q_i}(o, y) \land S_\sigma(o, y)) \rightarrow \\ (Q_{q_j}(o, y') \land S_{\sigma'}(o', y') \land \varphi(o, y) \land \psi(y')) ) $$ | |
| c) For every instruction $\delta(q_i, \sigma) = \langle q_j, \sigma', N \rangle$, the sentence: | |
| $$ \forall x \forall y ((Q_{q_i}(x, y) \land S_\sigma(x, y)) \rightarrow \\ (Q_{q_j}(x, y') \land S_{\sigma'}(x', y') \land \varphi(x, y) \land \psi(y')) ) $$ | |
| Let $\tau'(M, w)$ be the conjunction of all the above **sentences** for Turing ma- | |
| chine $M$ and input $w$. | |
| **Lemma und.1.** If $M$ started on input $w$ halts, then $\tau'(M, w) \land \alpha(M, w)$ has a finite model. | |
| *Proof.* Let $\mathfrak{M}'$ be as in the proof of ??, except | |
| $$ |\mathfrak{M}'| = \{0, \dots, n\} $$ | |
| $$ \mathfrak{M}'(x) = \begin{cases} x+1 & \text{if } x < n \\ n & \text{otherwise,} \end{cases} $$ | |
| where $n = \max(k, \text{len}(w))$ and $k$ is the least number such that $M$ started | |
| on input $w$ has halted after $k$ steps. We leave the verification that $\mathfrak{M}' \models$ | |
| $\tau(M, w) \land E(M, w)$ as an exercise. $\square$ | |
| **Problem und.1.** Complete the proof of **Lemma und.1** by proving that $\mathfrak{M}' \models \tau(M, w) \land E(M, w)$. | |
| **Lemma und.2.** If $\tau'(M, w) \land \alpha(M, w)$ has a finite model, then $M$ started on *input w halts*. | |
| *Proof.* We show the contrapositive. Suppose that $M$ started on $w$ does not halt. If $\tau'(M,w) \land \alpha(M,w)$ has no model at all, we are done. So assume $\mathfrak{M}$ is a model of $\tau(M,w) \land \alpha(M,w)$. We have to show that it cannot be finite. |