samples/texts/1364076/page_2.md

a) For every instruction $\delta(q_i, \sigma) = \langle q_j, \sigma', R \rangle$, the sentence:

$$\mathrm{\forall }x\mathrm{\forall }y((Q_{q_i}(x,y)\wedge S_{\sigma }(x,y))\to (Q_{q_j}(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\wedge S_{{\sigma }^{\mathrm{\prime }}}(x,y^{\mathrm{\prime }})\wedge \phi (x,y)\wedge \psi (y^{\mathrm{\prime }})))\backslash forall\; x\; \backslash forall\; y\; ((Q\_\{q\_i\}(x,\; y)\; \backslash land\; S\_\backslash sigma(x,\; y))\; \backslash rightarrow\; \backslash \backslash \; (Q\_\{q\_j\}(x\text{'},\; y\text{'})\; \backslash land\; S\_\{\backslash sigma\text{'}\}(x,\; y\text{'})\; \backslash land\; \backslash varphi(x,\; y)\; \backslash land\; \backslash psi(y\text{'}))\; )$$

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:

$$\mathrm{\forall }x\mathrm{\forall }y((Q_{q_i}(x,y)\wedge S_{\sigma }(x,y))\to (Q_{q_j}(x,y^{\mathrm{\prime }})\wedge S_{{\sigma }^{\mathrm{\prime }}}(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\wedge \phi (x,y)))\wedge \mathrm{\forall }y((Q_{q_i}(o,y)\wedge S_{\sigma }(o,y))\to (Q_{q_j}(o,y^{\mathrm{\prime }})\wedge S_{{\sigma }^{\mathrm{\prime }}}(o^{\mathrm{\prime }},y^{\mathrm{\prime }})\wedge \phi (o,y)\wedge \psi (y^{\mathrm{\prime }})))\backslash forall\; x\; \backslash forall\; y\; ((Q\_\{q\_i\}(x,\; y)\; \backslash land\; S\_\backslash sigma(x,\; y))\; \backslash rightarrow\; \backslash \backslash \; (Q\_\{q\_j\}(x,\; y\text{'})\; \backslash land\; S\_\{\backslash sigma\text{'}\}(x\text{'},\; y\text{'})\; \backslash land\; \backslash varphi(x,\; y)))\; \backslash land\; \backslash \backslash \; \backslash forall\; y\; ((Q\_\{q\_i\}(o,\; y)\; \backslash land\; S\_\backslash sigma(o,\; y))\; \backslash rightarrow\; \backslash \backslash \; (Q\_\{q\_j\}(o,\; y\text{'})\; \backslash land\; S\_\{\backslash sigma\text{'}\}(o\text{'},\; y\text{'})\; \backslash land\; \backslash varphi(o,\; y)\; \backslash land\; \backslash psi(y\text{'}))\; )$$

c) For every instruction $\delta(q_i, \sigma) = \langle q_j, \sigma', N \rangle$, the sentence:

$$\mathrm{\forall }x\mathrm{\forall }y((Q_{q_i}(x,y)\wedge S_{\sigma }(x,y))\to (Q_{q_j}(x,y^{\mathrm{\prime }})\wedge S_{{\sigma }^{\mathrm{\prime }}}(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\wedge \phi (x,y)\wedge \psi (y^{\mathrm{\prime }})))\backslash forall\; x\; \backslash forall\; y\; ((Q\_\{q\_i\}(x,\; y)\; \backslash land\; S\_\backslash sigma(x,\; y))\; \backslash rightarrow\; \backslash \backslash \; (Q\_\{q\_j\}(x,\; y\text{'})\; \backslash land\; S\_\{\backslash sigma\text{'}\}(x\text{'},\; y\text{'})\; \backslash land\; \backslash varphi(x,\; y)\; \backslash land\; \backslash psi(y\text{'}))\; )$$

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

$$\mathrm{\mid }{\mathfrak{M}}^{\mathrm{\prime }}\mathrm{\mid }=\{0,\dots ,n\}|\backslash mathfrak\{M\}\text{'}|\; =\; \backslash \{0,\; \backslash dots,\; n\backslash \}$$

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.