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	We can prove, just like in ??, that if $M$, started on input $w$, has not halted after $n$ steps, then $\tau'(M, w) \models \chi(M, w, n) \wedge \psi(\bar{n})$. Since $M$ started on input $w$ does not halt, $\tau'(M, w) \models \chi(M, w, n) \wedge \psi(\bar{n})$ for all $n \in \mathbb{N}$. Note that by ??, $\tau'(M, w) \models \bar{k} < \bar{n}$ for all $k < n$. Also $\psi(\bar{n}) \models \bar{k} < \bar{n} \rightarrow \bar{k} \neq \bar{n}$. So, $\mathfrak{M} \models \bar{k} \neq \bar{n}$ for all $k < n$, i.e., the infinitely many terms $\bar{k}$ must all have different values in $\mathfrak{M}$. But this requires that $\|\mathfrak{M}\|$ be infinite, so $\mathfrak{M}$ cannot be a finite model of $\tau(M, w) \wedge \alpha(M, w)$. $\square$

	Problem und.2. Complete the proof of Lemma und.2 by proving that if $M$, started on input $w$, has not halted after $n$ steps, then $\tau'(M, w) \models \psi(\bar{n})$.

	Theorem und.3 (Trakthenbrot's Theorem). It is undecidable if an arbitrary sentence of first-order logic has a finite model (i.e., is finitely satisfiable).

	Proof. Suppose there were a Turing machine F that decides the finite satisfiability problem. Then given any Turing machine M and input w, we could compute the sentence $\tau'(M, w) \wedge \alpha(M, w)$, and use F to decide if it has a finite model. By Lemmata und.1 and und.2, it does iff M started on input w halts. So we could use F to solve the halting problem, which we know is unsolvable. $\square$

	Corollary und.4. There can be no derivation system that is sound and complete for finite validity, i.e., a derivation system which has $\vdash \psi$ iff $\mathfrak{M} \models \psi$ for every finite structure $\mathfrak{M}$.

	Proof. Exercise. $\square$

	Problem und.3. Prove Corollary und.4. Observe that $\psi$ is satisfied in every finite structure iff $\neg\psi$ is not finitely satisfiable. Explain why finite satisfiability is semi-decidable in the sense of ???. Use this to argue that if there were a derivation system for finite validity, then finite satisfiability would be decidable.

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	We can prove, just like in ??, that if $M$, started on input $w$, has not halted after $n$ steps, then $\tau'(M, w) \models \chi(M, w, n) \wedge \psi(\bar{n})$. Since $M$ started on input $w$ does not halt, $\tau'(M, w) \models \chi(M, w, n) \wedge \psi(\bar{n})$ for all $n \in \mathbb{N}$. Note that by ??, $\tau'(M, w) \models \bar{k} < \bar{n}$ for all $k < n$. Also $\psi(\bar{n}) \models \bar{k} < \bar{n} \rightarrow \bar{k} \neq \bar{n}$. So, $\mathfrak{M} \models \bar{k} \neq \bar{n}$ for all $k < n$, i.e., the infinitely many terms $\bar{k}$ must all have different values in $\mathfrak{M}$. But this requires that $\|\mathfrak{M}\|$ be infinite, so $\mathfrak{M}$ cannot be a finite model of $\tau(M, w) \wedge \alpha(M, w)$. $\square$

	Problem und.2. Complete the proof of Lemma und.2 by proving that if $M$, started on input $w$, has not halted after $n$ steps, then $\tau'(M, w) \models \psi(\bar{n})$.

	Theorem und.3 (Trakthenbrot's Theorem). It is undecidable if an arbitrary sentence of first-order logic has a finite model (i.e., is finitely satisfiable).

	Proof. Suppose there were a Turing machine F that decides the finite satisfiability problem. Then given any Turing machine M and input w, we could compute the sentence $\tau'(M, w) \wedge \alpha(M, w)$, and use F to decide if it has a finite model. By Lemmata und.1 and und.2, it does iff M started on input w halts. So we could use F to solve the halting problem, which we know is unsolvable. $\square$

	Corollary und.4. There can be no derivation system that is sound and complete for finite validity, i.e., a derivation system which has $\vdash \psi$ iff $\mathfrak{M} \models \psi$ for every finite structure $\mathfrak{M}$.

	Proof. Exercise. $\square$

	Problem und.3. Prove Corollary und.4. Observe that $\psi$ is satisfied in every finite structure iff $\neg\psi$ is not finitely satisfiable. Explain why finite satisfiability is semi-decidable in the sense of ???. Use this to argue that if there were a derivation system for finite validity, then finite satisfiability would be decidable.

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