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Proposition 18.14 (Monotony). If \( \Gamma \subseteq \Delta \) and \( \Gamma \vdash \varphi \), then \( \Delta \vdash \varphi \) . | Proof. Any derivation of \( \varphi \) from \( \Gamma \) is also a derivation of \( \varphi \) from \( \Delta \) . | Yes |
Proposition 18.15 (Transitivity). If \( \Gamma \vdash \varphi \) and \( \{ \varphi \} \cup \Delta \vdash \psi \), then \( \Gamma \cup \Delta \vdash \psi \) . | Proof. If \( \Gamma \vdash \varphi \), there is a derivation \( {\delta }_{0} \) of \( \varphi \) with all undischarged assumptions in \( \Gamma \) . If \( \{ \varphi \} \cup \Delta \vdash \psi \), then there is a derivation \( {\delta }_{1} \) of \( \psi \) with all undischarged assumptions in \( \{ \varphi \} \cup \D... | Yes |
Proposition 18.16. \( \Gamma \) is inconsistent iff \( \Gamma \vdash \varphi \) for every sentence \( \varphi \) . | Proof. Exercise. | No |
Proposition 18.17 (Compactness). 1. If \( \Gamma \vdash \varphi \) then there is a finite subset \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \vdash \varphi \). | 1. If \( \Gamma \vdash \varphi \), then there is a derivation \( \delta \) of \( \varphi \) from \( \Gamma \) . Let \( {\Gamma }_{0} \) be the set of undischarged assumptions of \( \delta \) . Since any derivation is finite, \( {\Gamma }_{0} \) can only contain finitely many sentences. So, \( \delta \) is a derivation ... | Yes |
Proposition 18.18. If \( \Gamma \vdash \varphi \) and \( \Gamma \cup \{ \varphi \} \) is inconsistent, then \( \Gamma \) is inconsistent. | Proof. Let the derivation of \( \varphi \) from \( \Gamma \) be \( {\delta }_{1} \) and the derivation of \( \bot \) from \( \Gamma \cup \) \( \{ \varphi \} \) be \( {\delta }_{2} \) . We can then derive:\n\n\n\nIn t... | No |
Proposition 18.19. \( \Gamma \vdash \varphi \) iff \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent. | Proof. First suppose \( \Gamma \vdash \varphi \), i.e., there is a derivation \( {\delta }_{0} \) of \( \varphi \) from undischarged assumptions \( \Gamma \) . We obtain a derivation of \( \bot \) from \( \Gamma \cup \{ \neg \varphi \} \) as follows:\n\n and \( \neg \varphi \in \Gamma \), then \( \Gamma \) is inconsistent. | Proof. Suppose \( \Gamma \vdash \varphi \) and \( \neg \varphi \in \Gamma \) . Then there is a derivation \( \delta \) of \( \varphi \) from \( \Gamma \) . Consider this simple application of the \( \neg \) Elim rule:\n\n and \( \Gamma \cup \{ \neg \varphi \} \) are both inconsistent, then \( \Gamma \) is inconsistent. | Proof. There are derivations \( {\delta }_{1} \) and \( {\delta }_{2} \) of \( \bot \) from \( \Gamma \cup \{ \varphi \} \) and \( \bot \) from \( \Gamma \cup \{ \neg \varphi \} \) , respectively. We can then derive\n\n and \( \varphi \land \psi \vdash \psi \)\n\n2. \( \varphi ,\psi \vdash \varphi \land \psi \) . | Proof. 1. We can derive both\n\n\[ \frac{\varphi \land \psi }{\varphi } \land \operatorname{Elim}\;\frac{\varphi \land \psi }{\psi } \land \operatorname{Elim} \]\n\n2. We can derive:\n\n\[ \frac{\varphi }{\varphi \land \psi } \land \text{ Intro } \]\n\n\( ▱ \) | Yes |
Proposition 18.23. 1. \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \) is inconsistent. | Proof. 1. Consider the following derivation:\n\n\[ \frac{\varphi \vee \psi \;\frac{\neg \varphi \;{\left\lbrack \varphi \right\rbrack }^{1}}{\bot }\neg \text{Elim}\;\frac{\neg \psi \;{\left\lbrack \psi \right\rbrack }^{1}}{\bot }\neg \text{Elim}}{\bot }\text{VElim} \]\n\nThis is a derivation of \( \bot \) from undischa... | Yes |
Proposition 18.24. 1. \( \varphi ,\varphi \rightarrow \psi \vdash \psi \) . | 1. We can derive:\n\n\[ \frac{\varphi \rightarrow \psi \;\varphi }{\psi } \rightarrow \text{Elim} \] | Yes |
Theorem 18.25. If \( c \) is a constant not occurring in \( \Gamma \) or \( \varphi \left( x\right) \) and \( \Gamma \vdash \varphi \left( c\right) \), then \( \Gamma \vdash \forall {x\varphi }\left( x\right) \) . | Proof. Let \( \delta \) be a derivation of \( \varphi \left( c\right) \) from \( \Gamma \) . By adding a \( \forall \) Intro inference, we obtain a proof of \( \forall {x\varphi }\left( x\right) \) . Since \( c \) does not occur in \( \Gamma \) or \( \varphi \left( x\right) \), the eigenvariable condition is satisfied. | Yes |
Proposition 18.26. 1. \( \varphi \left( t\right) \vdash \exists {x\varphi }\left( x\right) \) . 2. \( \forall {x\varphi }\left( x\right) \vdash \varphi \left( t\right) \) . | Proof. 1. The following is a derivation of \( \exists {x\varphi }\left( x\right) \) from \( \varphi \left( t\right) \) : \[ \frac{\varphi \left( t\right) }{\exists {x\varphi }\left( x\right) }\exists \text{ Intro } \] 2. The following is a derivation of \( \varphi \left( t\right) \) from \( \forall {x\varphi }\left( x\... | Yes |
Corollary 18.29. If \( \Gamma \) is satisfiable, then it is consistent. | Proof. We prove the contrapositive. Suppose that \( \Gamma \) is not consistent. Then \( \Gamma \vdash \bot \), i.e., there is a derivation of \( \bot \) from undischarged assumptions in \( \Gamma \). By Theorem 18.27, any structure \( \mathfrak{M} \) that satisfies \( \Gamma \) must satisfy \( \bot \). Since \( \mathf... | Yes |
If \( s \) and \( t \) are closed terms, then \( \varphi \left( s\right), s = t \vdash \varphi \left( t\right) \) | \[ \frac{s = t\;\varphi \left( s\right) }{\varphi \left( t\right) } = \text{ Elim } \] | Yes |
We derive the sentence\n\n\[ \forall x\forall y\left( {\left( {\varphi \left( x\right) \land \varphi \left( y\right) }\right) \rightarrow x = y}\right) \]\n\nfrom the sentence\n\n\[ \exists x\forall y\left( {\varphi \left( y\right) \rightarrow y = x}\right) \] | We develop the derivation backwards:\n\n\[ \exists x\forall y\left( {\varphi \left( y\right) \rightarrow y = x}\right) \;{\left\lbrack \varphi \left( a\right) \land \varphi \left( b\right) \right\rbrack }^{1} \]\n\n\[ \begin{matrix} \vdots \\ \vdots \\ \frac{a = b}{b\left( {\left( {\varphi \left( a\right) \land \varphi... | Yes |
Proposition 18.32. Natural deduction with rules for \( = \) is sound. | Proof. Any formula of the form \( t = t \) is valid, since for every structure \( \mathfrak{M} \) , \( \mathfrak{M} \vDash t = t \) . (Note that we assume the term \( t \) to be ground, i.e., it contains no variables, so variable assignments are irrelevant).\n\nSuppose the last inference in a derivation is \( = \) Elim... | Yes |
Every set of assumptions on its own is a tableau, but it will generally not be closed. (Obviously, it is closed only if the assumptions already contain a pair of signed formulas \( \mathbb{T}\varphi \) and \( \mathbb{F}\varphi \) .) | From a tableau (open or closed) we can obtain a new, larger one by applying one of the rules of inference to a signed formula \( \varphi \) in it. The rule will append one or more signed formulas to the end of any branch containing the occurrence of \( \varphi \) to which we apply the rule. For instance, consider the a... | Yes |
Let’s find a closed tableau for the sentence \( \left( {\varphi \land \psi }\right) \rightarrow \varphi \) . | We begin by writing the corresponding assumption at the top of the tableau.\n\n\[ \text{1.}\mathbb{F}\left( {\varphi \land \psi }\right) \rightarrow \varphi \;\text{Assumption} \]\n\nThere is only one assumption, so only one signed formula to which we can apply a rule. (For every signed formula, there is always at most... | Yes |
Now let’s find a closed tableau for \( \left( {\neg \varphi \vee \psi }\right) \rightarrow \left( {\varphi \rightarrow \psi }\right) \). | We begin with the corresponding assumption:\n\n\[ \text{1.}\mathbb{F}\left( {\neg \varphi \vee \psi }\right) \rightarrow \left( {\varphi \rightarrow \psi }\right) \;\text{Assumption} \]\n\nThe one signed formula in this tableau has main operator \( \rightarrow \) and sign \( \mathbb{F} \), so we apply the \( \rightarro... | Yes |
We can give tableaux for any number of signed formulas as assumptions. Often it is also necessary to apply more than one rule that allows branching; and in general a tableau can have any number of branches. For instance, consider a tableau for \( \{ \mathbb{T}\varphi \vee \left( {\psi \land \chi }\right) ,\mathbb{F}\le... | We start by applying the \( \vee \mathbb{T} \) to the first assumption: \n\nNow we can apply the \( \land \mathbb{F} \) rule to line 2 . We do this on both branches simultaneously, and can therefore check off line 2:... | Yes |
Let’s see how we’d give a tableau for the sentence \( \exists x\neg \varphi \left( x\right) \rightarrow \neg \forall {x\varphi }\left( x\right) \) . | Starting as usual, we start by recording the assumption,\n\n\[ \text{1.}\;\mathbb{F}\exists x\neg \varphi \left( x\right) \rightarrow \neg \forall {x\varphi }\left( x\right) \;\text{Assumption} \]\n\nSince the main operator is \( \rightarrow \), we apply the \( \rightarrow \mathbb{F} \) :\n\n1. \( \;\mathbb{F}\exists x... | Yes |
We construct a tableau for the set\n\n\\[ \n\\mathbb{T}\\forall {x\\varphi }\\left( x\\right) ,\\mathbb{T}\\forall {x\\varphi }\\left( x\\right) \\rightarrow \\exists {y\\psi }\\left( y\\right) ,\\mathbb{T}\\neg \\exists {y\\psi }\\left( y\\right) .\n\\] | Starting as usual, we write down the assumptions:\n\n\\[ \n\\text{1.}\\;\\mathbb{T}\\forall {x\\varphi }\\left( x\\right) \\;\\text{Assumption}\n\\]\n\n\\[ \n\\text{2.}\\;\\mathbb{T}\\forall {x\\varphi }\\left( x\\right) \\rightarrow \\exists {y\\psi }\\left( y\\right) \\;\\text{Assumption}\n\\]\n\n\\[ \n\\text{3.}\\;\... | Yes |
Proposition 19.13 (Reflexivity). If \( \varphi \in \Gamma \), then \( \Gamma \vdash \varphi \) . | Proof. If \( \varphi \in \Gamma ,\{ \varphi \} \) is a finite subset of \( \Gamma \) and the tableau\n\n1. \( \mathbb{F}\varphi \; \) Assumption\n\n2. \( \mathbb{T}\varphi \; \) Assumption \( \otimes \)\n\nis closed. | No |
Proposition 19.15 (Transitivity). If \( \Gamma \vdash \varphi \) and \( \{ \varphi \} \cup \Delta \vdash \psi \), then \( \Gamma \cup \Delta \vdash \psi \) . | Proof. If \( \{ \varphi \} \cup \Delta \vdash \psi \), then there is a finite subset \( {\Delta }_{0} = \left\{ {{\chi }_{1},\ldots ,{\chi }_{n}}\right\} \subseteq \Delta \) such that\n\n\[ \left\{ {\mathbb{F}\psi ,\mathbb{T}\varphi ,\mathbb{T}{\chi }_{1},\ldots ,\mathbb{T}{\chi }_{n}}\right\} \]\n\nhas a closed tablea... | Yes |
Proposition 19.17 (Compactness). 1. If \( \Gamma \vdash \varphi \) then there is a finite subset \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \vdash \varphi \). | 1. If \( \Gamma \vdash \varphi \), then there is a finite subset \( {\Gamma }_{0} = \left\{ {{\psi }_{1},\ldots ,{\psi }_{n}}\right\} \) and a closed tableau for\n\n\[ \mathbb{F}\varphi ,\mathbb{T}{\psi }_{1},\cdots \mathbb{T}{\psi }_{n} \]\n\nThis tableau also shows \( {\Gamma }_{0} \vdash \varphi \). | Yes |
Proposition 19.18. If \( \Gamma \vdash \varphi \) and \( \Gamma \cup \{ \varphi \} \) is inconsistent, then \( \Gamma \) is inconsistent. | Proof. There are finite \( {\Gamma }_{0} = \left\{ {{\psi }_{1},\ldots ,{\psi }_{n}}\right\} \) and \( {\Gamma }_{1} = \left\{ {{\chi }_{1},\ldots ,{\chi }_{n}}\right\} \subseteq \Gamma \) such that\n\n\[ \left\{ {\mathbb{F}\varphi ,\mathbb{T}{\psi }_{1},\ldots ,\mathbb{T}{\psi }_{n}}\right\} \]\n\n\[ \left\{ {\mathbb{... | Yes |
Proposition 19.19. \( \Gamma \vdash \varphi \) iff \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent. | Proof. First suppose \( \Gamma \vdash \varphi \), i.e., there is a closed tableau for\n\n\[ \left\{ {\mathbb{F}\varphi ,\mathbb{T}{\psi }_{1},\ldots ,\mathbb{T}{\psi }_{n}}\right\} \]\n\nUsing the \( \neg \mathbb{T} \) rule, this can be turned into a closed tableau for\n\n\[ \left\{ {\mathbb{T}\neg \varphi ,\mathbb{T}{... | Yes |
Proposition 19.20. If \( \Gamma \vdash \varphi \) and \( \neg \varphi \in \Gamma \), then \( \Gamma \) is inconsistent. | Proof. Suppose \( \Gamma \vdash \varphi \) and \( \neg \varphi \in \Gamma \) . Then there are \( {\psi }_{1},\ldots ,{\psi }_{n} \in \Gamma \) such that\n\n\[ \left\{ {\mathbb{F}\varphi ,\mathbb{T}{\psi }_{1},\ldots ,\mathbb{T}{\psi }_{n}}\right\} \]\n\nhas a closed tableau. Replace the assumption \( \mathbb{F}\varphi ... | Yes |
Proposition 19.21. If \( \Gamma \cup \{ \varphi \} \) and \( \Gamma \cup \{ \neg \varphi \} \) are both inconsistent, then \( \Gamma \) is inconsistent. | Proof. If there are \( {\psi }_{1},\ldots ,{\psi }_{n} \in \Gamma \) and \( {\chi }_{1},\ldots ,{\chi }_{m} \in \Gamma \) such that\n\n\[ \left\{ {\mathbb{T}\varphi ,\mathbb{T}{\psi }_{1},\ldots ,\mathbb{T}{\psi }_{n}}\right\} \]\n\n\[ \left\{ {\mathbb{T}\neg \varphi ,\mathbb{T}{\chi }_{1},\ldots ,\mathbb{T}{\chi }_{m}... | Yes |
Proposition 19.22. 1. Both \( \varphi \land \psi \vdash \varphi \) and \( \varphi \land \psi \vdash \psi \) . | Proof. 1. Both \( \{ \mathbb{F}\varphi ,\mathbb{T}\varphi \land \psi \} \) and \( \{ \mathbb{F}\psi ,\mathbb{T}\varphi \land \psi \} \) have closed tableaux\n\n1. \( \;\mathbb{F}\varphi \; \) Assumption\n\n2. \( \mathbb{T}\varphi \land \psi \; \) Assumption\n\n3. \( \mathbb{T}\varphi \; \land \mathbb{T}2 \)\n\n4. \( \m... | Yes |
Proposition 19.23. 1. \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \) is inconsistent. | Proof. 1. We give a closed tableau of \( \{ \mathbb{T}\varphi \vee \psi ,\mathbb{T}\neg \varphi ,\mathbb{T}\neg \psi \} \) :\n\n | Yes |
Proposition 19.24. 1. \( \varphi ,\varphi \rightarrow \psi \vdash \psi \) . | Proof. 1. \( \{ \mathbb{F}\psi ,\mathbb{T}\varphi \rightarrow \psi ,\mathbb{T}\varphi \} \) has a closed tableau:\n\n | Yes |
Theorem 19.25. If \( c \) is a constant not occurring in \( \Gamma \) or \( \varphi \left( x\right) \) and \( \Gamma \vdash \varphi \left( c\right) \), then \( \Gamma \vdash \forall {x\varphi }\left( x\right) \) . | Proof. Suppose \( \Gamma \vdash \varphi \left( c\right) \), i.e., there are \( {\psi }_{1},\ldots ,{\psi }_{n} \in \Gamma \) and a closed tableau for\n\n\[ \left\{ {\mathbb{F}\varphi \left( c\right) ,\mathbb{T}{\psi }_{1},\ldots ,\mathbb{T}{\psi }_{n}}\right\} \]\n\nWe have to show that there is also a closed tableau f... | Yes |
Proposition 19.26. 1. \( \varphi \left( t\right) \vdash \exists {x\varphi }\left( x\right) \) . | 1. A closed tableau for \( \mathbb{F}\exists {x\varphi }\left( x\right) ,\mathbb{T}\varphi \left( t\right) \) is:\n\n\[ \begin{matrix} \text{ 1. } & \mathbb{F}\exists {x\varphi }\left( x\right) & \text{ Assumption } \\ \text{ 2. } & \mathbb{T}\varphi \left( t\right) & \text{ Assumption } \\ \text{ 3. } & \mathbb{F}\var... | Yes |
Corollary 19.31. If \( \Gamma \) is satisfiable, then it is consistent. | Proof. We prove the contrapositive. Suppose that \( \Gamma \) is not consistent. Then there are \( {\psi }_{1},\ldots ,{\psi }_{n} \in \Gamma \) and a closed tableau for \( \{ \mathbb{T}\psi ,\ldots ,\mathbb{T}\psi \} \) . By Theorem 19.28, there is no \( \mathfrak{M} \) such that \( \mathfrak{M} \vDash {\psi }_{i} \) ... | Yes |
If \( s \) and \( t \) are closed terms, then \( s = t,\varphi \left( s\right) \vdash \varphi \left( t\right) \) | \[ \text{1.}\mathbb{F}\varphi \left( t\right) \;\text{Assumption} \] \[ \text{2.}\mathbb{T}s = t\;\text{Assumption} \] \[ \text{3.}\mathbb{T}\varphi \left( s\right) \;\text{Assumption} \] 4. \( \;\mathbb{T}\varphi \left( t\right) \; = \mathbb{T}2,3 \) \( \otimes \) | No |
Proposition 19.33. Tableaux with rules for identity are sound: no closed tableau is satisfiable. | Proof. We just have to show as before that if a tableau has a satisfiable branch, the branch resulting from applying one of the rules for \( = \) to it is also satisfiable. Let \( \Gamma \) be the set of signed formulas on the branch, and let \( \mathfrak{M} \) be a structure satisfying \( \Gamma \) . Suppose the branc... | Yes |
Suppose we want to prove \( \left( {\neg \theta \vee \alpha }\right) \rightarrow \left( {\theta \rightarrow \alpha }\right) \) . | Clearly, this is not an instance of any of our axioms, so we have to use the MP rule to derive it. Our only rule is MP, which given \( \varphi \) and \( \varphi \rightarrow \psi \) allows us to justify \( \psi \) . One strategy would be to use eq. (20.6) with \( \varphi \) being \( \neg \theta ,\psi \) being \( \alpha ... | Yes |
Let’s try to find a derivation of \( \theta \rightarrow \theta \). | 1. \( \theta \rightarrow \left( {\left( {\theta \rightarrow \theta }\right) \rightarrow \theta }\right) \) eq. (20.7)\n\n2. \( \left( {\left( {\theta \rightarrow \left( {\theta \rightarrow \theta }\right) }\right) \rightarrow \left( {\theta \rightarrow \theta }\right) }\right) \) eq. (20.8)\n\n3. \( \left( {\theta \rig... | Yes |
Sometimes we want to show that there is a derivation of some formula from some other formulas \( \Gamma \) . For instance, let’s show that we can derive \( \varphi \rightarrow \chi \) from \( \Gamma = \{ \varphi \rightarrow \psi ,\psi \rightarrow \chi \} \) . | 1. \( \varphi \rightarrow \psi \; \) HYP\n2. \( \psi \rightarrow \chi \;\mathrm{{HYP}} \)\n3. \( \left( {\psi \rightarrow \chi }\right) \rightarrow \left( {\varphi \rightarrow \left( {\psi \rightarrow \chi }\right) }\right) \) eq. (20.7)\n4. \( \varphi \rightarrow \left( {\psi \rightarrow \chi }\right) \;2,3,\mathrm{{M... | Yes |
Proposition 20.12. If \( \Gamma \vdash \varphi \rightarrow \psi \) and \( \Gamma \vdash \psi \rightarrow \chi \), then \( \Gamma \vdash \varphi \rightarrow \chi \) | Proof. Suppose \( \Gamma \vdash \varphi \rightarrow \psi \) and \( \Gamma \vdash \psi \rightarrow \chi \) . Then there is a derivation of \( \varphi \rightarrow \psi \) from \( \Gamma \) ; and a derivation of \( \psi \rightarrow \chi \) from \( \Gamma \) as well. Combine these into a single derivation by concatenating ... | Yes |
Let us give a derivation of \( \left( {\forall {x\varphi }\left( x\right) \land \forall {y\psi }\left( y\right) }\right) \rightarrow \forall x(\varphi \left( x\right) \land \psi \left( x\right) ) \) . | First, note that\n\n\[ \left( {\forall {x\varphi }\left( x\right) \land \forall {y\psi }\left( y\right) }\right) \rightarrow \forall {x\varphi }\left( x\right) \]\n\nis an instance of eq. (20.1), and\n\n\[ \forall {x\varphi }\left( x\right) \rightarrow \varphi \left( a\right) \]\n\nof eq. (20.15). So, by Proposition 20... | Yes |
Proposition 20.17 (Reflexivity). If \( \varphi \in \Gamma \), then \( \Gamma \vdash \varphi \) . | Proof. The formula \( \varphi \) by itself is a derivation of \( \varphi \) from \( \Gamma \) . | Yes |
Proposition 20.18 (Monotony). If \( \Gamma \subseteq \Delta \) and \( \Gamma \vdash \varphi \), then \( \Delta \vdash \varphi \) . | Proof. Any derivation of \( \varphi \) from \( \Gamma \) is also a derivation of \( \varphi \) from \( \Delta \) . | Yes |
Proposition 20.19 (Transitivity). If \( \Gamma \vdash \varphi \) and \( \{ \varphi \} \cup \Delta \vdash \psi \), then \( \Gamma \cup \Delta \vdash \psi \) . | Proof. Suppose \( \{ \varphi \} \cup \Delta \vdash \psi \) . Then there is a derivation \( {\psi }_{1},\ldots ,{\psi }_{l} = \psi \) from \( \{ \varphi \} \cup \Delta \) . Some of the steps in that derivation will be correct because of a rule which refers to a prior line \( {\psi }_{i} = \varphi \) . By hypothesis, the... | Yes |
Proposition 20.21 (Compactness). 1. If \( \Gamma \vdash \varphi \) then there is a finite subset \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \vdash \varphi \). | 1. If \( \Gamma \vdash \varphi \), then there is a finite sequence of formulas \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) so that \( \varphi \equiv {\varphi }_{n} \) and each \( {\varphi }_{i} \) is either a logical axiom, an element of \( \Gamma \) or follows from previous formulas by modus ponens. Take \( {\Gamma }_... | Yes |
Proposition 20.22. If \( \Gamma \vdash \varphi \) and \( \Gamma \vdash \varphi \rightarrow \psi \), then \( \Gamma \vdash \psi \) . | Proof. We have that \( \{ \varphi ,\varphi \rightarrow \psi \} \vdash \psi \) :\n\n1. \( \varphi \; \) Hyp.\n\n2. \( \varphi \rightarrow \psi \; \) Hyp.\n\n3. \( \psi \;1,2,\mathrm{{MP}} \)\n\nBy Proposition 20.19, \( \Gamma \vdash \psi \) . | Yes |
Theorem 20.23 (Deduction Theorem). \( \Gamma \cup \{ \varphi \} \vdash \psi \) if and only if \( \Gamma \vdash \varphi \rightarrow \psi \) . | Proof. The \ | No |
Theorem 20.25 (Deduction Theorem). If \( \Gamma \cup \{ \varphi \} \vdash \psi \), then \( \Gamma \vdash \varphi \rightarrow \psi \) . | Proof. We again proceed by induction on the length of the derivation of \( \psi \) from \( \Gamma \cup \{ \varphi \} \). The proof of the induction basis is identical to that in the proof of Theorem 20.23. For the inductive step, suppose again that the derivation of \( \psi \) from \( \Gamma \cup \{ \varphi \} \) ends ... | No |
Proposition 20.26. If \( \Gamma \vdash \varphi \) and \( \Gamma \cup \{ \varphi \} \) is inconsistent, then \( \Gamma \) is inconsistent. | Proof. If \( \Gamma \cup \{ \varphi \} \) is inconsistent, then \( \Gamma \cup \{ \varphi \} \vdash \bot \) . By Proposition 20.17, \( \Gamma \vdash \psi \) for every \( \psi \in \Gamma \) . Since also \( \Gamma \vdash \varphi \) by hypothesis, \( \Gamma \vdash \psi \) for every \( \psi \in \Gamma \cup \{ \varphi \} \)... | Yes |
Proposition 20.27. \( \Gamma \vdash \varphi \) iff \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent. | Proof. First suppose \( \Gamma \vdash \varphi \) . Then \( \Gamma \cup \{ \neg \varphi \} \vdash \varphi \) by Proposition 20.18. \( \Gamma \cup \) \( \{ \neg \varphi \} \vdash \neg \varphi \) by Proposition 20.17. We also have \( \vdash \neg \varphi \rightarrow \left( {\varphi \rightarrow \bot }\right) \) by eq. (20.1... | Yes |
Proposition 20.28. If \( \Gamma \vdash \varphi \) and \( \neg \varphi \in \Gamma \), then \( \Gamma \) is inconsistent. | Proof. \( \Gamma \vdash \neg \varphi \rightarrow \left( {\varphi \rightarrow \bot }\right) \) by eq. (20.10). \( \Gamma \vdash \bot \) by two applications of Proposition 20.22. | Yes |
Proposition 20.29. If \( \Gamma \cup \{ \varphi \} \) and \( \Gamma \cup \{ \neg \varphi \} \) are both inconsistent, then \( \Gamma \) is inconsistent. | Proof. Exercise. | No |
Proposition 20.31. 1. \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \) is inconsistent. | Proof. 1. From eq. (20.9) we get \( \vdash \neg \varphi \rightarrow \left( {\varphi \rightarrow \bot }\right) \) and \( \vdash \neg \varphi \rightarrow \left( {\varphi \rightarrow \bot }\right) \) . So by the deduction theorem, we have \( \{ \neg \varphi \} \vdash \varphi \rightarrow \bot \) and \( \{ \neg \psi \} \vda... | Yes |
Proposition 20.32. 1. \( \varphi ,\varphi \rightarrow \psi \vdash \psi \) . | Proof. 1. We can derive:\n\n1. \( \varphi \;\mathrm{{HYP}} \)\n\n2. \( \varphi \rightarrow \psi \;\mathrm{{HYP}} \)\n\n3. \( \psi \;1,2,\mathrm{{MP}} \) | Yes |
Theorem 20.33. If \( c \) is a constant symbol not occurring in \( \Gamma \) or \( \varphi \left( x\right) \) and \( \Gamma \vdash \varphi \left( c\right) \) , then \( \Gamma \vdash \forall {x\varphi }\left( x\right) \) . | Proof. By the deduction theorem, \( \Gamma \vdash \top \rightarrow \varphi \left( c\right) \) . Since \( c \) does not occur in \( \Gamma \) or \( \top \), we get \( \Gamma \vdash \top \rightarrow \varphi \left( c\right) \) . By the deduction theorem again, \( \Gamma \vdash \forall {x\varphi }\left( x\right) \) . | No |
Proposition 20.34. 1. \( \varphi \left( t\right) \vdash \exists {x\varphi }\left( x\right) \) . | 1. By eq. (20.16) and the deduction theorem. | No |
Proposition 20.35. If \( \varphi \) is an axiom, then \( \mathfrak{M}, s \vDash \varphi \) for each structure \( \mathfrak{M} \) and assignment \( s \) . | Proof. We have to verify that all the axioms are valid. For instance, here is the case for eq. (20.15): suppose \( t \) is free for \( x \) in \( \varphi \), and assume \( \mathfrak{M}, s \vDash \forall {x\varphi } \) . Then by definition of satisfaction, for each \( {s}^{\prime }{ \sim }_{x}s \), also \( \mathfrak{M},... | Yes |
Corollary 20.38. If \( \Gamma \) is satisfiable, then it is consistent. | Proof. We prove the contrapositive. Suppose that \( \Gamma \) is not consistent. Then \( \Gamma \vdash \bot \), i.e., there is a derivation of \( \bot \) from \( \Gamma \) . By Theorem 20.36, any structure \( \mathfrak{M} \) that satisfies \( \Gamma \) must satisfy \( \bot \) . Since \( \mathfrak{M} \mathrel{\text{\vDa... | Yes |
Proposition 20.40. The axioms eq. (20.17) and eq. (20.18) are valid. | Proof. Exercise. | No |
Proposition 20.42. If \( \Gamma \vdash \varphi \left( {t}_{1}\right) \) and \( \Gamma \vdash {t}_{1} = {t}_{2} \), then \( \Gamma \vdash \varphi \left( {t}_{2}\right) \) . | Proof. The formula\n\n\[ \left( {{t}_{1} = {t}_{2} \rightarrow \left( {\varphi \left( {t}_{1}\right) \rightarrow \varphi \left( {t}_{2}\right) }\right) }\right) \]\n\n is an instance of eq. (20.18). The conclusion follows by two applications of MP. \( ▱ \) | Yes |
Proposition 21.2. Suppose \( \Gamma \) is complete and consistent. Then:\n\n1. If \( \Gamma \vdash \varphi \), then \( \varphi \in \Gamma \) . | Proof. Let us suppose for all of the following that \( \Gamma \) is complete and consistent.\n\n1. If \( \Gamma \vdash \varphi \), then \( \varphi \in \Gamma \).\n\nSuppose that \( \Gamma \vdash \varphi \) . Suppose to the contrary that \( \varphi \notin \Gamma \) . Since \( \Gamma \) is complete, \( \neg \varphi \in \... | Yes |
Lemma 21.6. Every consistent set \( \Gamma \) can be extended to a saturated consistent set \( {\Gamma }^{\prime } \) . | Proof. Given a consistent set of sentences \( \Gamma \) in a language \( \mathcal{L} \), expand the language by adding a denumerable set of new constant symbols to form \( {\mathcal{L}}^{\prime } \). By Proposition 21.3, \( \Gamma \) is still consistent in the richer language. Further, let \( {\theta }_{i} \) be as in ... | Yes |
Proposition 21.7. Suppose \( \Gamma \) is complete, consistent, and saturated.\n\n1. \( \exists {x\varphi }\left( x\right) \in \Gamma \) iff \( \varphi \left( t\right) \in \Gamma \) for at least one closed term \( t \) . | Proof. 1. First suppose that \( \exists {x\varphi }\left( x\right) \in \Gamma \) . Because \( \Gamma \) is saturated, \( (\exists {x\varphi }\left( x\right) \rightarrow \) \( \varphi \left( c\right) ) \in \Gamma \) for some constant symbol \( c \) . By Propositions 17.24 to 19.24 and 20.32, item (1), and Proposition 21... | Yes |
Lemma 21.8 (Lindenbaum’s Lemma). Every consistent set \( \Gamma \) in a language \( \mathcal{L} \) can be extended to a complete and consistent set \( {\Gamma }^{ * } \) . | Proof. Let \( \Gamma \) be consistent. Let \( {\varphi }_{0},{\varphi }_{1},\ldots \) be an enumeration of all the sentences of \( \mathcal{L} \) . Define \( {\Gamma }_{0} = \Gamma \), and\n\n\[ \n{\Gamma }_{n + 1} = \left\{ \begin{array}{ll} {\Gamma }_{n} \cup \left\{ {\varphi }_{n}\right\} & \text{ if }{\Gamma }_{n} ... | Yes |
1. \( \mathfrak{M}\left( {\Gamma }^{ * }\right) \vDash \exists {x\varphi }\left( x\right) \) iff \( \mathfrak{M} \vDash \varphi \left( t\right) \) for at least one term \( t \) . | 1. By Proposition 14.42, \( \mathfrak{M}\left( {\Gamma }^{ * }\right) \vDash \exists {x\varphi }\left( x\right) \) iff for at least one variable assignment \( s,\mathfrak{M}\left( {\Gamma }^{ * }\right), s \vDash \varphi \left( x\right) \) . As \( \left| {\mathfrak{M}\left( {\Gamma }^{ * }\right) }\right| \) consists o... | Yes |
Lemma 21.11 (Truth Lemma). Suppose \( \varphi \) does not contain \( = \) . Then \( \mathfrak{M}\left( {\Gamma }^{ * }\right) \vDash \varphi \) iff \( \varphi \in {\Gamma }^{ * } \) . | Proof. We prove both directions simultaneously, and by induction on \( \varphi \) .\n\n1. \( \varphi \equiv \bot : \mathfrak{M}\left( {\Gamma }^{ * }\right) \nvDash \bot \) by definition of satisfaction. On the other hand, \( \bot \notin {\Gamma }^{ * } \) since \( {\Gamma }^{ * } \) is consistent.\n\n2. \( \varphi \eq... | No |
Proposition 21.13. The relation \( \approx \) has the following properties:\n\n1. \( \approx \) is reflexive.\n\n2. \( \approx \) is symmetric.\n\n3. \( \approx \) is transitive.\n\n4. If \( t \approx {t}^{\prime } \) , \( f \) is a function symbol, and \( {t}_{1},\ldots ,{t}_{i - 1},{t}_{i + 1},\ldots ,{t}_{n} \) are ... | Proof. Since \( {\Gamma }^{ * } \) is consistent and complete, \( t = {t}^{\prime } \in {\Gamma }^{ * } \) iff \( {\Gamma }^{ * } \vdash t = {t}^{\prime } \) . Thus it is enough to show the following:\n\n1. \( {\Gamma }^{ * } \vdash t = t \) for all terms \( t \) .\n\n2. If \( {\Gamma }^{ * } \vdash t = {t}^{\prime } \... | Yes |
Proposition 21.16. \( \mathfrak{M}/ \approx \) is well defined, i.e., if \( {t}_{1},\ldots ,{t}_{n},{t}_{1}^{\prime },\ldots ,{t}_{n}^{\prime } \) are terms, and \( {t}_{i} \approx {t}_{i}^{\prime } \) then\n\n\[ \text{1.}{\left\lbrack f\left( {t}_{1},\ldots ,{t}_{n}\right) \right\rbrack }_{ \approx } = {\left\lbrack f... | Proof. Follows from Proposition 21.13 by induction on \( n \) . | No |
Lemma 21.17. \( \mathfrak{M}/ \approx \vDash \varphi \) iff \( \varphi \in {\Gamma }^{ * } \) for all sentences \( \varphi \) . | Proof. By induction on \( \varphi \), just as in the proof of Lemma 21.11. The only case that needs additional attention is when \( \varphi \equiv t = {t}^{\prime } \). \n\n\[ \mathfrak{M}/ \approx \vDash t = {t}^{\prime }\text{iff}{\left\lbrack t\right\rbrack }_{ \approx } = {\left\lbrack {t}^{\prime }\right\rbrack }_... | Yes |
Theorem 21.18 (Completeness Theorem). Let \( \Gamma \) be a set of sentences. If \( \Gamma \) is consistent, it is satisfiable. | Proof. Suppose \( \Gamma \) is consistent. By Lemma 21.6, there is a saturated consistent set \( {\Gamma }^{\prime } \supseteq \Gamma \) . By Lemma 21.8, there is a \( {\Gamma }^{ * } \supseteq {\Gamma }^{\prime } \) which is consistent and complete. Since \( {\Gamma }^{\prime } \subseteq {\Gamma }^{ * } \), for each f... | Yes |
Corollary 21.19 (Completeness Theorem, Second Version). For all \( \Gamma \) and sentences \( \varphi \) : if \( \Gamma \vDash \varphi \) then \( \Gamma \vdash \varphi \) . | Proof. Note that the \( {\Gamma }^{\prime } \) s in Corollary 21.19 and Theorem 21.18 are universally quantified. To make sure we do not confuse ourselves, let us restate Theorem 21.18 using a different variable: for any set of sentences \( \Delta \), if \( \Delta \) is consistent, it is satisfiable. By contraposition,... | Yes |
Theorem 21.21 (Compactness Theorem). The following hold for any sentences \( \Gamma \) and \( \varphi \) :\n\n1. \( \Gamma \vDash \varphi \) iff there is a finite \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \vDash \varphi \) .\n\n2. \( \Gamma \) is satisfiable if and only if it is finitely satisfiab... | Proof. We prove (2). If \( \Gamma \) is satisfiable, then there is a structure \( \mathfrak{M} \) such that \( \mathfrak{M} \vDash \varphi \) for all \( \varphi \in \Gamma \) . Of course, this \( \mathfrak{M} \) also satisfies every finite subset of \( \Gamma \) , so \( \Gamma \) is finitely satisfiable.\n\nNow suppose... | Yes |
In every model \( \mathfrak{M} \) of a theory \( \Gamma \), each term \( t \) of course picks out an element of \( \left| \mathfrak{M}\right| \) . Can we guarantee that it is also true that every element of \( \left| \mathfrak{M}\right| \) is picked out by some term or other? In other words, are there theories \( \Gamm... | The compactness theorem shows that this is not the case if \( \Gamma \) has infinite models. Here’s how to see this: Let \( \mathfrak{M} \) be an infinite model of \( \Gamma \), and let \( c \) be a constant symbol not in the language of \( \Gamma \) . Let \( \Delta \) be the set of all sentences \( c \neq t \) for \( ... | Yes |
Consider a language \( \mathcal{L} \) containing the predicate symbol \( < \) , constant symbols \( 0,1 \), and function symbols \( + , \times , - , \div \) . Let \( \Gamma \) be the set of all sentences in this language true in \( \mathfrak{Q} \) with domain \( \mathbb{Q} \) and the obvious interpretations. \( \Gamma ... | Let \( \Delta \) be \( \{ 0 < \) \( c\} \cup \left\{ {c < \left( {1 \div \bar{k}}\right) : k \in {\mathbb{Z}}^{ + }}\right\} \) (where \( \bar{k} = \left( {1 + \left( {1 + \cdots + \left( {1 + 1}\right) \ldots }\right) }\right) \) with \( k \) 1’s). For any finite subset \( {\Delta }_{0} \) of \( \Delta \) there is a \... | No |
Theorem 21.29 (Compactness). \( \Gamma \) is satisfiable if and only if it is finitely satisfiable. | Proof. If \( \Gamma \) is satisfiable, then there is a structure \( \mathfrak{M} \) such that \( \mathfrak{M} \vDash \varphi \) for all \( \varphi \in \Gamma \) . Of course, this \( \mathfrak{M} \) also satisfies every finite subset of \( \Gamma \), so \( \Gamma \) is finitely satisfiable.\n\nNow suppose that \( \Gamma... | Yes |
Theorem 21.30. If \( \Gamma \) is consistent then it has an enumerable model, i.e., it is satisfiable in a structure whose domain is either finite or denumerable. | Proof. If \( \Gamma \) is consistent, the structure \( \mathfrak{M} \) delivered by the proof of the completeness theorem has a domain \( \left| \mathfrak{M}\right| \) that is no larger than the set of the terms of the language \( \mathcal{L} \) . So \( \mathfrak{M} \) is at most denumerable. | Yes |
Theorem 21.31. If \( \Gamma \) is a consistent set of sentences in the language of first-order logic without identity, then it has a denumerable model, i.e., it is satisfiable in a structure whose domain is infinite and enumerable. | Proof. If \( \Gamma \) is consistent and contains no sentences in which identity appears, then the structure \( \mathfrak{M} \) delivered by the proof of the completness theorem has a domain \( \left| \mathfrak{M}\right| \) identical to the set of terms of the language \( {\mathcal{L}}^{\prime } \) . So \( \mathfrak{M}... | Yes |
Theorem 22.1. There are irrational numbers \( a \) and \( b \) such that \( {a}^{b} \) is rational. | Proof. Consider \( {\sqrt{2}}^{\sqrt{2}} \) . If this is rational, we are done: we can let \( a = b = \sqrt{2} \) . Otherwise, it is irrational. Then we have\n\n\[ \n{\left( {\sqrt{2}}^{\sqrt{2}}\right) }^{\sqrt{2}} = {\sqrt{2}}^{\sqrt{2} \cdot \sqrt{2}} = {\sqrt{2}}^{2} = 2, \n\]\n\nwhich is certainly rational. So, in... | Yes |
Proposition 23.2. If an \( \mathcal{L} \) -structure \( \mathfrak{M} \) is a reduct of an \( {\mathcal{L}}^{\prime } \) -structure \( {\mathfrak{M}}^{\prime } \), then for all \( \mathcal{L} \) -sentences \( \varphi \) , \[ \mathfrak{M} \vDash \varphi \text{iff}{\mathfrak{M}}^{\prime } \vDash \varphi \text{.} \] | Proof. Exercise. | No |
Theorem 23.5. If a set \( \Gamma \) of sentences has arbitrarily large finite models, then it has an infinite model. | Proof. Expand the language of \( \Gamma \) by adding countably many new constants \( {c}_{0} \) , \( {c}_{1},\ldots \) and consider the set \( \Gamma \cup \left\{ {{c}_{i} \neq {c}_{j} : i \neq j}\right\} \) . To say that \( \Gamma \) has arbitrarily large finite models means that for every \( m > 0 \) there is \( n \g... | Yes |
Proposition 23.6. There is no sentence \( \varphi \) of any first-order language that is true in a structure \( \mathfrak{M} \) if and only if the domain \( \left| \mathfrak{M}\right| \) of the structure is infinite. | Proof. If there were such a \( \varphi \), its negation \( \neg \varphi \) would be true in all and only the finite structures, and it would therefore have arbitrarily large finite models but it would lack an infinite model, contradicting Theorem 23.5. | Yes |
Proposition 23.12. For any \( \mathfrak{M},\operatorname{Th}\left( \mathfrak{M}\right) \) is complete. | Proof. For any sentence \( \varphi \) either \( \mathfrak{M} \vDash \varphi \) or \( \mathfrak{M} \vDash \neg \varphi \), so either \( \varphi \in \operatorname{Th}\left( \mathfrak{M}\right) \) or \( \neg \varphi \in \operatorname{Th}\left( \mathfrak{M}\right) \) . | Yes |
Proposition 23.13. If \( \mathfrak{N} \vDash \varphi \) for every \( \varphi \in \operatorname{Th}\left( \mathfrak{M}\right) \), then \( \mathfrak{M} \equiv \mathfrak{N} \) . | Proof. Since \( \mathfrak{N} \vDash \varphi \) for all \( \varphi \in \operatorname{Th}\left( \mathfrak{M}\right) ,\operatorname{Th}\left( \mathfrak{M}\right) \subseteq \operatorname{Th}\left( \mathfrak{N}\right) \) . If \( \mathfrak{N} \vDash \varphi \), then \( \mathfrak{N} \mathrel{\text{\vDash \not{} }} \neg \varph... | Yes |
Theorem 23.16. If \( \mathfrak{M}{ \simeq }_{p}\mathfrak{N} \) and \( \mathfrak{M} \) and \( \mathfrak{N} \) are enumerable, then \( \mathfrak{M} \simeq \mathfrak{N} \) . | Proof. Since \( \mathfrak{M} \) and \( \mathfrak{N} \) are enumerable, let \( \left| \mathfrak{M}\right| = \left\{ {{a}_{0},{a}_{1},\ldots }\right\} \) and \( \left| \mathfrak{N}\right| = \left\{ {{b}_{0},{b}_{1},\ldots }\right\} \) . Starting with an arbitrary \( {p}_{0} \in I \), we define an increasing sequence of p... | Yes |
Theorem 23.17. Suppose \( \mathfrak{M} \) and \( \mathfrak{N} \) are structures for a purely relational language (a language containing only predicate symbols, and no function symbols or constants). Then if \( \mathfrak{M}{ \simeq }_{p}\mathfrak{N} \), also \( \mathfrak{M} \equiv \mathfrak{N} \) . | Proof. By induction on formulas, one shows that if \( {a}_{1},\ldots ,{a}_{n} \) and \( {b}_{1},\ldots ,{b}_{n} \) are such that there is a partial isomorphism \( p \) mapping each \( {a}_{i} \) to \( {b}_{i} \) and \( {s}_{1}\left( {x}_{i}\right) = {a}_{i} \) and \( {s}_{2}\left( {x}_{i}\right) = {b}_{i} \) (for \( i ... | Yes |
Proposition 23.19. Let \( \mathcal{L} \) be a finite purely relational language, i.e., a language containing finitely many predicate symbols and constant symbols, and no function symbols. Then for each \( n \in \mathbb{N} \) there are only finitely many first-order sentences in the language \( \mathcal{L} \) that have ... | Proof. By induction on \( n \) . | No |
Theorem 23.23. Let \( \mathcal{L} \) be a purely relational language. Then \( {I}_{n}\left( {\mathbf{a},\mathbf{b}}\right) \) implies that for every \( \varphi \) such that \( \operatorname{qr}\left( \varphi \right) \leq n \), we have \( \mathfrak{M},\mathbf{a} \vDash \varphi \) if and only if \( \mathfrak{N},\mathbf{b... | Proof. The proof that \( {I}_{n}\left( {\mathbf{a},\mathbf{b}}\right) \) implies that \( \mathbf{a} \) and \( \mathbf{b} \) satisfy the same formulas of quantifier rank no greater than \( n \) is by an easy induction on \( \varphi \) . For the converse we proceed by induction on \( n \), using Proposition 23.19, which ... | Yes |
Theorem 23.26. Any two enumerable dense linear orderings without endpoints are isomorphic. | Proof. Let \( {\mathfrak{M}}_{1} \) and \( {\mathfrak{M}}_{2} \) be enumerable dense linear orderings without endpoints, with \( { < }_{1} = { < }^{{\mathfrak{M}}_{1}} \) and \( { < }_{2} = { < }^{{\mathfrak{M}}_{2}} \), and let \( \mathcal{I} \) be the set of all partial isomorphisms between them. \( \mathcal{I} \) is... | Yes |
Proposition 24.2. If a structure \( \mathfrak{M} \) standard, its domain is the set of values of the standard numerals, i.e., \[ \left| \mathfrak{M}\right| = \left\{ {{\operatorname{Val}}^{\mathfrak{M}}\left( \bar{n}\right) : n \in \mathbb{N}}\right\} \] | Proof. Clearly, every \( {\operatorname{Val}}^{\mathfrak{M}}\left( \bar{n}\right) \in \left| \mathfrak{M}\right| \) . We just have to show that every \( x \in \) \( \left| \mathfrak{M}\right| \) is equal to \( {\operatorname{Val}}^{\mathfrak{M}}\left( \bar{n}\right) \) for some \( n \) . Since \( \mathfrak{M} \) is sta... | Yes |
Proposition 24.4. If \( \mathfrak{M} \) is standard, then \( g \) from the proof of Proposition 24.3 is the only isomorphism from \( \mathfrak{N} \) to \( \mathfrak{M} \) . | Proof. Suppose \( h : \mathbb{N} \rightarrow \left| \mathfrak{M}\right| \) is an isomorphism between \( \mathfrak{N} \) and \( \mathfrak{M} \) . We show that \( g = h \) by induction on \( n \) . If \( n = 0 \), then \( g\left( 0\right) = {\mathrm{o}}^{\mathfrak{M}} \) by definition of \( g \) . But since \( h \) is an... | Yes |
Proposition 24.6. If a structure \( \mathfrak{M} \) for \( {\mathcal{L}}_{A} \) contains a non-standard number, \( \mathfrak{M} \) is non-standard. | Proof. Suppose not, i.e., suppose \( \mathfrak{M} \) standard but contains a non-standard number \( x \) . Let \( g : \mathbb{N} \rightarrow \left| \mathfrak{M}\right| \) be an isomorphism. It is easy to see (by induction on \( n \) ) that \( g\left( {{\operatorname{Val}}^{\mathfrak{N}}\left( \bar{n}\right) }\right) = ... | Yes |
Proposition 24.7. Let \( \mathrm{TA} = \{ \varphi : \mathfrak{N} \vDash \varphi \} \) be the theory of \( \mathfrak{N} \) . TA has an enumerable non-standard model. | Proof. Expand \( {\mathcal{L}}_{A} \) by a new constant symbol \( c \) and consider the set of sentences\n\n\[ \Gamma = \mathbf{TA} \cup \{ c \neq \overline{0}, c \neq \overline{1}, c \neq \overline{2},\ldots \}\]\n\nAny model \( {\mathfrak{M}}^{c} \) of \( \Gamma \) would contain an element \( x = {c}^{\mathfrak{M}} \... | Yes |
Example 24.8. Consider the structure \( \mathfrak{K} \) with domain \( \left| \mathfrak{K}\right| = \mathbb{N} \cup \{ a\} \) and interpretations\n\n\[{\mathrm{o}}^{\mathfrak{K}} = 0\]\n\n\[{\prime }^{\mathfrak{K}}\left( x\right) = \left\{ \begin{array}{ll} x + 1 & \text{ if }x \in \mathbb{N} \\ a & \text{ if }x = a \e... | \( \mathfrak{K} \vDash \forall x\forall y\left( {{x}^{\prime } = {y}^{\prime } \rightarrow x = y}\right) \) since \( * \) is injective. \( \mathfrak{K} \vDash \forall x \) o \( \neq {x}^{\prime } \) since 0 is not a \( * \) -successor in \( \mathfrak{K} \) . \( \mathfrak{K} \vDash \forall x\left( {x = 0\vee \exists {yx... | Yes |
Consider the structure \( \mathfrak{L} \) with domain \( \left| \mathfrak{L}\right| = \mathbb{N} \cup \{ a, b\} \) and interpretations \( {\prime }^{\mathfrak{L}} = * ,{ + }^{\mathfrak{L}} = \oplus \) given by\n\n<table><thead><tr><th>\( x \)</th><th>\( {x}^{ * } \)</th><th>\( x \oplus y \)</th><th>\( m \)</th><th>a</t... | Since \( * \) is injective,0 is not in its range, and every \( x \in \left| \mathfrak{L}\right| \) other than 0 is, axioms \( {Q}_{1} - {Q}_{3} \) are true in \( \mathfrak{L} \) . For any \( x, x \oplus 0 = x \), so \( {Q}_{4} \) is true as well. For \( {Q}_{5} \), consider \( x \oplus {y}^{ * } \) and \( {\left( x \op... | Yes |
Proposition 24.10. In \( \mathfrak{M} \) , \( \otimes \) is a linear strict order, i.e., it satisfies:\n\n1. Not \( x \otimes x \) for any \( x \in \left| \mathfrak{M}\right| \) .\n\n2. If \( x \otimes y \) and \( y \otimes z \) then \( x \otimes z \) .\n\n3. For any \( x \neq y, x \otimes y \) or \( y \otimes x \) | Proof. PA proves:\n\n1. \( \forall x\neg x < x \)\n\n2. \( \forall x\forall y\forall z\left( {\left( {x < y \land y < z}\right) \rightarrow x < z}\right) \)\n\n3. \( \forall x\forall y\left( {\left( {x < y \vee y < x}\right) \vee x = y}\right) \) | Yes |
Proposition 24.12. All standard elements of \( \mathfrak{M} \) are less than (according to \( \otimes \) ) all non-standard elements. | Proof. We’ll use \( n \) as short for \( {\operatorname{Val}}^{\mathfrak{M}}\left( \bar{n}\right) \), a standard element of \( \mathfrak{M} \). Already \( \mathbf{Q} \) proves that, for any \( n \in \mathbb{N},\forall x\left( {x < {\bar{n}}^{\prime } \rightarrow \left( {x = \overline{0} \vee x = \overline{1} \vee \cdot... | Yes |
Proposition 24.13. Every nonstandard element \( x \) of \( \left| \mathfrak{M}\right| \) is an element of the subset \[ \ldots {}^{* * * }x{ \ominus }^{* * }x{ \ominus }^{ * }x \ominus x \ominus {x}^{ * } \ominus {x}^{* * } \ominus {x}^{* * * } \ominus \ldots \] We call this subset the block of \( x \) and write it as ... | Proof. Clearly, such a set \( \left\lbrack x\right\rbrack \) always exists since every element \( y \) of \( \left| \mathfrak{M}\right| \) has a unique successor \( {y}^{ * } \) and unique predecessor \( {}^{ * }y \) . For successive elements \( y \) , \( {y}^{ * } \) we have \( y \otimes {y}^{ * } \) and \( {y}^{ * } ... | Yes |
Proposition 24.14. If \( \left\lbrack x\right\rbrack \neq \left\lbrack y\right\rbrack \) and \( x \otimes y \), then for any \( u \in \left\lbrack x\right\rbrack \) and any \( v \in \left\lbrack y\right\rbrack \) , \( u \otimes v \) . | Proof. Note that \( \mathbf{{PA}} \vdash \forall x\forall y\left( {x < y \rightarrow \left( {{x}^{\prime } < y \vee {x}^{\prime } = y}\right) }\right) \) . Thus, if \( u \otimes v \), we also have \( u \oplus {n}^{ * } \otimes v \) for any \( n \) if \( \left\lbrack u\right\rbrack \neq \left\lbrack v\right\rbrack \) .\... | Yes |
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