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Proposition 10.27. Let \( \Gamma \) be a set of formulas such that \( {\mathcal{g}}^{ + }{}^{ * }\Gamma \) is recursive. Then \( {\mathcal{J}}^{+ + }\left( {\Gamma \text{-Prf }}\right) \) is recursive. | Proof. For any \( x \in \omega \) we have: \( x \in {\mathcal{g}}^{+ + * }\left( {\Gamma \text{-Prf }}\right) \) iff \( x > 1 \) and for every \( i \leq {1x} \) one of the following conditions holds:\n\n(1) \( {\left( x\right) }_{i} \in {\mathcal{J}}^{+ * } \) Axm,\n\n(2) \( {\left( x\right) }_{i} \in {\mathcal{g}}^{+ ... | Yes |
Theorem 10.29. Let \( \Gamma \) be a set of formulas such that \( {\mathcal{g}}^{+ * }\Gamma \) is recursively enumerable. Then \( {\mathcal{g}}^{+ * }\left( {\Gamma \text{-Thm }}\right) \) is recursively enumerable. | Proof. For any \( x \in \omega \), we have \( x \in {\mathcal{g}}^{+ * }\left( {\Gamma \text{-Thm }}\right) \) iff \( \exists y\left\lbrack {y \in {\mathcal{g}}^{+ + * }\left( {\Gamma \text{-Prf }}\right) }\right. \) and \( \left. {{\left( y\right) }_{1y} = x}\right\rbrack \). | Yes |
Proposition 10.30. For any \( m \in \omega \), if \( \alpha \in {}^{m}\operatorname{Rng}v \) and \( \varphi \) and \( \psi \) are formulas, then\n\n\[ \vdash \forall {\alpha }_{0}\cdots \forall {\alpha }_{m - 1}\left( {\varphi \rightarrow \psi }\right) \rightarrow \left( {\forall {\alpha }_{0}\cdots \forall {\alpha }_{... | Proof. We proceed by induction on \( m \) . The case \( m = 0 \) is trivial, since \( \left( {\varphi \rightarrow \psi }\right) \rightarrow \left( {\varphi \rightarrow \psi }\right) \) is a tautology. Now assume the result for \( m \), and suppose that \( \varphi \in {}^{m + 1} \) Rng \( v \) . Then\n\n\[ \vdash \foral... | Yes |
Theorem 10.32. \( \Gamma \vdash \varphi \) iff \( \Gamma { \vdash }^{\prime }\varphi \) . | Proof. Γ-Thm clearly satisfies the conditions 10.31(i) and 10.31(ii). Hence \( \Gamma - {\operatorname{Thm}}^{\prime } \subseteq \Gamma - \operatorname{Thm} \), i.e., \( \Gamma { \vdash }^{\prime }\varphi \Rightarrow \Gamma \vdash \varphi \) . To prove the converse, let\n\n\( \Delta = \left\{ {\varphi \in \text{Fmla} :... | Yes |
Proposition 10.34. \( \; \vdash \sigma = \sigma \) . | Proof. Let \( \alpha \) be a variable not occurring in \( \sigma \) . Then\n\nby \( {10.23}\left( 5\right) \)\n\n\( \vdash \neg \left( {\sigma \equiv \sigma }\right) \rightarrow \neg \left( {\alpha \equiv \sigma }\right) \; \) by a tautology, detachment\n\n\( \vdash \forall \alpha \left\lbrack {\neg \left( {\sigma \equ... | Yes |
Proposition 10.35. \( {\tau \sigma } = \tau \rightarrow \tau = \sigma \) . | Proof.\n\n\[ \vdash \sigma \equiv \tau \rightarrow \left( {\sigma \equiv \sigma \rightarrow \tau \equiv \sigma }\right) \]\n\n10.23(5)\n\n\[ \vdash \sigma \mathrel{\text{:=}} \tau \rightarrow \tau \mathrel{\text{:=}} \sigma \] using a tautology | Yes |
Proposition 10.36. \( \; \vdash \sigma \equiv \tau \rightarrow \left( {\tau \equiv \rho \rightarrow \sigma \equiv \rho }\right) \) . | \[ \vdash \tau \equiv \sigma \rightarrow \left( {\tau \equiv \rho \rightarrow \sigma \equiv \rho }\right) \] 10.23(5) \[ \vdash \sigma \equiv \tau \rightarrow \left( {\tau \equiv \rho \rightarrow \sigma \equiv \rho }\right) \] using 10.35 and a tautology | Yes |
Lemma 10.37. If \( \mathbf{O} \) is an operation symbol of rank \( m, i < m,{\sigma }_{0},\ldots ,{\sigma }_{m - 1} \) are terms, and \( \tau \) is a term, then\n\n\[ \vdash {\sigma }_{i} \equiv \tau \rightarrow \mathbf{O}{\sigma }_{0}\cdots {\sigma }_{m - 1} = \mathbf{O}{\sigma }_{0}\cdots {\sigma }_{i - 1}\tau {\sigm... | Proof. The following formula is an instance of 10.23:\n\n\( {\sigma }_{i} = \tau \rightarrow \)\n\n\[ \left( {\mathbf{O}{\sigma }_{0}\cdots {\sigma }_{m - 1} = \mathbf{O}{\sigma }_{0}\cdots {\sigma }_{m - 1} \rightarrow \mathbf{O}{\sigma }_{0}\cdots {\sigma }_{m - 1} = \mathbf{O}{\sigma }_{0}\cdots {\sigma }_{i - 1}\ta... | Yes |
Theorem 10.38. If \( \mathbf{O} \) is an operation symbol of rank \( m \), and \( {\sigma }_{0},\ldots ,{\sigma }_{m - 1} \) , \( {\tau }_{0},\ldots ,{\tau }_{m - 1} \) are terms, then\n\n\[ \vdash {\sigma }_{0} \equiv {\tau }_{0} \land \cdots \land {\sigma }_{m - 1} \equiv {\tau }_{m - 1} \rightarrow \mathbf{O}{\sigma... | Proof. By 10.37 we have, for each \( i < m \) ,\n\n\[ \vdash {\sigma }_{i} \equiv {\tau }_{i} \rightarrow \mathbf{O}{\tau }_{0}\cdots {\tau }_{i - 1}{\sigma }_{i}{\sigma }_{i + 1}\cdots {\sigma }_{m - 1} \equiv \mathbf{O}{\tau }_{0}\cdots {\tau }_{i}{\sigma }_{i + 1}\cdots {\sigma }_{m - 1}. \]\n\nHence 10.36 and an ea... | Yes |
If \( \vdash \sigma \equiv \tau \rightarrow \left( {\varphi \rightarrow \psi }\right) \) and \( \alpha \in \) Rng \( v \) does not occur in \( \sigma \) or in \( \tau \) , then \( \vdash \sigma \equiv \tau \rightarrow \left( {\forall {\alpha \varphi } \rightarrow \forall {\alpha \psi }}\right) \) . | \[ F\xrightarrow[]{\sigma \equiv \tau \rightarrow \left( {\varphi \rightarrow \psi }\right) } \] hypothesis \[ \vdash \forall \alpha \left( {\sigma \rightrightarrows \tau }\right) \rightarrow \forall \alpha \left( {\varphi \rightarrow \psi }\right) \] generalization, 10.23(2) \[ \vdash \sigma \equiv \tau \rightarrow \f... | Yes |
Proposition 10.42. Let \( R = \{ \left( {m, i, x, y}\right) : m = {g\alpha } \) for some variable \( \alpha, x = \) \( {\mathcal{g}}^{ + }\varphi \) and \( y = {\mathcal{g}}^{ + }\psi \) for some formulas \( \varphi ,\psi, i < \operatorname{Dmn}\varphi \), and \( \forall \) is a quantifier on \( \alpha \) at the ith pl... | Proof. \( \left( {m, i, x, y}\right) \in R \) iff \( m \in \operatorname{Rng}\left( {g \circ v}\right) \) and \( x, y \in {\mathcal{g}}^{+ * } \) Fmla and \( i \leq {1x} \) and \( {\left( x\right) }_{i} = g{L}_{3} + 1 \) and \( {\left( x\right) }_{i + 1} = m + 1 \) and \( i + {1y} \leq {1x} \) and \( \forall j \leq \) ... | Yes |
Proposition 10.43. Let \( S = \{ \left( {m, k, x}\right) : m = {g\alpha } \) for some variable \( \alpha, x = \) \( {\mathcal{f}}^{ + }\varphi \) for some formula \( \varphi, k < \operatorname{Dmn}\varphi \), and \( \alpha \) occurs bound at the \( k \) th place of \( \varphi \} \) . Then \( S \) is recursive. | Proof. \( \left( {m, k, x}\right) \in S \) iff \( m \in \operatorname{Rng}\left( {g \circ v}\right), x \in {\mathcal{g}}^{+ * } \) Fmla, \( k \leq \mathrm{l}x,{\left( x\right) }_{k} = m + \) 1 and \( \exists i \leq \operatorname{lx}\exists y \leq x\left\lbrack {y \in {\mathcal{A}}^{ + }\text{Fmla,}i < k \leq i + \opera... | Yes |
Proposition 10.44. Let \( T = \left\{ {\left( {m, k, x}\right) : m = {\mathcal{J}}^{\alpha }}\right. \) for some variable \( \alpha, x = {\mathcal{J}}^{ + }\varphi \) for some formula \( \varphi, k < \operatorname{Dmn}\varphi \), and \( \alpha \) occurs free at the \( k \) th place of \( \varphi \) \}. Then \( T \) is ... | Proof. \( \;\left( {m, k, x}\right) \in T \) iff \( m \in \operatorname{Rng}\left( {g \circ v}\right), x \in {g}^{+ * } \) Fmla \(, k \leq \mathrm{l}x,{\left( x\right) }_{k} = m + 1 \) , and \( \left( {m, k, x}\right) \notin S \), where \( S \) is as in 10.43. | No |
Proposition 10.47. Let \( U = \left\{ {\left( {x, i, y}\right) : x = {\mathcal{g}}^{ + }\sigma }\right. \) for some term \( \sigma, y = {\mathcal{g}}^{ + }\varphi \) for some formula \( \varphi, i < \operatorname{Dmn}\varphi \), and \( \sigma \) occurs free at the ith place in \( \varphi \} \) . Then \( U \) is recursi... | Proof. \( \left( {x, i, y}\right) \in U \) iff \( x \in {\mathcal{g}}^{+ * }\operatorname{Trm}, y \in {\mathcal{g}}^{+ * } \) Fmla, \( i \leq \operatorname{ly} \), and \( \exists j \leq \) ly \( \left\lbrack {i \leq j}\right. \) and \( \forall k \leq j\left\lbrack {i \leq k \Rightarrow {\left( y\right) }_{k} = {\left( ... | Yes |
Lemma 10.48. If \( \varphi \) and \( \psi \) are formulas, and \( \psi \) is obtained from \( \varphi \) by replacing one free occurrence of \( \sigma \) in \( \varphi \) by a free occurrence of \( \tau \) in \( \psi \), then \( \vdash \sigma \equiv \tau \rightarrow \left( {\varphi \leftrightarrow \psi }\right) \). | Proof. We proceed by induction on \( \varphi \). For \( \varphi \) atomic,\n\n\[ \vdash \sigma = \tau \rightarrow \left( {\varphi \rightarrow \psi }\right) \]\n\n10.23(5)\n\n\[ \vdash \tau = \sigma \rightarrow \left( {\psi \rightarrow \varphi }\right) \]\n\n10.23(5)\n\nHence by 10.35 we have\n\n\[ \vdash \sigma = \tau ... | Yes |
Theorem 10.49 (Substitution of equals for equals). If \( \varphi \) and \( \psi \) are formulas, and \( \psi \) is obtained from \( \varphi \) by replacing zero or more free occurrences of \( \sigma \) in \( \varphi \) by free occurrences of \( \tau \) in \( \psi \), then \( \vdash \sigma \equiv \tau \rightarrow \left(... | This theorem is obtained from 10.48 by induction on the number of free occurrences of \( \sigma \) in \( \varphi \) which are replaced to obtain \( \psi \) . | Yes |
Corollary 10.50. If \( \rho \) and \( \xi \) are terms, and \( \xi \) is obtained from \( \rho \) by replacing zero or more occurrences of \( \sigma \) in \( \rho \) by \( \tau \), then \( \vdash \sigma \equiv \tau \rightarrow \rho \equiv \xi \) . | Proof. By 10.49, \( \vdash \sigma \equiv \tau \rightarrow \left( {\rho \equiv \rho \leftrightarrow \rho \equiv \xi }\right) \) . Hence the desired result follows by 10.34. | Yes |
Proposition 10.52. If \( m \) is the Gödel number of a variable \( \alpha, x \) is the Gödel number of a term \( \sigma \), and \( y \) is the Gödel number of a formula \( \varphi \), let \( f\left( {m, x, y}\right) \) be the Gödel number of \( {\operatorname{Subf}}_{\sigma }^{\alpha }\varphi \) ; otherwise, let \( f\l... | Proof. We will indicate a simple procedure for obtaining Subf \( {}_{\sigma }^{\alpha }\varphi \) which makes the effectiveness very clear. First, it is clearly an effective matter to take the first free occurrence of \( \alpha \) in \( \varphi \) (if any) and replace it by \( \sigma \) . Formally, for any \( m, x, y \... | Yes |
Lemma 10.53. If \( \alpha \) is a variable which does not occur bound in \( \varphi \) and does not occur in \( \sigma \), and if no free occurrence of \( \alpha \) in \( \varphi \) is within the scope of a quantifier on a variable occurring in \( \sigma \), then \( \vdash \forall {\alpha \varphi } \rightarrow {\operat... | Proof\n\n\[ \vdash \alpha \equiv \sigma \rightarrow \left( {\varphi \rightarrow {\operatorname{Subf}}_{\sigma }^{\alpha }\varphi }\right) \]\n\n10.49\n\n\[ \vdash \varphi \rightarrow \left( {\neg {\text{Subf }}_{\sigma }^{\alpha }\varphi \rightarrow \neg \alpha \equiv \sigma }\right) \;\text{suitable tautology} \]\n\n\... | Yes |
Lemma 10.54. If \( \psi \) is obtained from \( \varphi \) by replacing an occurrence of \( \chi \) in \( \varphi \) by \( \theta \), and if \( \mathrm{F}\chi \leftrightarrow \theta \), then \( \mathrm{F}\varphi \leftrightarrow \psi \) . | Proof. Induction on \( \varphi \) . | No |
Lemma 10.55. If \( \alpha \) and \( \beta \) are distinct variables, \( \alpha \) does not occur bound in \( \varphi \) , and \( \beta \) does not occur in \( \varphi \) at all, then \( \vdash \forall {\alpha \varphi } \leftrightarrow \forall \beta {\operatorname{Subf}}_{\sigma }^{\alpha }\varphi \) . | Proof\n\n\[ \vdash \forall {\alpha \varphi } \rightarrow {\operatorname{Subf}}_{\beta }^{\alpha }\varphi \]\n\n10.53\n\n\[ \vdash \forall \beta \forall {\alpha \varphi } \rightarrow \forall \beta {\operatorname{Subf}}_{\beta }^{\alpha }\varphi \]\nusing \( {10.23}\left( 2\right) \)\n\n\[ \vdash \forall {\alpha \varphi ... | Yes |
Lemma 10.58. If \( \alpha \) occurs bound in \( \varphi \), then there is a formula \( \forall {\alpha \psi } \) which occurs in \( \varphi \) such that \( \alpha \) does not occur bound in \( \psi \) . | Proof. Induction on \( \varphi \) . | No |
Theorem 10.59 (Change of bound variables). If \( \beta \) does not occur in \( \varphi \), then \( \vdash \varphi \leftrightarrow {\mathrm{{Subb}}}_{\beta }^{\alpha }\varphi . \) | Proof. We proceed by induction on the number \( m \) of bound occurrences of \( \alpha \) in \( \varphi \) . If \( m = 0 \), the desired conclusion is trivial. We now assume that \( m > 0 \), and that our result is known for all formulas with fewer than \( m \) bound occurrences of \( \alpha \) . By Lemma 10.58, let \(... | Yes |
Lemma 10.60. If the variable \( \alpha \) does not occur in \( \sigma \), and if no free occurrence of \( \alpha \) in \( \varphi \) is within the scope of a quantifier on a variable occurring in \( \sigma \), then \( \vdash \forall {\alpha \varphi } \rightarrow {\operatorname{Subf}}_{\sigma }^{\alpha }\varphi . \) | Proof. Let \( \beta \) be a variable not occurring in \( \varphi \), not occurring in \( \sigma \), and different from \( \alpha \) . Then by change of bound variables,\n\n\[ \vdash \varphi \leftrightarrow {\operatorname{Subb}}_{\beta }^{\alpha }\varphi \]\n\nHence using \( {10.23}\left( 2\right) \) we infer that\n\n(1... | Yes |
Theorem 10.61 (Universal specification). If no free occurrence in \( \varphi \) of the variable \( \alpha \) is within the scope of a quantifier on a variable occurring in \( \sigma \) , then \( \vdash \forall {\alpha \varphi } \rightarrow {\operatorname{Subf}}_{\sigma }^{\alpha }\varphi \) . | Proof. Let \( \beta \) be a variable not occurring in \( \varphi \) or in \( \sigma \), and distinct from \( \alpha \) . Then\n\n\[ \vdash \forall {\alpha \varphi } \rightarrow {\mathrm{{Subf}}}_{\beta }^{\alpha }\varphi \]\n\n10.60\n\n(1)\n\n\[ \vdash \forall \beta \forall {\alpha \varphi } \rightarrow \forall \beta {... | Yes |
Corollary 10.63. If the variable \( \alpha \) does not occur free in \( \varphi \), then \( \vdash \varphi \leftrightarrow \forall {\alpha \varphi } \) . | Proof. By 10.62,\n\n(1)\n\n\[ \vdash \forall {\alpha \varphi } \rightarrow \varphi \text{.} \]\n\nFor the other direction, let \( \beta \) be a variable not occurring in \( \varphi \) . Then by change of bound variable,\n\n(2)\n\n\[ \vdash \varphi \leftrightarrow {\mathrm{{Subb}}}_{\beta }^{\alpha }\varphi . \]\n\nHenc... | Yes |
Proposition 10.64. If the variable \( \alpha \) does not occur free in \( \varphi \), then \( \vdash \forall \alpha \left( {\varphi \rightarrow \psi }\right) \rightarrow \) \( \left( {\varphi \rightarrow \forall {\alpha \psi }}\right) \) . | Proof. \( \; \vdash \varphi \leftrightarrow \forall {\alpha \varphi } \) by 10.63; hence use \( {10.23}\left( 2\right) \) . | No |
Proposition 10.65. \( \; \vdash \forall \alpha \forall {\beta \varphi } \rightarrow \forall \beta \forall {\alpha \varphi } \) . | Proof\n\n\[ \nH\xrightarrow[]{\alpha }\forall \beta \xrightarrow[]{\varphi }\varphi \n\] \n\n10.62 (twice) \n\n\[ \n\mathcal{H}\alpha \forall \alpha \forall \beta \vDash \varphi \rightarrow \forall {\alpha \varphi } \n\] \n\n10.23(2) \n\n\[ \n\mathcal{H}\alpha \nabla \beta ▱\varphi \rightarrow \nabla \alpha \nabla \alp... | No |
Corollary 10.67. If no free occurrence of \( \alpha \) in \( \varphi \) is within the scope of a quantifier on a variable occurring in \( \sigma \), then \( {\mathrm{{FSubf}}}_{\sigma }^{\alpha }\varphi \rightarrow \exists {\alpha \varphi } \) . | Proof\n\n\[ \vdash \forall \alpha \sqsupset \varphi \rightarrow {\operatorname{Subf}}_{\sigma }^{\alpha } \sqsupset \varphi \;\text{ universal specification } \]\n\n\[ { \vdash }^{ \Vdash }{\mathrm{{Subf}}}_{\sigma }^{\alpha }\varphi \rightarrow \exists {\alpha \varphi } \]\n\nsuitable tautology | Yes |
Proposition 10.70. \( \; \vdash \exists \alpha \forall {\beta \varphi } \rightarrow \forall \beta \exists {\alpha \varphi } \) . | Proof\n\n\[ \vdash \varphi \rightarrow \exists {\alpha \varphi } \]\n\n\[ \mathcal{H}\beta \xrightarrow[]{\varphi }\xrightarrow[]{\varphi \rightarrow \forall \beta }\exists {\alpha \varphi } \]\n\n10.23(2)\n\n\[ \vdash \neg \forall \beta \exists {\alpha \varphi } \rightarrow \neg \forall {\beta \varphi } \]\ntautology\... | Yes |
Theorem 10.71 (Substitutivity of equivalence). Let \( \\varphi ,\\psi ,\\chi \) be formulas and \( \\alpha \\in {}^{m}\\operatorname{Rng}v \) . Suppose that if \( \\beta \) occurs free in \( \\varphi \) or in \( \\psi \) but bound in \( \\chi \) then \( \\beta \\in \\left\\{ {{\\alpha }_{i} : i < m}\\right\\} \) . Let ... | Proof. We proceed by induction on \( \\chi \) . We may assume that \( \\theta \\neq \\chi \) . If \( \\chi \) is atomic, then \( \\chi = \\varphi \) and \( \\psi = \\theta \) ; this case is trivial. Suppose \( \\chi \) is \( \\neg {\\chi }^{\\prime } \) . Then \( \\theta \) is of the form \( \\neg {\\theta }^{\\prime }... | Yes |
Corollary 10.74. \( \;\mathrm{h}\varphi \left( {{\sigma }_{0},\ldots ,{\sigma }_{m - 1}}\right) \rightarrow \exists {v}_{0}\cdots \exists {v}_{m - 1}\varphi \) . | Both of these corollaries are immediate consequences of earlier results, upon noticing that simultaneous substitution can be obtained by iterated ordinary substitution; in the notation of 10.72 , \[ \varphi \left( {{\sigma }_{0},\ldots ,{\sigma }_{m - 1}}\right) = {\operatorname{Subf}}_{▞}^{v\left( {k + n + 1}\right) }... | Yes |
If \( \alpha \) does not occur free in \( \psi \), then \( \vdash \forall {\alpha \varphi } \vee \psi \leftrightarrow \forall \alpha \left( {\varphi \vee \psi }\right) \) . | \[ \vdash \forall {\alpha \varphi } \rightarrow \varphi \] 10.62 \[ \vdash \forall {\alpha \varphi }\;\mathbf{v}\;\psi \rightarrow \varphi \;\mathbf{v}\;\psi \] \[ \vdash \forall \alpha \left( {\forall {\alpha \varphi } \vee \psi \rightarrow \varphi \vee \psi }\right) \] (1) \[ \vdash \forall {\alpha \varphi }\;\mathbf... | Yes |
Lemma 10.78. If \( \alpha \) does not occur free in \( \psi \), then \( \vdash \exists {\alpha \varphi } \vee \psi \leftrightarrow \exists \alpha \left( {\varphi \vee \psi }\right) \) . | \[ \vdash \exists \alpha \left( {\varphi \vee \psi }\right) \leftrightarrow \neg \forall \alpha \;\neg \left( {\varphi \vee \psi }\right) \]\n\n\[ \vdash \neg \forall \alpha \;\neg \left( {\varphi \vee \psi }\right) \leftrightarrow \neg \forall \alpha \left( {\;\neg \varphi \; \land \;\neg \psi }\right) \]\n\n\[ \vdash... | Yes |
Theorem 10.81 (Prenex normal form theorem). For any formula \( \varphi \) there is a formula \( \psi \) in prenex normal form such that \( \vdash \varphi \leftrightarrow \psi \), and such that a variable occurs free in \( \varphi \) iff it occurs free in \( \psi \) . | Proof. By induction on \( \varphi \) . The only difficult step is the passage from \( {\varphi }_{1} \) and \( {\varphi }_{2} \) to \( {\varphi }_{1} \vee {\varphi }_{2} \) (or to \( {\varphi }_{1} \land {\varphi }_{2} \) ). Since these cases are symmetric, we deal only with the first. Thus assume, with obvious notatio... | Yes |
Theorem 10.85. For \( \Gamma \cup \{ \varphi \} \subseteq {\operatorname{Sent}}_{\mathcal{L}},\Gamma { \vdash }_{\varphi } \) iff \( \Gamma { \vdash }^{\prime \prime }\varphi \) . | Proof. \( \left( {\Gamma - \operatorname{Thm}}\right) \cap \) Sent in place of \( \Delta \) clearly satisfies 10.84 \( \left( i\right) - \left( {ii}\right) \) . Hence \( \Gamma - {\operatorname{Thm}}^{\prime \prime } \subseteq \left( {\Gamma - \operatorname{Thm}}\right) \cap \) Sent, so \( \Gamma { \vdash }^{\prime \pr... | Yes |
Theorem 10.86 (Deduction theorem). If \( \varphi \) is a sentence and \( \Gamma \cup \{ \varphi \} \vdash \psi \) , then \( \Gamma \vdash \varphi \rightarrow \psi \) . | Proof. Let \( \Delta = \{ \chi : \Gamma \vdash \varphi \rightarrow \psi \} \) . If \( \chi \) is an axiom or a member of \( \Gamma \), then\n\n\[ \begin{matrix} \chi \\ \chi \rightarrow \left( {\varphi \rightarrow \chi }\right) \\ \varphi \rightarrow \chi \end{matrix} \]\n\nis a \( \Gamma \) -formal proof of \( \varphi... | Yes |
Corollary 10.87. Assume \( \Gamma \subseteq \) Sent, \( m \in \omega \), and \( \varphi \in {}^{m} \) Sent. Then the following conditions are equivalent:\n\n(i) \( \Gamma \cup \left\{ {{\varphi }_{0},\ldots ,{\varphi }_{m - 1}}\right\} \vdash \psi \) ;\n\n(ii) \( \Gamma \vdash {\varphi }_{0} \land \cdots \land {\varphi... | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . We proceed by induction on \( m \), the case \( m = 0 \) being trivial (condition (ii) is then to be interpreted as \( \Gamma \vdash \psi \), just like \( \left( i\right) \) ). Assume that the implication holds for \( m \), and suppose that \( \Gamma \cup \l... | Yes |
Theorem 10.91. 0 is consistent. | Proof. For any formula \( \varphi \), let \( {\varphi }^{\prime } \) be obtained from \( \varphi \) by first simultaneously replacing all atomic formulas in \( \varphi \) by \( {v}_{0} = {v}_{0} \rightarrow {v}_{0} = {v}_{0} \) and then deleting all quantifiers. Let \( \Delta = \left\{ {\varphi : {\varphi }^{\prime }}\... | Yes |
Proposition 10.92. Assume \( \Gamma \cup \Delta \subseteq \) Sent and \( \Delta \neq 0 \) . Then the following conditions are equivalent:\n\n(i) \( \Gamma \cup \Delta \) is inconsistent;\n\n(ii) there exist an \( m \in \omega \sim 1 \) and a \( \varphi \in {}^{m}\Delta \) such that \( \Gamma \vdash \neg {\varphi }_{0} ... | Proof\n\n\( \left( i\right) \Rightarrow \left( {ii}\right) \) . By 10.88 we obtain \( m \in \omega \) and \( \varphi \in {}^{m}\Delta \) such that\n\n\[ \Gamma \vdash {\varphi }_{0} \land \cdots \land {\varphi }_{m - 1} \rightarrow \neg \forall {v}_{0}\left( {{v}_{0} = {v}_{0}}\right) . \]\n\nWe may assume \( m \neq 0 ... | Yes |
Proposition 11.3. If every variable occurring in \( \sigma \) is in the set \( \left\{ {{v}_{i} : i \in \Gamma }\right\} \), where \( \Gamma \subseteq \omega \), then \( {\sigma }^{\mathfrak{A}}x = {\sigma }^{\mathfrak{A}}y \) whenever \( x \upharpoonright \Gamma = y \upharpoonright \Gamma \) . | This proposition is easily established by induction on \( \sigma \) . | No |
Proposition 11.6. If every variable occurring free in a formula \( \varphi \) is in the set \( \left\{ {{v}_{i} : i \in \Gamma }\right\} \), where \( \Gamma \subseteq \omega \), and if \( x \upharpoonright \Gamma = y \upharpoonright \Gamma \), then \( x \in {\varphi }^{\mathfrak{A}} \) iff \( y \in {\varphi }^{\mathfra... | This proposition is easily established by induction on \( \varphi \), and justifies the following definition. | No |
Proposition 11.11. For any \( \Gamma \subseteq {\operatorname{Sent}}_{\mathcal{L}} \) the following conditions are equivalent:\n\n(i) \( \Gamma \) is complete and consistent;\n\n(ii) \( \left\{ {\varphi : \varphi \in {\operatorname{Sent}}_{\mathcal{L}},\Gamma \vdash \varphi }\right\} \) is a maximal consistent set of s... | Proof\n\n\( \left( i\right) \Rightarrow \left( {ii}\right) \) . Obviously \( \Delta = \left\{ {\varphi : \varphi \in {\operatorname{Sent}}_{\mathcal{L}},\Gamma \vdash \varphi }\right\} \) is consistent. Suppose \( \Delta \subset \Theta \) ; say \( \varphi \in \Theta \sim \Delta \) . Thus \( \Gamma \nvdash \varphi \), s... | Yes |
Theorem 11.13 (Lindenbaum). If \( \Gamma \) is a consistent set of sentences of \( \mathcal{L} \), then there is a complete consistent set \( \Delta \) of sentences in \( \mathcal{L} \) such that \( \Gamma \subseteq \Delta \) . | Proof. Let \( \mathcal{A} = \left\{ {\Delta : \Gamma \subseteq \Delta \subseteq {\operatorname{Sent}}_{\mathcal{L}},\Delta }\right. \) consistent \( \} \) . If \( \mathcal{B} \) is a nonempty subset of \( \mathcal{A} \) simply ordered by \( \subseteq \), then \( \Gamma \subseteq \bigcup \mathcal{B} \subseteq {\text{Sen... | Yes |
Proposition 11.15. Let \( \mathcal{L},{\mathcal{L}}^{\prime },\mathfrak{A},\mathfrak{B} \) be as in 11.14. Then:\n\n(i) \( {\operatorname{Trm}}_{\mathcal{L}} \subseteq {\operatorname{Trm}}_{{\mathcal{L}}^{\prime }},{\mathrm{{Fmla}}}_{\mathcal{L}} \subseteq {\mathrm{{Fmla}}}_{{\mathcal{L}}^{\prime }},{\mathrm{{Axm}}}_{\... | This proposition is easy to prove; (i)-(iii) are proved by induction, and (iv) and (v) follow from (i)-(iii). By the completeness theorem which we will shortly prove, \( {11.15}\left( v\right) \) also holds for the notion \( \vdash \) . This statement is needed in our proof of the completeness theorem, however, so we w... | Yes |
Proposition 11.17 Let \( \mathrm{c} \) be an individual constant not occurring in any formula of \( \Gamma \cup \{ \varphi \} \) . Assume that \( \Gamma \vdash {\operatorname{Subf}}_{\mathbf{c}}^{\alpha }\varphi \) . Then \( \Gamma \vdash \varphi \) . | Proof. Let \( \left\langle {{\psi }_{0},\ldots ,{\psi }_{m - 1}}\right\rangle \) be a \( \Gamma \) -formal proof with \( {\psi }_{m - 1} = {\operatorname{Subf}}_{\mathrm{e}}^{\alpha }\varphi \) . Let \( \beta \) be a variable not occurring in any of the formulas \( {\psi }_{0},\ldots ,{\psi }_{m - 1} \) . For any \( i ... | Yes |
Lemma 11.18. Let \( \mathcal{L} \) be any first-order language and let \( \Gamma \) be a consistent set of sentences in \( \mathcal{L} \) . Let \( {\mathcal{L}}^{\prime } \) be an expansion of \( \mathcal{L} \) obtained by adjoining \( \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \) new individual constants. Then ther... | Proof. Let \( \left\langle {{\varphi }_{\alpha } : \alpha < \mathfrak{m}}\right\rangle \) be a list of all sentences of \( {\mathcal{L}}^{\prime } \) of the form \( \exists {\beta \psi } \) , where \( m \) is an infinite cardinal number. Note that \( m = \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \) . We now define ... | Yes |
Theorem 11.19 (Completeness theorem, first form). Any consistent set of sentences has a model. The model can be taken to have power \( \leq \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \) . | Proof. Let \( \Gamma \) be a consistent set of sentences in the language \( \mathcal{L} \) . Let \( {\mathcal{L}}^{\prime } \) be obtained from \( \mathcal{L} \) by adjoining \( \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \) new individual constants. By 11.18, let \( \Delta \) be a consistent rich set of sentences in... | Yes |
Theorem 11.20 (Completeness theorem, second form). \( \Gamma \vdash \varphi \) iff \( \Gamma \vDash \varphi \) . | Proof. \( \Rightarrow : \) by 11.8. \( \Leftarrow : \) Assume \( \Gamma \nvdash \varphi \) . Thus by \( {10.95}\left( {iii}\right) ,\{ \left\lbrack \left\lbrack \psi \right\rbrack \right\rbrack : \psi \in \Gamma \} \nvdash \) \( \left\lbrack \left\lbrack \varphi \right\rbrack \right\rbrack \) so by \( {10.92},\Delta = ... | Yes |
Theorem 11.21 (Weak completeness theorem). If \( \mathcal{L} \) is an effectivized first-order language, then \( \left\{ {{g}^{ + }\varphi : \vDash \varphi }\right\} \) is recursively enumerable. | Proof. By the completeness theorem, the given set is identical with \( \left\{ {{g}^{ + }\varphi : \vdash \varphi }\right\} \) . This set is r.e. by Theorem 10.29. | Yes |
Theorem 11.22 (Compactness theorem). If \( \Gamma \) is a set of sentences such that every finite subset of \( \Gamma \) has a model, then \( \Gamma \) has a model. | Proof. By 11.19 it suffices to show that \( \Gamma \) is consistent. Suppose not: say \( \Gamma \vdash \varphi \land \neg \varphi \) . By 10.33, \( \Delta \vdash \varphi \land \neg \varphi \) for some finite subset \( \Delta \) of \( \Gamma \) . By 11.20 (in fact, the easy part of 11.20 given in 11.8), \( \Delta \vDash... | Yes |
Corollary 11.23. Let \( \mathcal{L} \) be any first-order language. Then there is no set \( \Gamma \) of sentences of \( \mathcal{L} \) such that for any \( \mathcal{L} \) -structure \( \mathfrak{A} \) , \( \mathfrak{A} \) is a model of \( \Gamma \) iff \( \mathfrak{A} \) is finite. | Proof. Assume the contrary. Expand \( \mathcal{L} \) to \( {\mathcal{L}}^{\prime } \) by adjoining new individual constants \( {\mathbf{c}}_{m} \) for \( m \in \omega \) . Let \( \Delta \) be \( \Gamma \) together with all sentences \( \neg \left( {{\mathbf{c}}_{i} = {\mathbf{c}}_{j}}\right) \) for \( i \neq j \) . Eve... | Yes |
Corollary 11.24. Let \( \mathcal{L} \) be a first-order language with just one nonlogical constant, a binary relation symbol \( \leq \) . Let \( \mathbf{K} \) be the class of all \( \mathcal{L} \) -structures \( \mathfrak{A} = \left( {A, \leq }\right) \) such that \( \leq \) is a well-ordering of \( A \) . Then there i... | Proof. Assume the contrary. Again adjoin new individual constants \( {\mathbf{c}}_{m} \) for \( m \in \omega \) . Let \( \Delta \) be \( \Gamma \) together with all sentences \( {\mathbf{c}}_{m + 1} < {\mathbf{c}}_{m} \), for \( m \in \omega \) . Every finite subset of \( \Delta \) clearly has a model. Hence by the com... | Yes |
Corollary 11.25. If a sentence \( \varphi \) holds in every infinite \( \mathcal{L} \) -structure, then there is an \( m \in \omega \) such that \( \varphi \) holds in every finite \( \mathcal{L} \) -structure of power \( > m \) . | Proof. Suppose the conclusion fails. Thus for every \( m \in \omega \) there is a finite \( \mathcal{L} \) -structure of power \( > m \) in which \( \varphi \) fails to hold. Adjoin new individual constants \( {\mathbf{c}}_{m} \) for \( m \in \omega \) . Let\n\n\[ \Gamma = \{ \neg \varphi \} \cup \left\{ {\neg \left( {... | Yes |
Theorem 11.28. Let \( {\mathcal{L}}^{\prime } \) be a relational version of \( \mathcal{L} \) with translation, \( T \) , notation as in \( {11.26}\left( i\right) \). (i) If \( \varphi \) is a formula of \( \mathcal{L} \) in which no operation symbol occurs, then \( {\varphi }^{\prime } = \varphi \). (ii) Let \( \mathf... | Proof. (i) and the first part of (ii) are obvious. We prove the second part of (ii) by induction on \( \varphi \), following \( {11.26}\left( {ii}\right) \). If \( \varphi \) is \( {v}_{j} = {v}_{i} \), obviously \( \mathfrak{A} \vDash \) \( \varphi \left\lbrack x\right\rbrack \) iff \( {\mathfrak{A}}^{\prime } \vDash ... | Yes |
Theorem 11.30. Let \( \left( {{\Gamma }^{\prime },{\mathcal{L}}^{\prime }}\right) \) be a definitional extension of \( \left( {\Gamma ,\mathcal{L}}\right) \), with notation as in 11.29. Then for any formula \( \psi \) of \( {\mathcal{L}}^{\prime } \) there is a formula \( {\psi }^{\prime } \) of \( \mathcal{L} \) with ... | Proof. We construct \( {\psi }^{\prime } \) by induction on \( \psi \) . In each step the desired properties of \( {\psi }^{\prime } \) are easy to prove, and we prove the desired result in only one step as an illustration. To begin with, we construct \( {\psi }^{\prime } \) for \( \psi \) of the form \( \sigma = {v}_{... | Yes |
Theorem 11.31. Let \( \left( {{\Gamma }^{\prime },{\mathcal{L}}^{\prime }}\right) \) be a definitional expansion of \( \left( {\Gamma ,\mathcal{L}}\right) \). Then \( \left( {{\Gamma }^{\prime },{\mathcal{L}}^{\prime }}\right) \) is a conservative extension of \( \left( {\Gamma ,\mathcal{L}}\right) \). | Proof. Again we take all notation as in 11.29. Suppose \( \psi \) is a formula of \( \mathcal{L} \) and \( {\Gamma }^{\prime } \vDash \psi \); we must show that \( \Gamma \vDash \psi \). Let \( \mathfrak{A} \) be any model of \( \Gamma \) ( \( \mathfrak{A} \) is an \( \mathcal{L} \)-structure). For \( \mathbf{C} \) a n... | Yes |
Theorem 11.32. Let \( \Gamma \) be a theory in a language \( \mathcal{L} \), and \( \varphi \) a formula of \( \mathcal{L} \) with free variables among \( {v}_{0},\ldots ,{v}_{m} \) . Assume that \( \Gamma \vDash \forall {v}_{0}\cdots \forall {v}_{m - 1}\exists {v}_{m}\varphi \) . Let \( {\mathcal{L}}^{\prime } \) be a... | Proof. Assume that \( {\Gamma }^{\prime } \vDash \psi \), where \( \psi \) is a formula of \( \mathcal{L} \) . To prove \( \Gamma \vDash \psi \) , let \( \mathfrak{A} \) be any model of \( \Gamma \) . By the axiom of choice, there is an \( m \) -ary function \( g \) on \( A \) such that for all \( {x}_{0},\ldots ,{x}_{... | Yes |
Theorem 11.38 (Skolem normal form theorem). Let \( {\mathcal{L}}^{\prime } \) be a Skolem expansion of \( \mathcal{L} \). For every prenex formula \( \varphi \) of \( {\mathcal{L}}^{\prime } \), the formula \( {\varphi }^{\mathrm{S}} \) is universal and the same variables occur free in \( \varphi \) as do in \( {\varph... | Proof. Clearly \( {\varphi }^{\mathrm{S}} \) is universal and \( \mathrm{{Fv}}\varphi = \mathrm{{Fv}}{\varphi }^{\mathrm{S}} \). Condition \( \left( i\right) \) is easily proved by induction on \( \varphi \), as is \( \left( {ii}\right) \). To prove \( \left( {iii}\right) \left( a\right) \), assume its hypothesis and l... | Yes |
Theorem 11.40. Let \( {\mathcal{L}}^{\prime } \) be a Skolem expansion of \( \mathcal{L} \), and let \( \varphi \) be a prenex formula of \( {\mathcal{L}}^{\prime } \). Then:\n\n(i) \( {\varphi }^{\mathrm{H}} \) is an existential formula;\n\n(ii) \( \vDash {\varphi }^{\mathrm{H}} \leftrightarrow \neg {\varphi }^{\mathr... | Proof. (i) is obvious from the definitions. To prove (ii), first note that \( \vdash \neg \varphi \leftrightarrow {\varphi }^{\mathrm{n}} \). Next we prove (ii) by induction on the length of \( \varphi \) ; it is clear if \( \varphi \) is quantifier free. Now \( {\left( \exists \alpha \varphi \right) }^{\mathrm{H}} = \... | Yes |
Theorem 11.41. If \( \varphi \) is a quantifier-free formula not involving equality and \( \vDash \varphi \), then \( \varphi \) is a tautology. | Proof. Assume that \( \varphi \) is not a tautology; let \( f \) be a truth valuation (10.19) such that \( {f\varphi } = 0 \) . Let \( A = {\operatorname{Trm}}_{\mathcal{L}} \) . For any relation symbol \( \mathbf{R} \), say \( \mathbf{R} \) of rank \( m \), let\n\n\[{\mathbf{R}}^{\mathfrak{A}} = \left\{ {\left( {{\tau... | Yes |
Theorem 11.42 (Herbrand). Let \( {\mathcal{L}}^{\prime } \) be a Skolem expansion of \( \mathcal{L} \), and let \( \varphi \) be a prenex sentence of \( {\mathcal{L}}^{\prime } \) not involving equality. Say \( {\varphi }^{\mathrm{H}} = \exists {\alpha }_{0}\cdots \exists {\alpha }_{m - 1}\psi \) with \( \psi \) quanti... | Proof. \( \Rightarrow \) . Assume \( \vDash \varphi \) . Thus by \( {11.40}, \vDash {\varphi }^{\mathrm{H}} \) . Now let \( \Gamma \) be the set of all sentences \[ \neg {\mathrm{{Subf}}}_{▞}^{⓪}\cdots {\mathrm{{Subf}}}_{\sigma \left( {m - 1}\right) }^{\alpha \left( {m - 1}\right) }\psi \] where \( {\sigma }_{0},\ldots... | Yes |
Corollary 11.46. Let \( \Gamma \) be a theory in \( \mathcal{L} \), and let \( \mathfrak{A} = \left( {\chi, f, R,{\Gamma }^{\prime }}\right) \) be an interpretation of \( \Gamma \) in \( {\mathcal{L}}^{\prime } \) . Then for any formula \( \varphi \) of \( \mathcal{L} \), the condition \( \Gamma \vDash \varphi \) impli... | Proof. Assume that \( \Gamma \vDash \varphi \), and let \( \mathfrak{B} \) be any model of \( {\Gamma }^{\prime } \) (thus \( \mathfrak{B} \) is an \( \left. {{\mathcal{L}}^{\prime }\text{-structure}}\right) \) . Since \( {\Gamma }^{\prime } \vDash {\psi }^{\mathfrak{A}} \) for each \( \psi \in \Gamma \), it follows th... | Yes |
(i) If \( {\mathrm{c}}_{\kappa }x = 0 \), then \( x = 0 \) . | Proof. \( \left( i\right) \) ,(ii): immediate by \( \left( {\mathrm{C}}_{2}\right) \) . | No |
Proposition 12.15. Let \( I \) be an ideal in a \( {\mathrm{{CA}}}_{\alpha }\mathfrak{A} \), and let \( R = \{ \left( {x, y}\right) : x \cdot - y + - x \cdot y \in I\} \) (cf. 9.16). Then for any \( \kappa < \alpha \) and \( x, y \in A \), if \( {xRy} \) then \( {\mathrm{c}}_{\kappa }{xR}{\mathrm{c}}_{\kappa }y \) . | Proof. Assume that \( {xRy} \) and \( \kappa < \alpha \) . Thus \( x \cdot - y + - x \cdot y \in I \) . Now\n\n\[{\mathrm{c}}_{\kappa }x = {\mathrm{c}}_{\kappa }\left( {x \cdot - y + x \cdot y}\right) = {\mathrm{c}}_{\kappa }\left( {x \cdot - y}\right) + {\mathrm{c}}_{\kappa }\left( {x \cdot y}\right) \leq {\mathrm{c}}... | Yes |
Proposition 12.21. If \( X \subseteq A,\mathfrak{A}a{\mathrm{{CA}}}_{\alpha } \), then the ideal generated by \( X \) is the collection of all \( y \in A \) such that there exist \( m, n \in \omega \) and \( x \in {}^{m}X,\kappa \in {}^{n}\alpha \) with \( y \leq {\mathrm{c}}_{╄}\cdots {\mathrm{c}}_{\kappa \left( {n - ... | Proof. Let \( I \) be the collection of all \( y \in A \) such that such \( m, n, x,\kappa \) exist. Clearly \( I \) is contained in the ideal generated by \( X \) . Thus it is enough to show that \( X \subseteq I \) and \( I \) is an ideal. Taking \( m = 1 \) and \( n = 0 \) we easily see that \( X \subseteq I \) . Ta... | Yes |
Proposition 12.24. Let \( \Gamma \) be a set of sentences in a first-order language \( \mathcal{L} \), and let \( \mathfrak{A} \) be a model of \( \Gamma \). Then \( \left\{ {{\varphi }^{\mathfrak{A}} : \varphi \in {\mathrm{{Fmla}}}_{\mathcal{L}}}\right\} \) is an \( \omega \)-dimensional field of sets. Let \( \mathfra... | This proposition can be routinely checked. | No |
Theorem 12.30. If \( \mathfrak{A} \) is a locally finite dimensional \( {\mathrm{{CA}}}_{\omega },\left| A\right| > 1 \), then there is a homomorphism of \( \mathfrak{A} \) onto a cylindric set algebra. | Proof. By 12.29 we may assume that \( \mathfrak{A} = {\mathfrak{M}}_{T}^{\mathcal{L}} \) for some \( \mathcal{L},\Gamma \) . Then \( \Gamma \) is consistent, since \( \left| A\right| > 1 \) . Let \( \mathcal{L} \) be a model of \( \Gamma \) . The desired conclusion now follows from 12.24. | No |
Theorem 13.3 (Disjunctive normal form theorem). Let \( \Gamma \subseteq {\mathrm{{Fmla}}}_{\mathcal{L}} \). Then for any \( \varphi \in \mathrm{{Qf}}\Gamma \) such that \( \neg \varphi \) is not a tautology there exist \( p, m \in \omega \) and a function \( \psi \) such that:\n\n(i) the domain of \( \psi \) is \( \lef... | For the proof, see 8.38. | No |
Theorem 13.5. \( {\Gamma }_{1} \) is decidable. | Proof. Let \( \varphi \) be any sentence in our language. Let \( \psi \) be found from \( \varphi \) by 13.4: \( \psi \) is a quantifier-free combination of basic formulas and \( \mathrm{{Fv}}\psi \subseteq \mathrm{{Fv}}\varphi \) , therefore \( \psi \) is a sentence. Now by 13.3, whose proof is obviously effective, th... | Yes |
Corollary 13.6. The theory of \( \left( {\omega, s}\right) \) is decidable. | Proof. There is an effective method for recognizing when a sentence does not involve \( \mathbf{O} \) . Such a sentence holds in \( \left( {\omega, s}\right) \) iff it holds in \( \left( {\omega, s,0}\right) \) . | Yes |
Corollary 13.7. A set \( A \subseteq \omega \) is elementarily definable in \( \left( {\omega, s}\right) \) or in \( \left( {\omega, s,0}\right) \) , iff it is finite or cofinite. | Proof. First we treat \( \left( {\omega, s,0}\right) \) . \( \Rightarrow \) . Suppose \( \varphi \) elementarily defines \( A \) . Thus \( \mathrm{{Fv}}\varphi \subseteq \left\{ {v}_{0}\right\} \) and \( {}^{1}{\varphi }^{\mathfrak{A}} = A \), where \( \mathfrak{A} = \left( {\omega, s,0}\right) \) . By 13.4 let \( \psi... | Yes |
Theorem 13.9. \( {\Gamma }_{2} \) is decidable. | Proof. By the method of proof of 13.5 we see that it suffices to describe a method for determining the truth in \( \mathfrak{A} \) of basic sentences. The basic sentences are easily seen to be effectively equivalent to sentences of the forms\n\n\[ \mathbf{m} = \mathbf{n}, \]\n\n\[ \mathbf{m} < \mathbf{n}, \]\n\n\[ \mat... | Yes |
Theorem 13.12. \( {\Gamma }_{3} \) is decidable. | Proof. The only basic sentences are \( {\varepsilon }_{m}, m \in \omega \sim 1 \) . Clearly \( {\varepsilon }_{1} \in {\Gamma }_{3} \) while \( {\varepsilon }_{m} \notin {\Gamma }_{3} \) if \( m \neq 1 \) . Let \( \Delta \) be the set of all sentential combinations of basic sentences. Note:\n\n(1)\n\n\[ m \leq \left| A... | Yes |
Proposition 14.2. Let \( R \subseteq {}^{m}\omega \) . Let \( \Gamma \) be a theory in \( {\mathcal{L}}_{\text{nos }} \), and assume that \( \neg \left( {\mathbf{0} = \mathbf{1}}\right) \in \Gamma \) . Then the following conditions are equivalent:\n\n(i) \( R \) is syntactically definable in \( \Gamma \) ;\n\n(ii) \( {... | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Let \( \varphi \) syntactically define \( R \) in \( \Gamma \) . Let \( \psi \) be the formula\n\n\[ \left( {\varphi \land {v}_{m} = 1}\right) \vee \left( {\neg \varphi \land {v}_{m} = 0}\right) . \]\n\nThus for any \( {x}_{0},\ldots ,{x}_{m - 1} \in \omega ... | Yes |
Theorem 14.5. Let \( f : {}^{m}\omega \rightarrow \omega \) and let \( \Gamma \) be a consistent recursively axiomatiz-able theory in \( {\mathcal{L}}_{\text{nos }} \) such that \( \neg \left( {\mathbf{n} = \mathbf{p}}\right) \in \Gamma \) whenever \( n \neq p \) . If \( f \) is syntactically definable in \( \Gamma \) ... | Proof. Say \( \Gamma \) is recursively axiomatized by \( \Delta \) . Let\n\n\( {\mathbf{T}}_{m}^{\Delta } = \left\{ {\left( {e,{x}_{0},\ldots ,{x}_{m - 1}, z, y}\right) : e}\right. \) is the Gödel number of a formula \( \varphi \) with free variables among \( {v}_{0},\ldots ,{v}_{m} \), and \( y \) is the Gödel number ... | Yes |
Theorem 14.6. Let \( R \subseteq {}^{m}\omega \) and let \( \Gamma \) be a recursively axiomatizable theory in \( {\mathcal{L}}_{\text{nos }} \) . If \( R \) is weakly syntactically definable in \( \Gamma \), then \( R \) is recursively enumerable. | Proof. Say \( \Gamma \) is recursively axiomatized by \( \Delta \) . Let \( {\mathbf{T}}_{m}^{\Delta } \) be as in the above proof. Let \( R \) be weakly syntactically defined in \( \Gamma \) by \( \varphi \), and let \( e = {g}^{ + }\varphi \) . Clearly for all \( {x}_{0},\ldots ,{x}_{m - 1} \in \omega \) we have\n\n\... | Yes |
If \( \Gamma \) is an \( \omega \) -consistent theory in \( {\mathcal{L}}_{\text{nos }},\neg \left( {\mathbf{m} = \mathbf{n}}\right) \in \Gamma \) whenever \( m \neq n \), and every unary recursive function is syntactically definable in \( \Gamma \), then every r.e. set is weakly syntactically definable in \( \Gamma \)... | First note that every \( \omega \) -consistent theory is also consistent. Hence the formula \( \neg \left( {{v}_{0} = {v}_{0}}\right) \) clearly weakly syntactically defines the empty set. Now let \( A \) be any nonempty r.e. set, say \( A = \operatorname{Rng}f \) where \( f \) is recursive. By hypothesis, there is a f... | Yes |
Corollary 14.12. A function is recursive iff it is syntactically definable in \( \mathrm{R} \) . | By 14.3, 14.6, 14.8, 14.10, and 14.11, we have: | No |
Lemma 14.14. Let \( \mathfrak{A} = \left( {\omega ,+,\cdot, s,0}\right) \), and let \( \Gamma \) be a theory in \( {\mathcal{L}}_{\text{nos }} \) with \( \mathfrak{A} \) as a model. Then if a function or relation is syntactically definable in \( \Gamma \) it is also elementarily definable in \( \mathfrak{A} \) . | Proof. First suppose \( f \), an \( m \) -ary function, is syntactically defined by a formula \( \varphi \) in \( \Gamma \) . Then \( {}^{m + 1}{\varphi }^{\mathfrak{A}} = f \), i.e., \( f \) is elementarily defined by \( \varphi \) (see 11.7). In fact, first suppose \( f{x}_{0}\cdots {x}_{m - 1} = {x}_{m} \) . Then by... | Yes |
Theorem 14.19. \( \mathrm{R} \subseteq \mathrm{Q} \subseteq \mathrm{P} \subseteq \mathrm{N} \) | Proof. Obviously \( \mathrm{P} \subseteq \mathrm{N} \). To prove \( \mathrm{Q} \subseteq \mathrm{P} \) it is obviously enough to show that the sentence \( \forall {v}_{0}\left\lbrack {\neg \left( {{v}_{0} = \mathbf{0}}\right) \rightarrow \exists {v}_{1}\left( {\mathbf{s}{v}_{1} = {v}_{0}}\right) }\right\rbrack \) is in... | Yes |
Lemma 14.23. Any spectrally representable function is recursive. | Proof. We give an intuitive proof for this, thus appealing to the weak Church’s thesis (see p. 46). Let \( f, n \) -ary, be spectrally represented by \( \varphi \) . Let \( \Gamma = \left\{ {{\mathbf{R}}_{i}^{m} : m \in \omega \sim 1, i \in \omega ,{\mathbf{R}}_{i}^{m}}\right. \) occurs in \( \left. {\varphi \text{or}m... | Yes |
Lemma 14.24. Let \( f \) be a unary function spectrally represented by a sentence \( \varphi \) . Then there is a sentence \( \psi \) which also spectrally represents \( f \) such that for any model \( \mathfrak{A} \) of \( \psi \) the following conditions hold:\n\n(i) \( \left| {\mathbf{R}}_{2}^{1\mathfrak{A}}\right| ... | Proof. We first divide the symbols of \( {\mathcal{L}}_{\text{un }} \) different from \( {\mathbf{R}}_{0}^{1},{\mathbf{R}}_{1}^{1},{\mathbf{R}}_{2}^{1},{\mathbf{R}}_{0}^{2} \) , and \( {\mathbf{R}}_{1}^{2} \) into two disjoint classes, consisting of distinct symbols \( {}^{0}{\mathbf{R}}_{r}^{q} \) and \( {}^{1}{\mathb... | Yes |
Theorem 15.1 (Craig). For any theory \( \Gamma \) in an elementary effectivized first-order language the following conditions are equivalent:\n\n(i) \( \Gamma \) is axiomatizable by a set \( \Delta \) such that \( {\mathcal{g}}^{+ * }\Delta \) is elementary;\n\n(ii) \( \Gamma \) is recursively axiomatizable;\n\n(iii) \... | Proof. Obviously \( \left( i\right) \Rightarrow \left( {ii}\right) \), while \( \left( {ii}\right) \Rightarrow \left( {iii}\right) \) by 10.29. Now assume that (iii) holds. Let \( f \) be an elementary function with range \( {g}^{+ * }\Gamma \) . Set\n\n\( \Delta = \left\{ {\varphi : \varphi }\right. \) is a sentence, ... | Yes |
Theorem 15.2. For any complete theory \( \Gamma \) the following conditions are equivalent;\n\n(i) \( \Gamma \) is decidable:\n\n(ii) \( \Gamma \) is recursively axiomatizable. | Proof. \( \;\left( i\right) \Rightarrow \left( {ii}\right) \) . This is obvious, since \( \Gamma \) axiomatizes \( \Gamma \) . \( \left( {ii}\right) \Rightarrow \left( i\right) \) . Assume (ii), say \( \Delta \) axiomatizes \( \Gamma \), where \( {\mathcal{J}}^{+ * }\Delta \) is recursive. A decision procedure for \( \... | Yes |
Theorem 15.5. If \( \Gamma \) is consistent and decidable, then \( \Gamma \subseteq \Delta \) for some consistent, decidable complete theory \( \Delta \) . | Proof. Intuitively we proceed as follows, merely effectivizing a proof of Lindenbaum's Theorem 11.13. Effectively list out all sentences. Now extend \( \Gamma \) by an effective, recursive procedure: at the \( m \) th step, add the \( m \) th sequence if it is consistent to do so, otherwise add nothing. In this way we ... | Yes |
Let \( \Gamma \) be an inseparable theory. Then there is a recursive function \( h \) such that if \( \Delta \) is any consistent extension of \( \Gamma \) in \( \mathcal{L} \) and if \( {g}^{+ * }\Delta = \) Dmn \( {\mathbf{\varphi }}_{e}^{1} \), then he \( = {\mathcal{g}}^{ + }\varphi \) for some sentence \( \varphi ... | Proof. Let \( f \) be as above. We define a partial recursive function \( {k}^{\prime } \) by setting\n\n\[ \n{k}^{\prime }\left( {x, e}\right) \simeq {\mu y}\left\lbrack {\left( {e, x, y}\right) \in {\mathrm{T}}_{1}\text{ or }x \notin {\mathcal{g}}^{+ * }{\operatorname{Sent}}_{\mathcal{L}}}\right\rbrack , \n\] \n\nfor... | Yes |
Proposition 15.9. Every inseparable theory is essentially undecidable, and every essentially undecidable theory is undecidable. Every finitely inseparable theory is undecidable. If \( \Gamma \) is finitely inseparable, then \( \left\{ {{g}^{ + }\varphi : \varphi }\right. \) is a sentence which holds in every finite mod... | Proof. Let \( \Gamma \) be an inseparable theory in \( \mathcal{L} \). Thus \( {\mathcal{J}}^{+ * }\Gamma \) and \( A = \) \( {\mathcal{J}}^{+ * }\{ \varphi : \varphi \) is a sentence and \( \neg \varphi \in \Gamma \} \) are disjoint, by 6.21, so \( \Gamma \) is consistent. Suppose that \( \Delta \) is a consistent the... | Yes |
Proposition 15.10 Let \( \Gamma \) be a theory in \( \mathcal{L},\Delta \) a consistent theory in an effective expansion \( {\mathcal{L}}^{\prime } \) of \( \mathcal{L} \), and assume that \( \Gamma \subseteq \Delta \) . Then \( \Gamma \) essentially undecidable implies \( \Delta \) essentially undecidable, and \( \Gam... | Proof. Assume the first sentence of 15.10. Let \( \Gamma \) be essentially undecidable. Suppose that \( \Theta \) is a consistent decidable extension of \( \Delta \) in \( {\mathcal{L}}^{\prime } \) . Then \( \Theta \cap {\operatorname{Sent}}_{\mathcal{L}} \) is a consistent decidable extension of \( \Gamma \) in \( \m... | Yes |
Proposition 15.13. If \( \Gamma \) and \( \Delta \) are theories in \( \mathcal{L} \) and \( \Delta \) is a finite extension of \( \Gamma \), then \( \Delta \) undecidable implies \( \Gamma \) undecidable. | Proof. Let \( \Theta \) be a finite set of sentences such that \( \Gamma \cup \Theta \) axiomatizes \( \Delta \) . Then for any sentence \( \varphi \), we have \( \varphi \in \Delta \) iff \( \bigwedge \Theta \rightarrow \varphi \in \Gamma \), so \( \Gamma \) is undecidable. | Yes |
Proposition 15.14. Let \( {\mathcal{L}}^{\prime } \) be an effective expansion of a language \( \mathcal{L} \) . Suppose \( \Gamma \) is a theory in \( \mathcal{L},\Delta \) is a theory in \( {\mathcal{L}}^{\prime } \), and \( \Gamma \cup \Delta \) is consistent. If \( \Gamma \) is finitely axiomatizable and essentiall... | Proof. Let \( \Theta = \left\{ {\varphi : \varphi }\right. \) is a sentence of \( \left. {{\mathcal{L}}^{\prime }\text{and}\Gamma \cup \Delta \vDash \varphi }\right\} \) . Clearly \( \Theta \) is a theory in \( {\mathcal{L}}^{\prime } \) which is an extension of \( \Gamma \) and a finite extension of \( \Delta \) . By ... | Yes |
Proposition 15.15. Let \( \left( {{\Gamma }^{\prime },{\mathcal{L}}^{\prime }}\right) \) be an effective definitional expansion of \( \left( {\Gamma ,\mathcal{L}}\right) \) . Then:\n\n(i) \( \Gamma \) is decidable iff \( {\Gamma }^{\prime } \) is decidable; | Proof. Let the notation be as in 11.29 and 11.30. For any sentence \( \varphi \) of \( \mathcal{L} \), by 11.31 we know that \( \varphi \in \Gamma \) iff \( \varphi \in {\Gamma }^{\prime } \) . Hence \( {\Gamma }^{\prime } \) decidable \( \Rightarrow \Gamma \) decidable. By 11.30 we know that for any sentence \( \psi \... | Yes |
Proposition 15.16. Let \( \mathcal{L} \) and \( {\mathcal{L}}^{\prime } \) be two languages, \( \Gamma \) and \( \Delta \) consistent theories in \( \mathcal{L} \) and \( {\mathcal{L}}^{\prime } \) respectively, and suppose that \( \mathfrak{A} \) is an effective interpretation of \( \Gamma \) in \( \Delta \) . Then:\n... | (i) Let \( {\Delta }^{\prime } \) be any consistent theory in \( {\mathcal{L}}^{\prime } \) which extends \( \Delta \) . Set \( {\Gamma }^{\prime } = \) \( \left\{ {\varphi \in {\operatorname{Sent}}_{\mathcal{L}} : {\Delta }^{\prime } \vDash {\varphi }^{\mathfrak{A}}}\right\} \) . Thus \( \Gamma \subseteq {\Gamma }^{\p... | Yes |
Proposition 15.17. Let \( \mathcal{L} \) and \( {\mathcal{L}}^{\prime } \) be two languages, and let \( \Gamma \) and \( \Delta \) be consistent theories in \( \mathcal{L} \) and \( {\mathcal{L}}^{\prime } \) respectively. Assume that \( \mathcal{L} \) has only finitely many operation symbols (with no restriction on th... | Proof. By virtue of 6.25 it suffices to establish the following:\n\n(1) for any sentence \( \varphi \) of \( \mathcal{L}, \vDash \varphi \) implies \( \vDash \bigwedge {\Gamma }^{\prime } \rightarrow {\varphi }^{\mathfrak{A}} \) ;\n\nfor any sentence \( \varphi \) of \( \mathcal{L} \), if \( \neg \varphi \) holds in so... | Yes |
Theorem 15.18. If \( \Gamma \) is a theory in \( {\mathcal{L}}_{\text{nos }} \) in which every recursive set is weakly syntactically definable, then \( \Gamma \) is undecidable. | Proof. Suppose \( \Gamma \) is decidable. Let\n\n\( A = \left\{ {e : e}\right. \) is the Gödel number of a formula \( \psi \) having at most \( {v}_{0} \) free, and \( \psi \left( \mathbf{e}\right) \notin \Gamma \} \) .\n\nClearly \( A \) is recursive, by the hypothesis that \( \Gamma \) is decidable. Let \( \varphi \)... | Yes |
Theorem 15.20. Let \( \Gamma \) be a theory in \( {\mathcal{L}}_{\text{nos }} \) in which every unary recursive function is syntactically definable, and let \( \varphi \) be a formula of \( {\mathcal{L}}_{\text{nos }} \) with \( \mathrm{{Fv}}\varphi \subseteq \left\{ {v}_{0}\right\} \) . Then there is a sentence \( \ps... | Proof. For any \( x \in \omega \), let\n\n\[ \begin{array}{l} {fx} = {g}^{ + }\psi \left( \mathbf{x}\right) \;\text{ if }x = {g}^{ + }\psi \text{ for some formula }\psi , \\ {fx} = 0\;\text{ otherwise. } \end{array} \]\n\nThus \( f \) is recursive. By hypothesis, let \( \chi \) be a formula of \( {\mathcal{L}}_{\text{n... | Yes |
Theorem 15.21. (Tarski). Let \( \mathfrak{A} = \langle \omega , + , \cdot, s,0\rangle \) and let \( X = \left\{ {{g}^{ + }\psi : \psi }\right. \) is a sentence of \( {\mathcal{L}}_{\text{nos }} \) and \( \mathfrak{A} \vDash \psi \} \) . Then \( X \) is not elementarily definable in \( \mathfrak{A} \) . | Proof. Suppose \( X \) is elementarily definable by a formula \( \varphi \) in \( \mathfrak{A} \) . Thus \( \mathrm{{Fv}}\varphi \subseteq \left\{ {v}_{0}\right\} \) and \( X = {}^{1}{\varphi }^{\mathfrak{A}} \) . By 15.20 choose a sentence \( \psi \) such that \( \mathbf{R} \vDash \) \( \neg \varphi \left( {\Delta {\m... | Yes |
Lemma 16.4. For each prime \( p > 2 \) there is an \( m \) with \( 1 \leq m < p \) such that mp is a sum of four squares. | Proof. The members of \( \left\{ {{x}^{2} : 0 \leq x \leq \left( {p - 1}\right) /2}\right\} \) are pairwise incongruent \( {\;\operatorname{mod}\;p} \), as are the members of \( \left\{ {-1 - {y}^{2} : 0 \leq y \leq \left( {p - 1}\right) /2}\right\} \) . There are \( p + 1 \) numbers in the union of these two sets, so ... | Yes |
Lemma 16.5. For any positive prime \( p, p \) is a sum of four squares. | Proof. Obvious for \( p = 2\left( {2 = 1 + 1 + 0 + 0}\right) \) . Suppose \( p > 2 \) . Let \( m \) be the smallest positive integer such that \( {mp} \) is a sum of four squares. Thus \( 1 \leq m < p \) by 16.4. Say \( {mp} = {x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2} \) . Suppose \( m > 1 \) ; we shall get... | Yes |
Lemma 16.11. \( {\mathcal{L}}_{1} \vDash \mathbf{J}{v}_{0} \rightarrow \mathbf{{JOp}}\left( {{v}_{0},{v}_{1}}\right) \) . | Proof. Assume that \( \mathbf{J}x, y \) is arbitrary, \( z = \mathbf{{Op}}\left( {x, y}\right), w \subseteq z \), and \( v \) is arbitrary; we want to show that \( w{ \cap }_{z}v \) has its usual meaning, i.e., that \( \exists u\forall s\left( {s \in u \leftrightarrow s \in w\text{and}s \in v}\right) \) . Since \( \mat... | Yes |
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