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For instance, consider the assumption \( \mathbb{T}\varphi \land \neg \varphi \) . Here is the (open) tableau consisting of just that assumption:\n\n\[ \text{1.}\mathbb{T}\varphi \land \neg \varphi \;\text{Assumption} \] | We obtain a new tableau from it by applying the \( \land \mathbb{T} \) rule to the assumption. That rule allows us to add two new lines to the tableau, \( \mathbb{T}\varphi \) and \( \mathbb{T}\neg \varphi \) :\n\n\[ \begin{matrix} \text{ 1. } & \mathbb{T}\varphi \land \neg \varphi & \text{ Assumption } \\ \text{ 2. } ... | 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 \( \rightar... | Yes |
Example 11.6. 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) ... | We start by applying the \( \vee \mathbb{T} \) to the first assumption:\n\n\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... | Yes |
Proposition 11.10 (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\n\n\( \otimes \)\n\nis closed. | No |
Proposition 11.11 (Monotony). If \( \Gamma \subseteq \Delta \) and \( \Gamma \vdash \varphi \), then \( \Delta \vdash \varphi \) . | Proof. Any finite subset of \( \Gamma \) is also a finite subset of \( \Delta \) . | No |
Proposition 11.12 (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 11.13. \( \Gamma \) is inconsistent iff \( \Gamma \vdash \varphi \) for every sentence \( \varphi \) . | Proof. Exercise. | No |
Proposition 11.14 (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 11.15. 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 11.16. \( \Gamma \vdash \varphi \) 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 11.17. 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 11.18. 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 11.19. 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\n\[ \text{1.}\mathbb{F}\varphi \;\text{Assumption} \]\n\n2. \( \mathbb{T}\varphi \land \psi \; \) Assumption\n\n3. \( \mathbb{T}\varphi \; \land \mathbb{T}2 \) | No |
Proposition 11.20. 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 11.21. 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 |
Corollary 11.25. If \( \Gamma \vdash \varphi \) then \( \Gamma \vDash \varphi \) . | Proof. If \( \Gamma \vdash \varphi \) then for some \( {\psi }_{1},\ldots ,{\psi }_{n} \in \Gamma ,\left\{ {\mathbb{F}\varphi ,\mathbb{T}{\psi }_{1},\ldots ,\mathbb{T}{\psi }_{n}}\right\} \) has a closed tableau. By Theorem 11.23, every valuation \( \mathfrak{v} \) either makes some \( {\psi }_{i} \) false or makes \( ... | Yes |
Corollary 11.26. 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 11.23, there is no \( \mathfrak{v} \) such that \( \mathfrak{v} \vDash {\psi }_{i} \) ... | 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. (12.6) 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. (12.7)\n\n2. \( \left( {\theta \rightarrow \left( {\left( {\theta \rightarrow \theta }\right) \rightarrow \theta }\right) }\right) \rightarrow \)\n\n\[ \left( {\left( {\theta \rightarrow \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. \( \psi \rightarrow \chi \; \) HYP\n3. \( \left( {\psi \rightarrow \chi }\right) \rightarrow \left( {\varphi \rightarrow \left( {\psi \rightarrow \chi }\right) }\right) \) eq. (12.7)\n4. \( \varphi \rightarrow \left( {\psi \rightarrow \chi }\right) \;2,3,\mathrm{{MP}} \)\n5.... | Yes |
Proposition 12.10. 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 |
Proposition 12.14 (Reflexivity). If \( \varphi \in \Gamma \), then \( \Gamma \vdash \varphi \) . | Proof. The formula \( \varphi \) by itself is a derivation of \( \varphi \) from \( \Gamma \) . | Yes |
Proposition 12.15 (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 12.16 (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 12.17. \( \Gamma \) is inconsistent iff \( \Gamma \vdash \varphi \) for every \( \varphi \) . | Proof. Exercise. | No |
Proposition 12.18 (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 12.19. 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 12.16, \( \Gamma \vdash \psi \) . | Yes |
Theorem 12.20 (Deduction Theorem). \( \Gamma \cup \{ \varphi \} \vdash \psi \) if and only if \( \Gamma \vdash \varphi \rightarrow \psi \) . | Proof. The \ | No |
Proposition 12.22. 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 12.14, \( \Gamma \vdash \psi \) for every \( \psi \in \Gamma \) . Since also \( \Gamma \vdash \varphi \) by hypothesis, \( \Gamma \vdash \psi \) for every \( \psi \in \Gamma \cup \{ \varphi \} \)... | Yes |
Proposition 12.23. \( \Gamma \vdash \varphi \) iff \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent. | Proof. First suppose \( \Gamma \vdash \varphi \) . Then \( \Gamma \cup \{ \neg \varphi \} \vdash \varphi \) by Proposition 12.15. \( \Gamma \cup \) \( \{ \neg \varphi \} \vdash \neg \varphi \) by Proposition 12.14. We also have \( \vdash \neg \varphi \rightarrow \left( {\varphi \rightarrow \bot }\right) \) by eq. (12.1... | Yes |
Proposition 12.24. 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. (12.10). \( \Gamma \vdash \bot \) by two applications of Proposition 12.19. | Yes |
Proposition 12.25. If \( \Gamma \cup \{ \varphi \} \) and \( \Gamma \cup \{ \neg \varphi \} \) are both inconsistent, then \( \Gamma \) is inconsistent. | Proof. Exercise. | No |
Proposition 12.26. 1. Both \( \varphi \land \psi \vdash \varphi \) and \( \varphi \land \psi \vdash \psi \) | Proof. 1. From eq. (12.1) and eq. (12.1) by modus ponens. | No |
Proposition 12.27. 1. \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \) is inconsistent. | Proof. 1. From eq. (12.9) 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 12.28. 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 |
Proposition 12.29. If \( \varphi \) is an axiom, then \( \mathfrak{v} \vDash \varphi \) for each valuation \( \mathfrak{v} \) . | Proof. Do truth tables for each axiom to verify that they are tautologies. | No |
Theorem 12.30 (Soundness). If \( \Gamma \vdash \varphi \) then \( \Gamma \vDash \varphi \) . | Proof. By induction on the length of the derivation of \( \varphi \) from \( \Gamma \) . If there are no steps justified by inferences, then all formulas in the derivation are either instances of axioms or are in \( \Gamma \) . By the previous proposition, all the axioms are tautologies, and hence if \( \varphi \) is a... | Yes |
Corollary 12.32. 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 12.30, any valuation \( \mathfrak{v} \) that satisfies \( \Gamma \) must satisfy \( \bot \) . Since \( \mathfrak{v} \mathrel{\tex... | Yes |
1. If \( \Gamma \vdash \varphi \), then \( \varphi \in \Gamma \) . | Suppose that \( \Gamma \vdash \varphi \) . Suppose to the contrary that \( \varphi \notin \Gamma \) . Since \( \Gamma \) is complete, \( \neg \varphi \in \Gamma \) . By Propositions 10.17,11.17,9.19 and 12.24, \( \Gamma \) is inconsistent. This contradicts the assumption that \( \Gamma \) is consistent. Hence, it canno... | Yes |
Lemma 13.3 (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 |
Lemma 13.5 (Truth Lemma). \( \mathfrak{v}\left( {\Gamma }^{ * }\right) \vDash \varphi \) iff \( \varphi \in {\Gamma }^{ * } \) . | Proof. We prove both directions simultaneously, and by induction on \( \varphi \) .\n\n1. \( \varphi \equiv \bot : \mathfrak{v}\left( {\Gamma }^{ * }\right) \nvDash \bot \) by definition of satisfaction. On the other hand, \( \bot \notin {\Gamma }^{ * } \) since \( {\Gamma }^{ * } \) is consistent.\n\n2. \( \varphi \eq... | No |
Theorem 13.6 (Completeness Theorem). Let \( \Gamma \) be a set of sentences. If \( \Gamma \) is consistent, it is satisfiable. | Proof. Suppose \( \Gamma \) is consistent. By Lemma 13.3, there is a \( {\Gamma }^{ * } \supseteq \Gamma \) which is consistent and complete. By Lemma 13.5, \( \mathfrak{v}\left( {\Gamma }^{ * }\right) \vDash \varphi \) iff \( \varphi \in {\Gamma }^{ * } \) . From this it follows in particular that for all \( \varphi \... | Yes |
Corollary 13.7 (Completeness Theorem, Second Version). For all \( \Gamma \) and sentences \( \varphi \) : if \( \Gamma \vDash \varphi \) then \( \Gamma \vdash \varphi \) . | Proof. Note that the \( {\Gamma }^{\prime }\mathrm{s} \) in Corollary 13.7 and Theorem 13.6 are universally quantified. To make sure we do not confuse ourselves, let us restate Theorem 13.6 using a different variable: for any set of sentences \( \Delta \), if \( \Delta \) is consistent, it is satisfiable. By contraposi... | No |
Theorem 13.9 (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 satisfiabl... | Proof. We prove (2). If \( \Gamma \) is satisfiable, then there is a valuation \( \mathfrak{v} \) such that \( \mathfrak{v} \vDash \varphi \) for all \( \varphi \in \Gamma \) . Of course, this \( \mathfrak{v} \) also satisfies every finite subset of \( \Gamma \), so \( \Gamma \) is finitely satisfiable.\n\nNow suppose ... | Yes |
Theorem 13.12 (Compactness). \( \Gamma \) is satisfiable if and only if it is finitely satisfiable. | Proof. If \( \Gamma \) is satisfiable, then there is a valuation \( \mathfrak{v} \) such that \( \mathfrak{v} \vDash \varphi \) for all \( \varphi \in \Gamma \) . Of course, this \( \mathfrak{v} \) also satisfies every finite subset of \( \Gamma \), so \( \Gamma \) is finitely satisfiable.\n\nNow suppose that \( \Gamma... | No |
Lemma 14.8. The number of left and right parentheses in a formula \( \varphi \) are equal. | Proof. We prove this by induction on the way \( \varphi \) is constructed. This requires two things: (a) We have to prove first that all atomic formulas have the property in question (the induction basis). (b) Then we have to prove that when we construct new formulas out of given formulas, the new formulas have the pro... | No |
Lemma 14.10. If \( \varphi \) is a formula, and \( \psi \) is a proper prefix of \( \varphi \), then \( \psi \) is not a formula. | Proof. Exercise. | No |
Proposition 14.11. If \( \varphi \) is an atomic formula, then it satisfes one, and only one of the following conditions.\n\n1. \( \varphi \equiv \bot \) .\n\n2. \( \varphi \equiv R\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) where \( R \) is an \( n \) -place predicate symbol, \( {t}_{1},\ldots ,{t}_{n} \) are terms, an... | Proof. Exercise. | No |
Proposition 14.12 (Unique Readability). Every formula satisfies one, and only one of the following conditions.\n\n1. \( \varphi \) is atomic.\n\n2. \( \varphi \) is of the form \( \neg \psi \) .\n\n3. \( \varphi \) is of the form \( \left( {\psi \land \chi }\right) \) .\n\n4. \( \varphi \) is of the form \( \left( {\ps... | Proof. The formation rules require that if a formula is not atomic, it must start with an opening parenthesis (, -, or with a quantifier. On the other hand, every formula that start with one of the following symbols must be atomic: a predicate symbol, a function symbol, a constant symbol, \( \bot \) .\n\nSo we really o... | Yes |
Consider the following formula:\n\n\[ \exists {v}_{0}\underset{\psi }{\underbrace{{A}_{0}^{2}\left( {{v}_{0},{v}_{1}}\right) }} \] | \( \psi \) represents the scope of \( \exists {v}_{0} \) . The quantifier binds the occurence of \( {v}_{0} \) in \( \psi \) , but does not bind the occurence of \( {v}_{1} \) . So \( {v}_{1} \) is a free variable in this case. | Yes |
A structure \( \mathfrak{M} \) for the language of arithmetic consists of a set, an element of \( \left| \mathfrak{M}\right| ,{\mathrm{o}}^{\mathfrak{M}} \), as interpretation of the constant symbol \( \mathrm{o} \), a one-place function \( {\prime }^{\mathfrak{M}} : \left| \mathfrak{M}\right| \rightarrow \left| \mathf... | An obvious example of such a structure is the following:\n\n1. \( \left| \mathfrak{N}\right| = \mathbb{N} \)\n\n2. \( {\mathrm{o}}^{\mathfrak{N}} = 0 \)\n\n3. \( {\prime }^{\mathfrak{N}}\left( n\right) = n + 1 \) for all \( n \in \mathbb{N} \)\n\n4. \( { + }^{\mathfrak{N}}\left( {n, m}\right) = n + m \) for all \( n, m... | Yes |
Definition 14.30 (Covered structure). A structure is covered if every element of the domain is the value of some closed term. | Example 14.31. Let | No |
Let \( \mathcal{L} \) be the language with constant symbols zero, one, two, \( \ldots \), the binary predicate symbol \( < \), and the binary function symbols + and \( \times \) . Then a structure \( \mathfrak{M} \) for \( \mathcal{L} \) is the one with domain \( \left| \mathfrak{M}\right| = \{ 0,1,2,\ldots \} \) and a... | \[
{\operatorname{Val}}^{\mathfrak{M}}( \times \left( {\text{two,} + \left( \text{three, zero}\right) }\right) =
\]
\[
= { \times }^{\mathfrak{M}}\left( {{\mathrm{{Val}}}^{\mathfrak{M}}\left( \text{two}\right) ,{\mathrm{{Val}}}^{\mathfrak{M}}\left( {+\left( {\text{three},\text{zero}}\right) }\right) }\right)
\]
\[
= ... | Yes |
Example 14.36. Let \( \mathcal{L} = \{ a, b, f, R\} \) where \( a \) and \( b \) are constant symbols, \( f \) is a two-place function symbol, and \( R \) is a two-place predicate symbol. Consider the structure \( \mathfrak{M} \) defined by:\n\n1. \( \left| \mathfrak{M}\right| = \{ 1,2,3,4\} \)\n\n2. \( {a}^{\mathfrak{... | Since \( a \) and \( b \) are constant symbols, \( {\operatorname{Val}}_{s}^{\mathfrak{M}}\left( a\right) = {a}^{\mathfrak{M}} = 1 \) and \( {\operatorname{Val}}_{s}^{\mathfrak{M}}\left( b\right) = {b}^{\mathfrak{M}} = \) 2. So\n\n\[ \n{\operatorname{Val}}_{s}^{\mathfrak{M}}\left( {f\left( {a, b}\right) }\right) = {f}^... | Yes |
Proposition 14.37. If the variables in a term \( t \) are among \( {x}_{1},\ldots ,{x}_{n} \), and \( {s}_{1}\left( {x}_{i}\right) = \) \( {s}_{2}\left( {x}_{i}\right) \) for \( i = 1,\ldots, n \), then \( {\operatorname{Val}}_{{s}_{1}}^{\mathfrak{M}}\left( t\right) = {\operatorname{Val}}_{{s}_{2}}^{\mathfrak{M}}\left(... | Proof. By induction on the complexity of \( t \) . For the base case, \( t \) can be a constant symbol or one of the variables \( {x}_{1},\ldots ,{x}_{n} \) . If \( t = c \), then \( {\operatorname{Val}}_{{s}_{1}}^{\mathfrak{M}}\left( t\right) = {c}^{\mathfrak{M}} = \) \( {\operatorname{Val}}_{{s}_{2}}^{3n}\left( t\rig... | Yes |
Corollary 14.39. If \( \varphi \) is a sentence and \( s \) a variable assignment, then \( \mathfrak{M}, s \vDash \varphi \) iff \( \mathfrak{M},{s}^{\prime } \vDash \varphi \) for every variable assignment \( {s}^{\prime } \) . | Proof. Let \( {s}^{\prime } \) be any variable assignment. Since \( \varphi \) is a sentence, it has no free variables, and so every variable assignment \( {s}^{\prime } \) trivially assigns the same things to all free variables of \( \varphi \) as does \( s \) . So the condition of Proposition 14.38 is satisfied, and ... | Yes |
Proposition 14.41. Let \( \mathfrak{M} \) be a structure, \( \varphi \) be a sentence, and \( s \) a variable assignment. \( \mathfrak{M} \vDash \varphi \) iff \( \mathfrak{M}, s \vDash \varphi \) . | Proof. Exercise. | No |
Proposition 14.42. Suppose \( \varphi \left( x\right) \) only contains \( x \) free, and \( \mathfrak{M} \) is a structure. Then:\n\n1. \( \mathfrak{M} \vDash \exists {x\varphi }\left( x\right) \) iff \( \mathfrak{M}, s \vDash \varphi \left( x\right) \) for at least one variable assignment \( s \) .\n\n2. \( \mathfrak{... | Proof. Exercise. | No |
Proposition 14.43 (Extensionality). Let \( \varphi \) be a formula, and \( {\mathfrak{M}}_{1} \) and \( {\mathfrak{M}}_{2} \) be structures with \( \left| {\mathfrak{M}}_{1}\right| = \left| {\mathfrak{M}}_{2}\right| \), and \( s \) a variable assignment on \( \left| {\mathfrak{M}}_{1}\right| = \left| {\mathfrak{M}}_{2}... | Proof. First prove (by induction on \( t \) ) that for every term, \( {\operatorname{Val}}_{s}^{{\mathfrak{M}}_{1}}\left( t\right) = {\operatorname{Val}}_{s}^{{\mathfrak{M}}_{2}}\left( t\right) \) . Then prove the proposition by induction on \( \varphi \), making use of the claim just proved for the induction basis (wh... | No |
Corollary 14.44 (Extensionality for Sentences). Let \( \varphi \) be a sentence and \( {\mathfrak{M}}_{1},{\mathfrak{M}}_{2} \) as in Proposition 14.43. Then \( {\mathfrak{M}}_{1} \vDash \varphi \) iff \( {\mathfrak{M}}_{2} \vDash \varphi \) . | Proof. Follows from Proposition 14.43 by Corollary 14.39. | No |
Proposition 14.45. Let \( \mathfrak{M} \) be a structure, \( t \) and \( {t}^{\prime } \) terms, and \( s \) a variable assignment. Let \( {s}^{\prime }{ \sim }_{x}s \) be the \( x \) -variant of \( s \) given by \( {s}^{\prime }\left( x\right) = {\operatorname{Val}}_{s}^{\mathfrak{M}}\left( {t}^{\prime }\right) \) . T... | Proof. By induction on \( t \) . \n\n1. If \( t \) is a constant, say, \( t \equiv c \), then \( t\left\lbrack {{t}^{\prime }/x}\right\rbrack = c \), and \( {\operatorname{Val}}_{s}^{\mathfrak{M}}\left( c\right) = {c}^{\mathfrak{M}} = \) \( {\operatorname{Val}}_{{s}^{\prime }}^{\mathfrak{M}}\left( c\right) \). \n\n2. I... | Yes |
Proposition 14.46. Let \( \mathfrak{M} \) be a structure, \( \varphi \) a formula, \( t \) a term, and \( s \) a variable assignment. Let \( {s}^{\prime }{ \sim }_{x}s \) be the \( x \) -variant of \( s \) given by \( {s}^{\prime }\left( x\right) = {\operatorname{Val}}_{s}^{3\Re }\left( t\right) \) . Then \( \mathfrak{... | Proof. Exercise. | No |
Proposition 14.50. A sentence \( \varphi \) is valid iff \( \Gamma \vDash \varphi \) for every set of sentences \( \Gamma \) . | Proof. For the forward direction, let \( \varphi \) be valid, and let \( \Gamma \) be a set of sentences. Let \( \mathfrak{M} \) be a structure so that \( \mathfrak{M} \vDash \Gamma \) . Since \( \varphi \) is valid, \( \mathfrak{M} \vDash \varphi \), hence \( \Gamma \vDash \varphi \) .\n\nFor the contrapositive of the... | Yes |
Proposition 14.51. \( \Gamma \vDash \varphi \) iff \( \Gamma \cup \{ \neg \varphi \} \) is unsatisfiable. | Proof. For the forward direction, suppose \( \Gamma \vDash \varphi \) and suppose to the contrary that there is a structure \( \mathfrak{M} \) so that \( \mathfrak{M} \vDash \Gamma \cup \{ \neg \varphi \} \) . Since \( \mathfrak{M} \vDash \Gamma \) and \( \Gamma \vDash \varphi \) , \( \mathfrak{M} \vDash \varphi \) . A... | Yes |
Proposition 14.52. If \( \Gamma \subseteq {\Gamma }^{\prime } \) and \( \Gamma \vDash \varphi \), then \( {\Gamma }^{\prime } \vDash \varphi \) . | Proof. Suppose that \( \Gamma \subseteq {\Gamma }^{\prime } \) and \( \Gamma \vDash \varphi \) . Let \( \mathfrak{M} \) be such that \( \mathfrak{M} \vDash {\Gamma }^{\prime } \) ; then \( \mathfrak{M} \vDash \Gamma \), and since \( \Gamma \vDash \varphi \), we get that \( \mathfrak{M} \vDash \varphi \) . Hence, whenev... | Yes |
Theorem 14.53 (Semantic Deduction Theorem). \( \Gamma \cup \{ \varphi \} \vDash \psi \) iff \( \Gamma \vDash \varphi \rightarrow \psi \) . | Proof. For the forward direction, let \( \Gamma \cup \{ \varphi \} \vDash \psi \) and let \( \mathfrak{M} \) be a structure so that \( \mathfrak{M} \vDash \Gamma \) . If \( \mathfrak{M} \vDash \varphi \), then \( \mathfrak{M} \vDash \Gamma \cup \{ \varphi \} \), so since \( \Gamma \cup \{ \varphi \} \) entails \( \psi ... | Yes |
Proposition 14.54. Let \( \mathfrak{M} \) be a structure, and \( \varphi \left( x\right) \) a formula with one free variable \( x \), and \( t \) a closed term. Then:\n\n1. \( \varphi \left( t\right) \vDash \exists {x\varphi }\left( x\right) \)\n\n2. \( \forall {x\varphi }\left( x\right) \vDash \varphi \left( t\right) ... | Proof. 1. Suppose \( \mathfrak{M} \vDash \varphi \left( t\right) \) . Let \( s \) be a variable assignment with \( s\left( x\right) = \) \( {\operatorname{Val}}^{\mathfrak{M}}\left( t\right) \) . Then \( \mathfrak{M}, s \vDash \varphi \left( t\right) \) since \( \varphi \left( t\right) \) is a sentence. By Proposition ... | No |
The theory of strict linear orders in the language \( {\mathcal{L}}_{ < } \) is axiomatized by the set\n\n\[ \forall x\neg x < x \]\n\n\[ \forall x\forall y\left( {\left( {x < y \vee y < x}\right) \vee x = y}\right) ,\]\n\n\[ \forall x\forall y\forall z\left( {\left( {x < y \land y < z}\right) \rightarrow x < z}\right)... | It completely captures the intended structures: every strict linear order is a model of this axiom system, and vice versa, if \( R \) is a linear order on a set \( X \) , then the structure \( \mathfrak{M} \) with \( \left| \mathfrak{M}\right| = X \) and \( { < }^{\mathfrak{M}} = R \) is a model of this theory. | Yes |
The theory of Peano arithmetic is axiomatized by the following sentences in the language of arithmetic \( {\mathcal{L}}_{A} \) . | \[ \forall x\forall y\left( {{x}^{\prime } = {y}^{\prime } \rightarrow x = y}\right) \] \[ \forall x\mathrm{o} \neq {x}^{\prime } \] \[ \forall x\left( {x + 0}\right) = x \] \[ \forall x\forall y\left( {x + {y}^{\prime }}\right) = {\left( x + y\right) }^{\prime } \] \[ \forall x\left( {x \times 0}\right) = 0 \] \[ \for... | Yes |
Show that the comprehension principle is inconsistent by giving a derivation that shows\n\n\[ \exists y\forall x\left( {x \in y \leftrightarrow x \notin x}\right) \vdash \bot . \] | It may help to first show \( \left( {A \rightarrow \neg A}\right) \land \left( {\neg A \rightarrow A}\right) \vdash \bot \) . | No |
Every initial sequent, e.g., \( \chi \Rightarrow \chi \) is a derivation. | We can obtain a new derivation from this by applying, say, the WL rule,\n\n\[ \frac{\Gamma \Rightarrow \Delta }{\varphi ,\Gamma \Rightarrow \Delta }\mathrm{{WL}} \]\n\nThe rule, however, is meant to be general: we can replace the \( \varphi \) in the rule with any sentence, e.g., also with \( \theta \) . If the premise... | Yes |
Example 17.5. Give an LK-derivation for the sequent \( \varphi \land \psi \Rightarrow \varphi \) . | We begin by writing the desired end-sequent at the bottom of the derivation.\n\n\[ \varphi \land \psi \Rightarrow \varphi \]\n\nNext, we need to figure out what kind of inference could have a lower sequent of this form. This could be a structural rule, but it is a good idea to start by looking for a logical rule. The o... | Yes |
Suppose we want to prove \( \Rightarrow \varphi \vee \neg \varphi \) . | Applying VR backwards would give us one of these two derivations:\n\n\[ \n\begin{array}{l} \Rightarrow \varphi \\ \Rightarrow \varphi \vee \neg \varphi \vee \mathrm{R} \\ \end{array} \]\n\nNeither of these of course ends in an initial sequent. The trick is to realize that the contraction rule allows us to combine two c... | Yes |
Give an LK-derivation of the sequent \( \exists x\neg \varphi \left( x\right) \Rightarrow \neg \forall {x\varphi }\left( x\right) \) . | Starting as usual, we write\n\n\[ \exists x\neg \varphi \left( x\right) \Rightarrow \neg \forall {x\varphi }\left( x\right) \]\n\nWe could either carry out the \( \exists \mathrm{L} \) rule or the \( \neg \mathrm{R} \) rule. Since the \( \exists \mathrm{L} \) rule is subject to the eigenvariable condition, it's a good ... | Yes |
Proposition 17.13 (Reflexivity). If \( \varphi \in \Gamma \), then \( \Gamma \vdash \varphi \) . | Proof. The initial sequent \( \varphi \Rightarrow \varphi \) is derivable, and \( \{ \varphi \} \subseteq \Gamma \) . | Yes |
Proposition 17.14 (Monotony). If \( \Gamma \subseteq \Delta \) and \( \Gamma \vdash \varphi \), then \( \Delta \vdash \varphi \) . | Proof. Suppose \( \Gamma \vdash \varphi \), i.e., there is a finite \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \Rightarrow \varphi \) is derivable. Since \( \Gamma \subseteq \Delta \), then \( {\Gamma }_{0} \) is also a finite subset of \( \Delta \) . The derivation of \( {\Gamma }_{0} \Rightarrow ... | Yes |
Proposition 17.15 (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 finite \( {\Gamma }_{0} \subseteq \Gamma \) and a derivation \( {\pi }_{0} \) of \( {\Gamma }_{0} \Rightarrow \varphi \) . If \( \{ \varphi \} \cup \Delta \vdash \psi \), then for some finite subset \( {\Delta }_{0} \subseteq \Delta \), there is a derivation \( {\pi }_{... | Yes |
Proposition 17.16. \( \Gamma \) is inconsistent iff \( \Gamma \vdash \varphi \) for every sentence \( \varphi \) . | Proof. Exercise. | No |
Proposition 17.17 (Compactness). 1. If \( \Gamma \vdash \varphi \) then there is a finite subset \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \vdash \varphi \) . 2. If every finite subset of \( \Gamma \) is consistent, then \( \Gamma \) is consistent. | Proof. 1. If \( \Gamma \vdash \varphi \), then there is a finite subset \( {\Gamma }_{0} \subseteq \Gamma \) such that the sequent \( {\Gamma }_{0} \Rightarrow \varphi \) has a derivation. Consequently, \( {\Gamma }_{0} \vdash \varphi \) . 2. If \( \Gamma \) is inconsistent, there is a finite subset \( {\Gamma }_{0} \s... | Yes |
Proposition 17.18. If \( \Gamma \vdash \varphi \) and \( \Gamma \cup \{ \varphi \} \) is inconsistent, then \( \Gamma \) is inconsistent. | Proof. There are finite \( {\Gamma }_{0} \) and \( {\Gamma }_{1} \subseteq \Gamma \) such that \( \mathbf{{LK}} \) derives \( {\Gamma }_{0} \Rightarrow \varphi \) and \( \varphi ,{\Gamma }_{1} \Rightarrow \) . Let the LK-derivation of \( {\Gamma }_{0} \Rightarrow \varphi \) be \( {\pi }_{0} \) and the LK-derivation of ... | Yes |
Proposition 17.19. \( \Gamma \vdash \varphi \) iff \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent. | Proof. First suppose \( \Gamma \vdash \varphi \), i.e., there is a derivation \( {\pi }_{0} \) of \( \Gamma \Rightarrow \varphi \) . By adding a \( \neg \mathrm{L} \) rule, we obtain a derivation of \( \neg \varphi ,\Gamma \Rightarrow \), i.e., \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent.\n\nIf \( \Gamma \cup ... | Yes |
Proposition 17.20. 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 is a derivation \( \pi \) of a sequent \( {\Gamma }_{0} \Rightarrow \varphi \) . The sequent \( \neg \varphi ,{\Gamma }_{0} \Rightarrow \) is also derivable:\n\n\[ \begin{array}{l} {\Gamma }_{0} \Rightarrow \varphi \;\frac{\varphi... | Yes |
Proposition 17.21. If \( \Gamma \cup \{ \varphi \} \) and \( \Gamma \cup \{ \neg \varphi \} \) are both inconsistent, then \( \Gamma \) is inconsistent. | Proof. There are finite sets \( {\Gamma }_{0} \subseteq \Gamma \) and \( {\Gamma }_{1} \subseteq \Gamma \) and LK-derivations \( {\pi }_{0} \) and \( {\pi }_{1} \) of \( \varphi ,{\Gamma }_{0} \Rightarrow \) and \( \neg \varphi ,{\Gamma }_{1} \Rightarrow \), respectively. We can then derive ![c5962287-92b4-4003-ac67-b5... | Yes |
Proposition 17.22. 1. Both \( \varphi \land \psi \vdash \varphi \) and \( \varphi \land \psi \vdash \psi \) .\n\n2. \( \varphi ,\psi \vdash \varphi \land \psi \) . | Proof. 1. Both sequents \( \varphi \land \psi \Rightarrow \varphi \) and \( \varphi \land \psi \Rightarrow \psi \) are derivable:\n\n\[ \frac{\varphi \Rightarrow \varphi }{\varphi \land \psi \Rightarrow \varphi } \land \mathrm{L}\;\frac{\psi \Rightarrow \psi }{\varphi \land \psi \Rightarrow \psi } \land \mathrm{L} \]\n... | Yes |
Proposition 17.23. 1. \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \) is inconsistent. | Proof. 1. We give a derivation of the sequent \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \Rightarrow \) :\n\n\[ \n\frac{\varphi \;\Rightarrow \;\varphi }{{\neg \varphi },\varphi \;\Rightarrow \;}\neg \text{L}\;\frac{\psi \;\Rightarrow \;\psi }{{\neg \psi },\psi \;\Rightarrow \;}\neg \text{L} \]\n\n\[ \n\frac{\varphi... | Yes |
Proposition 17.24. 1. \( \varphi ,\varphi \rightarrow \psi \vdash \psi \) . 2. Both \( \neg \varphi \vdash \varphi \rightarrow \psi \) and \( \psi \vdash \varphi \rightarrow \psi \) . | Proof. 1. The sequent \( \varphi \rightarrow \psi ,\varphi \Rightarrow \psi \) is derivable: \[ \frac{\varphi \Rightarrow \varphi \;\psi \Rightarrow \psi }{\varphi \rightarrow \psi ,\varphi \Rightarrow \psi } \rightarrow \mathrm{L} \] 2. Both sequents \( \neg \varphi \Rightarrow \varphi \rightarrow \psi \) and \( \psi ... | Yes |
Theorem 17.25. 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 \( {\pi }_{0} \) be an LK-derivation of \( {\Gamma }_{0} \Rightarrow \varphi \left( c\right) \) for some finite \( {\Gamma }_{0} \subseteq \Gamma \) . By adding a \( \forall \mathrm{R} \) inference, we obtain a proof of \( {\Gamma }_{0} \Rightarrow \forall {x\varphi }\left( x\right) \), since \( c \) does no... | Yes |
Proposition 17.26. 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 sequent \( \varphi \left( t\right) \Rightarrow \exists {x\varphi }\left( x\right) \) is derivable: \[ \frac{\varphi \left( t\right) \Rightarrow \varphi \left( t\right) }{\varphi \left( t\right) \Rightarrow \exists {x\varphi }\left( x\right) }\exists \mathrm{R} \] 2. The sequent \( \forall {x\varphi }\left... | Yes |
Corollary 17.30. If \( \Gamma \vdash \varphi \) then \( \Gamma \vDash \varphi \) . | Proof. If \( \Gamma \vdash \varphi \) then for some finite subset \( {\Gamma }_{0} \subseteq \Gamma \), there is a derivation of \( {\Gamma }_{0} \Rightarrow \varphi \) . By Theorem 17.28, every structure \( \mathfrak{M} \) either makes some \( \psi \in {\Gamma }_{0} \) false or makes \( \varphi \) true. Hence, if \( \... | Yes |
Corollary 17.31. If \( \Gamma \) is satisfiable, then it is consistent. | Proof. We prove the contrapositive. Suppose that \( \Gamma \) is not consistent. Then there is a finite \( {\Gamma }_{0} \subseteq \Gamma \) and a derivation of \( {\Gamma }_{0} \Rightarrow \) . By Theorem 17.28, \( {\Gamma }_{0} \Rightarrow \) is valid. In other words, for every structure \( \mathfrak{M} \), there is ... | Yes |
If \( s \) and \( t \) are closed terms, then \( s = t,\varphi \left( s\right) \vdash \varphi \left( t\right) \) : | \n\[
\begin{aligned} \varphi \left( s\right) & \Rightarrow \varphi \left( s\right) \\ s = t,\varphi \left( s\right) & \Rightarrow \varphi \left( s\right) \\ s = t,\varphi \left( s\right) & \Rightarrow \varphi \left( t\right) \end{aligned}\text{ WL }
\] | Yes |
Proposition 17.34. LK with initial sequents and rules for identity is sound. | Proof. Initial sequents of the form \( \Rightarrow t = t \) are valid, since for every structure \( \mathfrak{M},\mathfrak{M} \vDash t = t \) . (Note that we assume the term \( t \) to be closed, i.e., it contains no variables, so variable assignments are irrelevant).\n\nSuppose the last inference in a derivation is \(... | Yes |
Every assumption on its own is a derivation. So, e.g., \( \chi \) by itself is a derivation, and so is \( \theta \) by itself. We can obtain a new derivation from these by applying, say, the \( \land \) Intro rule, | \[ \frac{\varphi }{\varphi \land \psi } \land \text{ Intro } \] These rules are meant to be general: we can replace the \( \varphi \) and \( \psi \) in it with any sentences, e.g., by \( \chi \) and \( \theta \) . Then the conclusion would be \( \chi \land \theta \), and so \[ \frac{\chi \;\theta }{\chi \land \theta }\... | Yes |
Let’s give a derivation of the sentence \( \left( {\varphi \land \psi }\right) \rightarrow \varphi \) . | \[ \frac{\frac{{\left\lbrack \varphi \land \psi \right\rbrack }^{1}}{\varphi } \land \text{Elim}}{1\frac{(\varphi \land \psi ) \rightarrow \varphi }{(\varphi \land \psi ) \rightarrow \varphi } \rightarrow \text{Intro}} \] | Yes |
For instance, suppose we want to derive \( \varphi \vee \neg \varphi \) . Our usual strategy would be to attempt to derive \( \varphi \vee \neg \varphi \) using VIntro. But this would require us to derive either \( \varphi \) or \( \neg \varphi \) from no assumptions, and this can’t be done. \( { \bot }_{C} \) to the r... | Now we’re looking for a derivation of \( \bot \) from \( \neg \left( {\varphi \vee \neg \varphi }\right) \) . Since \( \bot \) is the conclusion of \( \neg \) Elim we might try that:  Our strategy for finding a deriv... | Yes |
Let’s see how we’d give a derivation of the formula \( \exists x\neg \varphi \left( x\right) \rightarrow \neg \forall {x\varphi }\left( x\right) \) . | \[ \exists x\neg \varphi \left( x\right) \rightarrow \neg \forall {x\varphi }\left( x\right) \] We start by writing down what it would take to justify that last step using the \( \rightarrow \) Intro rule. \[ \frac{{\left\lbrack \exists x\neg \varphi \left( x\right) \right\rbrack }^{1}\;\neg \forall {x\varphi }\left( x... | Yes |
Let’s see how we’d give a derivation of the formula \( \exists {x\chi }\left( {x, b}\right) \) from the assumptions \( \exists x\left( {\varphi \left( x\right) \land \psi \left( x\right) }\right) \) and \( \forall x\left( {\psi \left( x\right) \rightarrow \chi \left( {x, b}\right) }\right) \). | \[ \exists {x\chi }\left( {x, b}\right) \] \n\nWe have two premises to work with. To use the first, i.e., try to find a derivation of \( \exists {x\chi }\left( {x, b}\right) \) from \( \exists x\left( {\varphi \left( x\right) \land \psi \left( x\right) }\right) \) we would use the \( \exists \) Elim rule. Since it has ... | Yes |
Give a derivation of the formula \( \neg \forall {x\varphi }\left( x\right) \) from the assumptions \( \forall {x\varphi }\left( x\right) \rightarrow \exists {y\psi }\left( y\right) \) and \( \neg \exists {y\psi }\left( y\right) \). | \[ \frac{\neg \exists {y\psi }\left( y\right) \;\frac{\forall {x\varphi }\left( x\right) \rightarrow \exists {y\psi }\left( y\right) \;{\left\lbrack \forall x\varphi \left( x\right) \right\rbrack }^{1}}{\exists {y\psi }\left( y\right) } \rightarrow \text{ Elim }}{\frac{1}{\neg \forall {x\varphi }\left( x\right) }\neg \... | Yes |
Proposition 18.13 (Reflexivity). If \( \varphi \in \Gamma \), then \( \Gamma \vdash \varphi \) . | Proof. The assumption \( \varphi \) by itself is a derivation of \( \varphi \) where every undischarged assumption (i.e., \( \varphi \) ) is in \( \Gamma \) . | Yes |
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