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Proposition 42.15. Every general recursive function is \( \lambda \) -definable. | Proof. By Lemma 42.8, all basic functions are \( \lambda \) -definable, and by Lemma 42.9, Lemma 42.10, and Lemma 42.14, the \( \lambda \) -definable functions are closed under composition, primitive recursion, and regular minimization. | Yes |
Theorem 42.17. If a partial function \( f \) is \( \lambda \) -definable, it is partial recursive. | Proof. We only sketch the proof. First, we arithmetize \( \lambda \) -terms, i.e., system-atially assign Gödel numbers to \( \lambda \) -terms as using the usual power-of-primes coding of sequences. Then we define a partial recursive function normalize \( \left( t\right) \) operating on the Gödel number \( t \) of a la... | No |
Proposition 43.11. 1. \( \varphi \) is a tautology if and only if \( \varnothing \vDash \varphi \) ; | Proof. Exercise. | No |
Proposition 43.12. Suppose that a many-valued logic \( \mathbf{L} \) contains the connectives \( \neg \) , \( \land , \vee , \rightarrow \) in its language, \( \mathbb{T},\mathbb{F} \in V \), and its truth functions satisfy:\n\n1. \( {\widetilde{\neg }}_{\mathbf{L}}\left( x\right) = {\widetilde{\neg }}_{\mathbf{C}}\lef... | Proof. By induction on \( \varphi \) .\n\n1. If \( \varphi \equiv p \) is atomic, we have \( {\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \varphi \right) = \mathfrak{v}\left( p\right) = {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \varphi \right) \) .\n\n2. If \( \varphi \equiv \neg B \), we have\n\n\[ {\overline{\math... | Yes |
Corollary 43.13. If a many-valued logic satisfies the conditions of Proposition 43.12, \( \mathbb{T} \in {V}^{ + } \) and \( \mathbb{F} \notin {V}^{ + } \), then \( { \vDash }_{\mathbf{L}} \subseteq { \vDash }_{\mathbf{C}} \), i.e., if \( \Gamma { \vDash }_{\mathbf{L}}\psi \) then \( \Gamma { \vDash }_{\mathbf{C}}\psi ... | Proof. We prove the contrapositive. Suppose \( \Gamma { \mathrel{\text{\vDash \not{} }} }_{\mathrm{C}}\psi \) . Then there is some valuation \( \mathfrak{v} : {\mathrm{{At}}}_{0} \rightarrow \{ \mathbb{T},\mathbb{F}\} \) such that \( {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \varphi \right) = \mathbb{T} \) for all \... | Yes |
Proposition 44.2. If \( \mathfrak{v}\left( p\right) \in \{ \mathbb{T},\mathbb{F}\} \) for all \( p \) in \( \varphi \), then \( {\overline{\mathfrak{v}}}_{{Ł}_{3}}\left( \varphi \right) = {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \varphi \right) \) . | Many classical tautologies are also tautologies in \( {\mathbf{L}}_{3} \), e.g, \( \neg p \rightarrow \left( {p \rightarrow q}\right) \) . Just like in classical logic, we can use truth tables to verify this:\n\n | No |
Proposition 44.5. Ks and Kw have no tautologies. | Proof. If \( \mathfrak{v}\left( p\right) = \mathbb{U} \) for all propositional variables \( p \), then any formula \( \varphi \) will have truth value \( \overline{\mathfrak{v}}\left( \varphi \right) = \mathbb{U} \), since\n\n\[ \widetilde{\neg }\left( \mathbb{U}\right) = \widetilde{ \vee }\left( {\mathbb{U},\mathbb{U}... | Yes |
Proposition 44.11. The matrix with \( V = \{ \mathbb{F},\mathbb{U},\mathbb{T}\} ,{V}^{ + } = \{ \mathbb{T},\mathbb{U}\} \), and the truth functions of 3-valued Gödel logic defines classical logic. | Proof. Exercise. | No |
Problem 44.6. Which of the following relations hold in (a) strong and (b) weak Kleene logic? Give a truth table for each. | \[ \text{1.}p, p \rightarrow q \vDash q \] | No |
Problem 44.10. Which of the following relations hold in Gödel logic? Give a truth table for each. | 1. \( p, p \rightarrow q \vDash q \)\n2. \( p \vee q,\neg p \vDash q \)\n3. \( p \land q \vDash p \)\n4. \( p \vDash p \land p \)\n5. \( p \vDash p \vee q \) | No |
Problem 44.13. Which of the following relations hold in (a) LP and in (b) Hal? Give a truth table for each. | 1. \( p, p \rightarrow q \vDash q \)\n\n2. \( \neg q, p \rightarrow q \vDash \neg p \)\n\n3. \( p \vee q,\neg p \vDash q \)\n\n4. \( \neg p, p \vDash q \)\n\n5. \( p \vDash p \vee q \)\n\n6. \( p \rightarrow q, q \rightarrow r \vDash p \rightarrow r \) | No |
Proposition 45.2. The logic \( {L}_{3} \) defined by Definition 44.1 is the same as \( {L}_{3} \) defined by Definition 45.1. | Proof. This can be seen by comparing the truth tables for the connectives given in Definition 44.1 with the truth tables determined by the equations in Definition 45.1:   | Yes |
Proposition 45.6. If \( \Gamma { \vDash }_{{\mathbf{G}}_{\infty }}\psi \) then \( \Gamma { \vDash }_{{\mathbf{G}}_{m}}\psi \) for all \( m \geq 2 \) . | Proof. Exercise. | No |
Suppose \( \varphi \) is \( {p}_{1} \rightarrow ▱\left( {{p}_{1} \land {p}_{2}}\right) ,{\theta }_{1} \) is \( \diamond \left( {{p}_{2} \rightarrow {p}_{3}}\right) \) and \( {\theta }_{2} \) is \( \neg ▱{p}_{1} \) . Then \( \varphi \left\lbrack {{\theta }_{1}/{p}_{1},{\theta }_{2}/{p}_{2}}\right\rbrack \) is | \[ \diamond \left( {{p}_{2} \rightarrow {p}_{3}}\right) \rightarrow ▱\left( {\diamond \left( {{p}_{2} \rightarrow {p}_{3}}\right) \land \neg ▱{p}_{1}}\right) \] | Yes |
Proposition 47.8. 1. \( \mathfrak{M}, w \Vdash ▱\varphi \) iff \( \mathfrak{M}, w \Vdash \neg \Diamond \neg \varphi \) . | Proof. 1. \( \mathfrak{M}, w \Vdash \neg \diamond \neg \varphi \) iff \( \mathfrak{M} \nVdash \diamond \neg \varphi \) by definition of \( \mathfrak{M}, w \Vdash .\mathfrak{M}, w \Vdash \diamond \neg \varphi \) iff for some \( {w}^{\prime } \) with \( {Rw}{w}^{\prime },\mathfrak{M},{w}^{\prime } \Vdash \neg \varphi \) ... | Yes |
1. If \( \mathfrak{M} \Vdash \varphi \) then \( \mathfrak{M} \nVdash \neg \varphi \), but not vice-versa. | 1. If \( \mathfrak{M} \Vdash \varphi \) then \( \varphi \) is true at all worlds in \( W \), and since \( W \neq \varnothing \), it can’t be that \( \mathfrak{M} \Vdash \neg \varphi \), or else \( \varphi \) would have to be both true and false at some world.\n\nOn the other hand, if \( \mathfrak{M} \nVdash \neg \varph... | Yes |
Proposition 47.13. If \( \varphi \) is valid, then so is \( ▱\varphi \) . | Proof. Assume \( \vDash \varphi \) . To show \( \vDash ▱\varphi \) let \( \mathfrak{M} = \langle W, R, V\rangle \) be a model and \( w \in W \) . If \( {Rw}{w}^{\prime } \) then \( \mathfrak{M},{w}^{\prime } \Vdash \varphi \), since \( \varphi \) is valid, and so also \( \mathfrak{M}, w \Vdash ▱\varphi \) . Since \( \m... | Yes |
Proposition 47.19. The following schema \( \mathrm{K} \) is valid\n\n\[ \n▱\left( {\varphi \rightarrow \psi }\right) \rightarrow \left( {▱\varphi \rightarrow ▱\psi }\right)\n\]\n\n(K) | Proof. We need to show that all instances of the schema are true at every world in every model. So let \( \mathfrak{M} = \langle W, R, V\rangle \) and \( w \in W \) be arbitrary. To show that a conditional is true at a world we assume the antecedent is true to show that consequent is true as well. In this case, let \( ... | Yes |
Proposition 47.20. The following schema DUAL is valid\n\n\[ \diamond \varphi \leftrightarrow \neg ▱\neg \varphi \] | Proof. Exercise. | No |
Proposition 47.22. A formula \( \varphi \) is valid iff all its substitution instances are. In other words, a schema is valid iff its characteristic formula is. | Proof. The \ | No |
To show \( p \rightarrow \diamond p \vDash ▱\neg p \rightarrow \neg p \) | Consider a model \( \mathfrak{M} = \langle W, R, V\rangle \) and \( w \in W \), and suppose \( \mathfrak{M}, w \Vdash p \rightarrow \diamond p \) . We have to show that \( \mathfrak{M}, w \Vdash \) \( ▱\neg p \rightarrow \neg p \) . Suppose not. Then \( \mathfrak{M}, w \Vdash ▱\neg p \) and \( \mathfrak{M}, w \nVdash \... | Yes |
Theorem 48.1. Let \( \mathfrak{M} = \langle W, R, V\rangle \) be a model. If \( R \) has the property on the left side of table 48.1, every instance of the formula on the right side is true in \( \mathfrak{M} \) . | Proof. Here is the case for B: to show that the schema is true in a model we need to show that all of its instances are true at all worlds in the model. So let \( \varphi \rightarrow ▱\diamond \varphi \) be a given instance of B, and let \( w \in W \) be an arbitrary world. Suppose the antecedent \( \varphi \) is true ... | Yes |
Proposition 48.2. Let \( \mathfrak{M} = \langle W, R, V\rangle \) be a model such that \( W = \{ u, v\} \), where worlds \( u \) and \( v \) are related by \( R \) : i.e., both Ruv and Rvu. Suppose that for all \( p \) : \( u \in V\left( p\right) \Leftrightarrow v \in V\left( p\right) \) . Then:\n\n1. For all \( \varph... | Since \( \mathfrak{M} \) is not reflexive (it is, in fact, irreflexive), the converse of Theorem 48.1 fails in the case of \( \mathrm{T} \) (similar arguments can be given for some-though not all-the other schemas mentioned in Theorem 48.1). | No |
Theorem 48.6. If the formula on the right side of table 48.1 is valid in a frame \( \mathfrak{F} \) , then \( \mathfrak{F} \) has the property on the left side. | Proof. 1. Suppose \( \mathrm{D} \) is valid in \( \mathfrak{F} = \langle W, R\rangle \), i.e., \( \mathfrak{F} \vDash ▱p \rightarrow \diamond p \) . Let \( \mathfrak{M} = \) \( \langle W, R, V\rangle \) be a model based on \( \mathfrak{F} \), and \( w \in W \) . We have to show that there is a \( v \) such that \( {Rwv... | Yes |
Corollary 48.8. Each formula on the right side of table 48.1 defines the class of frames which have the property on the left side. | Proof. In Theorem 48.1, we proved that if a model has the property on the left, the formula on the right is true in it. Thus, if a frame \( \mathfrak{F} \) has the property on the left, the formula on the right is valid in \( \mathfrak{F} \). In Theorem 48.6, we proved the converse implications: if a formula on the rig... | Yes |
Proposition 48.12. The following are equivalent:\n\n1. \( R \) is an equivalence relation;\n\n2. \( R \) is reflexive and euclidean;\n\n3. \( R \) is serial, symmetric, and euclidean;\n\n4. \( R \) is serial, symmetric, and transitive. | Proof. Exercise. | No |
Proposition 48.14. A formula \( \varphi \) is valid in all frames \( \mathfrak{F} = \langle W, R\rangle \) where \( R \) is an equivalence relation, if and only if it is valid in all frames \( \mathfrak{F} = \langle W, R\rangle \) where \( R \) is universal. Hence, the logic of universal frames is just \( \mathbf{{S5}}... | Proof. It’s immediate to verify that a universal relation \( R \) on \( W \) is an equivalence. Hence, if \( \varphi \) is valid in all frames where \( R \) is an equivalence it is valid in all universal frames. For the other direction, we argue contrapositively: suppose \( \psi \) is a formula that fails at a world \(... | Yes |
Proposition 48.16. Let \( \mathfrak{M} = \langle W, R, V\rangle ,{\mathfrak{M}}^{\prime } \) be the first-order structure with \( \left| {\mathfrak{M}}^{\prime }\right| = \) \( W,{Q}^{{\mathfrak{M}}^{\prime }} = R \), and \( {P}_{i}^{{\mathfrak{M}}^{\prime }} = V\left( {p}_{i}\right) \), and \( s\left( x\right) = w \) ... | Proof. By induction on \( \varphi \) . | No |
Proposition 48.17. Suppose \( \varphi \) is a modal formula and \( \mathfrak{F} = \langle W, R\rangle \) is a frame. Let \( {\mathfrak{F}}^{\prime } \) be the first-order structure with \( \left| {\mathfrak{F}}^{\prime }\right| = W \) and \( {Q}^{{\mathfrak{F}}^{\prime }} = R \), and let \( {\varphi }^{\prime } \) be t... | Proof. \( {\mathfrak{F}}^{\prime } \vDash {\varphi }^{\prime } \) iff for every structure \( {\mathfrak{M}}^{\prime } \) where \( {P}_{i}^{{\mathfrak{M}}^{\prime }} \subseteq W \) for \( i = 1,\ldots, n \), and for every \( s \) with \( s\left( x\right) \in W,{\mathfrak{M}}^{\prime }, s \vDash {\mathrm{{ST}}}_{x}\left(... | Yes |
Corollary 48.19. If a class of frames is definable by a formula \( \varphi \), the corresponding class of accessibility relations is definable by a monadic second-order sentence. | Proof. The monadic second-order sentence \( {\varphi }^{\prime } \) of the preceding proof has the required property. | No |
Problem 48.4. Show that if the formula on the right side of table 48.2 is valid in a frame \( \mathfrak{F} \), then \( \mathfrak{F} \) has the property on the left side. | To do this, consider a frame that does not satisfy the property on the left, and define a suitable \( V \) such that the formula on the right is false at some world. | No |
Proposition 49.6. Every normal modal logic is closed under rule RK, | Proof. By induction on \( n \) : If \( n = 1 \), then the rule is just NEC, and every normal modal logic is closed under NEC.\n\nNow suppose the result holds for \( n - 1 \) ; we show it holds for \( n \) .\n\nAssume\n\n\[{\varphi }_{1} \rightarrow \left( {{\varphi }_{2} \rightarrow \cdots \left( {{\varphi }_{n - 1} \r... | Yes |
Proposition 49.8. Let \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) be formulas. Then there is a smallest modal logic \( \sum \) containing all instances of \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) . | Proof. Given \( {\varphi }_{1},\ldots ,{\varphi }_{n} \), define \( \sum \) as the intersection of all normal modal logics containing all instances of \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) . The intersection is non-empty as \( \operatorname{Frm}\left( \mathcal{L}\right) \), the set of all formulas, is such a moda... | Yes |
Proposition 49.12. \( \mathbf{K} \vdash ▱\varphi \rightarrow ▱\left( {\psi \rightarrow \varphi }\right) \) | Proof.\n\n1. \( \varphi \rightarrow \left( {\psi \rightarrow \varphi }\right) \; \) TAUT\n\n2. \( ▱\left( {\varphi \rightarrow \left( {\psi \rightarrow \varphi }\right) }\right) \; \) NEC,1\n\n3. \( ▱\left( {\varphi \rightarrow \left( {\psi \rightarrow \varphi }\right) }\right) \rightarrow \left( {▱\varphi \rightarrow ... | Yes |
Proposition 49.13. \( \mathbf{K} \vdash ▱\left( {\varphi \land \psi }\right) \rightarrow \left( {▱\varphi \land ▱\psi }\right) \) | Proof.\n\n1. \( \left( {\varphi \land \psi }\right) \rightarrow \varphi \; \) TAUT\n\n2. \( ▱\left( {\left( {\varphi \land \psi }\right) \rightarrow \varphi }\right) \; \) NEC\n\n3. \( ▱\left( {\left( {\varphi \land \psi }\right) \rightarrow \varphi }\right) \rightarrow \left( {▱\left( {\varphi \land \psi }\right) \rig... | Yes |
Proposition 49.14. \( \mathbf{K} \vdash \left( {▱\varphi \land ▱\psi }\right) \rightarrow ▱\left( {\varphi \land \psi }\right) \) | Proof.\n\n1. \( \varphi \rightarrow \left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) \)\n\n2. \( ▱\left( {\varphi \rightarrow \left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) }\right) \)\n\nNEC, 1\n\n3. \( ▱\left( {\varphi \rightarrow \left( {\psi \rightarrow \left( {\varphi \l... | Yes |
Proposition 49.15. \( \mathbf{K} \vdash \neg ▱p \rightarrow \Diamond \neg p \) | Proof.\n\n1. \( \Diamond \neg p \leftrightarrow \neg ▱\neg \neg p \) DUAL\n\n2. \( \left( {\Diamond \neg p \leftrightarrow \neg ▱\neg \neg p}\right) \rightarrow \)\n\n\( \left( {\neg ▱\neg \neg p \rightarrow \Diamond \neg p}\right) \; \) TAUT\n\n3. \( \neg ▱\neg p \rightarrow \diamond \neg p\;\mathrm{{MP}},1,2 \)\n\n4.... | Yes |
Proposition 49.16. If \( \mathbf{K} \vdash {\varphi }_{1},\ldots ,\mathbf{K} \vdash {\varphi }_{n} \), and \( \psi \) follows from \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) by propositional logic, then \( \mathbf{K} \vdash \psi \) . | Proof. If \( \psi \) follows from \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) by propositional logic, then\n\n\[{\varphi }_{1} \rightarrow \left( {{\varphi }_{2} \rightarrow \cdots \left( {{\varphi }_{n} \rightarrow \psi }\right) \ldots }\right)\]\n\nis a tautological instance. Applying MP \( n \) times gives a derivat... | Yes |
Proposition 49.17. If \( \mathbf{K} \vdash {\varphi }_{1} \rightarrow \left( {{\varphi }_{2} \rightarrow \cdots \left( {{\varphi }_{n - 1} \rightarrow {\varphi }_{n}}\right) \ldots }\right) \) then \( \mathbf{K} \vdash ▱{\varphi }_{1} \rightarrow \left( {▱{\varphi }_{2} \rightarrow \cdots \left( {▱{\varphi }_{n - 1} \r... | Proof. By induction on \( n \), just as in the proof of Proposition 49.6. | No |
Proposition 49.18. \( \mathbf{K} \vdash \left( {▱\varphi \land ▱\psi }\right) \rightarrow ▱\left( {\varphi \land \psi }\right) \) | Proof.\n\n1. \( \mathbf{K} \vdash \varphi \rightarrow \left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) \; \) TAUT\n\n2. \( \;\left. {\mathbf{K} \vdash ▱\varphi \rightarrow \left( {▱\psi \rightarrow ▱\left( {\varphi \land \psi }\right) }\right) }\right) \;\mathrm{{RK}},1 \n\n3. \( \mathbf{K} \vdash ... | Yes |
Proposition 49.19. If \( \mathbf{K} \vdash \varphi \leftrightarrow \psi \) and \( \mathbf{K} \vdash \chi \left\lbrack {\varphi /q}\right\rbrack \) then \( \mathbf{K} \vdash \chi \left\lbrack {B/q}\right\rbrack \) | Proof. Exercise. | No |
Proposition 49.20. \( \mathbf{K} \vdash \neg ▱p \rightarrow \Diamond \neg p \) | Proof.\n\n1. \( \mathbf{K} \vdash \diamond \neg p \leftrightarrow \neg ▱\neg \neg p\; \) DUAL\n\n2. \( \mathbf{K} \vdash \neg ▱\neg \neg p \rightarrow \Diamond \neg p\; \) PL,1\n\n3. \( \mathbf{K} \vdash \neg ▱p \rightarrow \diamond \neg p\;p \) for \( \neg \neg p \)\n\nIn the above derivation, the final step \ | Yes |
Proposition 49.21. If \( \varphi \) is a substitution instance of \( \psi \) and \( \mathbf{K} \vdash \psi \), then \( \mathbf{K} \vdash \varphi \) . | Proof. It is tedious but routine to verify (by induction on the length of the derivation of \( \psi \) ) that applying a substitution to an entire derivation also results in a correct derivation. Specifically, substitution instances of tautological instances are themselves tautological instances, substitution instances... | No |
Proposition 49.22. \( \mathbf{K} \vdash ▱\left( {\varphi \rightarrow \psi }\right) \rightarrow \left( {\Diamond \varphi \rightarrow \Diamond \psi }\right) \) | Proof.\n\n1. \( \mathbf{K} \vdash \left( {\varphi \rightarrow \psi }\right) \rightarrow \left( {\neg \psi \rightarrow \neg \varphi }\right) \; \) PL\n\n2. \( \;\mathbf{K} \vdash ▱\left( {\varphi \rightarrow \psi }\right) \rightarrow \left( {▱\neg \psi \rightarrow ▱\neg \varphi }\right) \; \) RK,1\n\n3. \( \mathbf{K} \v... | Yes |
Proposition 49.23. \( \mathbf{K} \vdash ▱\varphi \rightarrow \left( {\Diamond \left( {\varphi \rightarrow \psi }\right) \rightarrow \Diamond \psi }\right) \) | Proof.\n\n1. \( \;\mathbf{K} \vdash \varphi \rightarrow \left( {\neg \psi \rightarrow \neg \left( {\varphi \rightarrow \psi }\right) }\right) \; \) TAUT\n\n2. \( \;\mathbf{K} \vdash ▱\varphi \rightarrow \left( {▱\neg \psi \rightarrow ▱\neg \left( {\varphi \rightarrow \psi }\right) }\right) \; \) RK,1\n\n3. \( \mathbf{K... | Yes |
Proposition 49.24. \( \mathbf{K} \vdash \left( {\Diamond \varphi \vee \Diamond \psi }\right) \rightarrow \Diamond \left( {\varphi \vee \psi }\right) \) | Proof.\n\n1. \( \mathbf{K} \vdash \neg \left( {\varphi \vee \psi }\right) \rightarrow \neg \varphi \; \) TAUT\n\n2. \( \mathbf{K} \vdash ▱\neg \left( {\varphi \vee \psi }\right) \rightarrow ▱\neg \varphi \; \) RK,1\n\n3. \( \mathbf{K} \vdash \neg ▱\neg \varphi \rightarrow \neg ▱\neg \left( {\varphi \vee \psi }\right) \... | Yes |
Proposition 49.25. \( \mathbf{K} \vdash \diamond \left( {\varphi \vee \psi }\right) \rightarrow \left( {\diamond \varphi \vee \diamond \psi }\right) \) | Proof.\n\n1. \( \mathbf{K} \vdash \neg \varphi \rightarrow (\neg \psi \rightarrow \neg \left( {\varphi \vee \psi }\right) \; \) TAUT\n\n2. \( \mathbf{K} \vdash ▱\neg \varphi \rightarrow (▱\neg \psi \rightarrow ▱\neg \left( {\varphi \vee \psi }\right) \)\n\n3. \( \;\mathbf{K} \vdash ▱\neg \varphi \rightarrow \left( {\ne... | Yes |
Proposition 49.27. The following provability results obtain:\n\n1. \( \mathrm{KT}5 \vdash \mathrm{B} \) ; | 1. KT5 \( \vdash \) B:\n\n1. \( \mathbf{KT}\mathbf{5} \vdash \Diamond \varphi \rightarrow ▱\Diamond \varphi \;5 \)\n\n2. \( \mathbf{KT}\mathbf{5} \vdash \varphi \rightarrow \diamond \varphi \;{\mathrm{T}}_{\diamond } \)\n\n3. \( \mathbf{KT}\mathbf{5} \vdash \varphi \rightarrow ▱\diamond \varphi \; \) PL. | Yes |
Proposition 49.29. KTB4 \( = \) KT5 \( = \) KDB4 \( = \) KDB5. | Proof. Exercise. | No |
Theorem 49.30 (Soundness Theorem). If every instance of \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) is valid in the classes of models \( {\mathcal{C}}_{1},\ldots ,{\mathcal{C}}_{n} \), respectively, then \( \mathbf{K}{\varphi }_{1}\ldots {\varphi }_{n} \vdash \psi \) implies that \( \psi \) is valid in the class of mod... | Proof. By induction on length of proofs. For brevity, put \( \mathcal{C} = {\mathcal{C}}_{n} \cap \cdots \cap {\mathcal{C}}_{n} \). \n\n1. Induction Basis: If \( \psi \) has a proof of length 1, then it is either a tautological instance, an instance of \( \mathrm{K} \), or of DUAL, or an instance of one of \( {\varphi ... | Yes |
Theorem 49.34. \( \mathrm{{KD}}5 \neq \mathrm{{KT}}4 = \mathrm{S}4 \) . | Proof. By Theorem 48.1 we know that all instances of D and 5 are true in all serial euclidean models. So it suffices to find a serial euclidean model containing a world at which some instance of 4 fails. Consider the model of Figure 49.3, and notice that \( \mathfrak{M},{w}_{1} \nVdash ▱p \rightarrow ▱▱p \) . | Yes |
Proposition 49.36. Let \( \sum \) be a modal system and \( \Gamma \) a set of modal formulas. The following properties hold:\n\n1. Monotony: If \( \Gamma { \vdash }_{\sum }\varphi \) and \( \Gamma \subseteq \Delta \) then \( \Delta { \vdash }_{\sum }\varphi \) ;\n\n2. Reflexivity: If \( \varphi \in \Gamma \) then \( \G... | The proof is an easy exercise. Part (5) of Proposition 49.36 gives us that, for instance, if \( \Gamma { \vdash }_{\sum }\varphi \vee \psi \) and \( \Gamma { \vdash }_{\sum }\neg \varphi \), then \( \Gamma { \vdash }_{\sum }\psi \) . Also, in what follows, we write \( \Gamma ,\varphi { \vdash }_{\sum }\psi \) instead o... | No |
Proposition 49.39. Let \( \Gamma \) be a set of formulas. Then:\n\n1. A set \( \Gamma \) is \( \sum \) -consistent if and only if there is some formula \( \varphi \) such that \( \Gamma { \nvdash }_{\sum }\varphi \) . | Proof. These facts follow easily using classical propositional logic. We give the argument for (3). Proceed contrapositively and suppose neither \( \Gamma \cup \{ \varphi \} \) nor \( \Gamma \cup \{ \neg \varphi \} \) is \( \sum \) -consistent. Then by (2), both \( \Gamma ,\varphi { \vdash }_{\sum } \bot \) and \( \Gam... | Yes |
Proposition 50.2. Suppose \( \Gamma \) is complete \( \sum \) -consistent. Then:\n\n1. \( \Gamma \) is deductively closed in \( \sum \) .\n\n2. \( \sum \subseteq \Gamma \) .\n\n3. \( \bot \notin \Gamma \)\n\n4. \( \neg \varphi \in \Gamma \) if and only if \( \varphi \notin \Gamma \) .\n\n5. \( \varphi \land \psi \in \G... | Proof. 1. Suppose \( \Gamma { \vdash }_{\sum }\varphi \) but \( \varphi \notin \Gamma \) . Then since \( \Gamma \) is complete \( \sum \) -consistent, \( \neg \varphi \in \Gamma \) . This would make \( \Gamma \) inconsistent, since \( \varphi ,\neg \varphi { \vdash }_{\sum } \bot \) .\n\n2. If \( \varphi \in \sum \) th... | No |
Corollary 50.4. \( \Gamma { \vdash }_{\sum }\varphi \) if and only if \( \varphi \in \Delta \) for each complete \( \sum \) -consistent set \( \Delta \) extending \( \Gamma \) (including when \( \Gamma = \varnothing \), in which case we get another characterization of the modal system \( \sum \) .) | Proof. Suppose \( \Gamma { \vdash }_{\sum }\varphi \), and let \( \Delta \) be any complete \( \sum \) -consistent set extending \( \Gamma \) . If \( \varphi \notin \Delta \) then by maximality \( \neg \varphi \in \Delta \) and so \( \Delta { \vdash }_{\sum }\varphi \) (by monotony) and \( \Delta { \vdash }_{\sum }\neg... | Yes |
Lemma 50.7. If \( {▱}^{-1}\Gamma { \vdash }_{\sum }\varphi \) then \( \Gamma { \vdash }_{\sum }▱\varphi \) . | Proof. Suppose \( {▱}^{-1}\Gamma { \vdash }_{\sum }\varphi \) ; then by Lemma 50.6, \( ▱{▱}^{-1}\Gamma \vdash ▱\varphi \) . But since \( ▱{▱}^{-1}\Gamma \subseteq \Gamma \), also \( \Gamma { \vdash }_{\sum }▱\varphi \) by Monotony. | Yes |
Proposition 50.8. If \( \Gamma \) is complete \( \sum \) -consistent, then \( ▱\varphi \in \Gamma \) if and only if for every complete \( \sum \) -consistent \( \Delta \) such that \( {▱}^{-1}\Gamma \subseteq \Delta \), it holds that \( \varphi \in \Delta \) . | Proof. Suppose \( \Gamma \) is complete \( \sum \) -consistent. The \ | No |
Lemma 50.9. Suppose \( \Gamma \) and \( \Delta \) are complete \( \sum \) -consistent. Then: \( {▱}^{-1}\Gamma \subseteq \Delta \) if and only if \( \Diamond \Delta \subseteq \Gamma \) . | Proof. \ | No |
Proposition 50.10. If \( \Gamma \) is complete \( \sum \) -consistent, then \( \diamond \varphi \in \Gamma \) if and only if for some complete \( \sum \) -consistent \( \Delta \) such that \( \diamond \Delta \subseteq \Gamma \), it holds that \( \varphi \in \Delta \) . | Proof. Suppose \( \Gamma \) is complete \( \sum \) -consistent. \( \diamond \varphi \in \Gamma \) iff \( \neg ▱\neg \varphi \in \Gamma \) by DUAL and closure. \( \neg ▱\neg \varphi \in \Gamma \) iff \( ▱\neg \varphi \notin \Gamma \) by Proposition 50.2(4) since \( \Gamma \) is complete \( \sum \) -consistent. By Propos... | Yes |
Proposition 50.12 (Truth Lemma). For every formula \( \varphi ,{\mathfrak{M}}^{\sum },\Delta \Vdash \varphi \) if and only if \( \varphi \in \Delta \) . | Proof. By induction on \( \varphi \) .\n\n1. \( \varphi \equiv \bot : {\mathfrak{M}}^{\sum },\Delta \nVdash \bot \) by Definition 47.7, and \( \bot \notin \Delta \) by Proposition 50.2(3).\n\n2. \( \varphi \equiv p : {\mathfrak{M}}^{\sum },\Delta \Vdash p \) iff \( \Delta \in {V}^{\sum }\left( p\right) \) by Definition... | No |
Theorem 50.14 (Determination). \( {\mathfrak{M}}^{\sum } \Vdash \varphi \) if and only if \( \sum \vdash \varphi \) . | Proof. If \( {\mathfrak{M}}^{\sum } \Vdash \varphi \), then for every complete \( \sum \) -consistent \( \Delta \), we have \( {\mathfrak{M}}^{\sum },\Delta \Vdash \varphi \) . Hence, by the Truth Lemma, \( \varphi \in \Delta \) for every complete \( \sum \) -consistent \( \Delta \), whence by Corollary 50.4 (with \( \... | Yes |
Corollary 50.15. The basic modal logic \( \mathbf{K} \) is complete with respect to the class of all models, i.e., if \( \vDash \varphi \) then \( \mathbf{K} \vdash \varphi \) . | Proof. Contrapositively, if \( \mathbf{K} \nvdash \varphi \) then by Determination \( {\mathfrak{M}}^{\mathbf{K}} \nVdash \varphi \) and hence \( \varphi \) is not valid. | Yes |
Theorem 50.16. If a normal modal logic \( \sum \) contains one of the formulas on the left-hand side of table 50.1, then the canonical model for \( \sum \) has the corresponding property on the right-hand side. | Proof. We take each of these up in turn.\n\nSuppose \( \sum \) contains \( \mathrm{D} \), and let \( \Delta \in {W}^{\sum } \) ; we need to show that there is a \( {\Delta }^{\prime } \) such that \( {R}^{\sum }\Delta {\Delta }^{\prime } \) . It suffices to show that \( {▱}^{-1}\Delta \) is \( \sum \) -consistent, for ... | No |
1. If \( \sum \) contains the schema \( \diamond \varphi \rightarrow ▱\varphi \) then the canonical model for \( \sum \) is partially functional. | 1. Suppose that \( \sum \) contains the schema \( \diamond \varphi \rightarrow ▱\varphi \), to show that \( {R}^{\sum } \) is partially functional we need to prove that for any \( {\Delta }_{1},{\Delta }_{2},{\Delta }_{3} \in {W}^{\sum } \), if \( {R}^{\sum }{\Delta }_{1}{\Delta }_{2} \) and \( {R}^{\sum }{\Delta }_{1}... | Yes |
\[ {\varphi }_{1},\ldots ,{\varphi }_{n}{ \vdash }_{\sum }\diamond \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \rightarrow \bot \] | \[ {\varphi }_{1},\ldots ,{\varphi }_{n}{ \vdash }_{\sum }\neg \diamond \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \]
\[ \text{by PL} \]
\[ {\varphi }_{1},\ldots ,{\varphi }_{n}{ \vdash }_{\sum }▱\neg \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \]
\[ ▱\neg \text{for}\neg \diamond \]
\... | Yes |
Proposition 51.3. Given \( \mathfrak{M} \) and \( \Gamma \) , \( \equiv \) as defined above is an equivalence relation, i.e., it is reflexive, symmetric, and transitive. | Proof. The relation \( \equiv \) is reflexive, since \( w \) makes exactly the same formulas from \( \Gamma \) true as itself. It is symmetric since if \( u \) makes the same formulas from \( \Gamma \) true as \( v \), the same holds for \( v \) and \( u \) . It is also transitive, since if \( u \) makes the same formu... | Yes |
Proposition 51.8. The finest filtration \( {\mathfrak{M}}^{ * } \) is indeed a filtration. | Proof. We need to check that \( {R}^{ * } \), so defined, satisfies Definition 51.4(2). We check the three conditions in turn.\n\nIf \( {Ruv} \) then since \( u \in \left\lbrack u\right\rbrack \) and \( v \in \left\lbrack v\right\rbrack \), also \( {R}^{ * }\left\lbrack u\right\rbrack \left\lbrack v\right\rbrack \), so... | No |
Proposition 51.10. The coarsest filtration \( {\mathfrak{M}}^{ * } \) is indeed a filtration. | Proof. Given the definition of \( {R}^{ * } \), the only condition that is left to verify is the implication from \( {Ruv} \) to \( {R}^{ * }\left\lbrack u\right\rbrack \left\lbrack v\right\rbrack \) . So assume \( {Ruv} \) . Suppose \( ▱\varphi \in \Gamma \) and \( \mathfrak{M}, u \Vdash ▱\varphi \) ; then obviously \... | No |
Let \( W = {\mathbb{Z}}^{ + },{Rnm} \) iff \( m = n + 1 \), and \( V\left( p\right) = \{ {2n} : n \in \mathbb{N}\} \) . The model \( \mathfrak{M} = \langle W, R, V\rangle \) is depicted in Figure 51.1. The worlds are 1,2, etc.; each world can access exactly one other world-its successor-and \( p \) is true at all and o... | Now let \( \Gamma \) be the set of sub-formulas of \( ▱p \rightarrow p \), i.e., \( \{ p,▱p,▱p \rightarrow p\} \) . \( p \) is true at all and only the even numbers, \( ▱p \) is true at all and only the odd numbers, so \( ▱p \rightarrow p \) is true at all and only the even numbers. In other words, every odd number mak... | Yes |
Proposition 51.12. If \( \Gamma \) is finite then any filtration \( {\mathfrak{M}}^{ * } \) of a model \( \mathfrak{M} \) through \( \Gamma \) is also finite. | Proof. The size of \( {W}^{ * } \) is the number of different classes \( \left\lbrack w\right\rbrack \) under the equivalence relation \( \equiv \) . Any two worlds \( u, v \) in such class-that is, any \( u \) and \( v \) such that \( u \equiv v \) -agree on all formulas \( \varphi \) in \( \Gamma ,\varphi \in \Gamma ... | Yes |
Proposition 51.14. K has the finite model property. | Proof. \( \mathbf{K} \) is the set of valid formulas, i.e., any model is a model of \( \mathbf{K} \) . By Theorem 51.5, if \( \mathfrak{M}, w \Vdash \varphi \), then \( {\mathfrak{M}}^{ * }, w \Vdash \varphi \) for any filtration of \( \mathfrak{M} \) through the set \( \Gamma \) of sub-formulas of \( \varphi \) . Any ... | Yes |
Proposition 51.15. Let \( \mathcal{U} \) be the class of universal models (see Proposition 48.14) and \( {\mathcal{U}}_{\text{Fin }} \) the class of all finite universal models. Then any formula \( \varphi \) is valid in \( \mathcal{U} \) if and only if it is valid in \( {\mathcal{U}}_{\text{Fin }} \) . | Proof. Finite universal models are universal models, so the left-to-right direction is trivial. For the right-to left direction, suppose that \( \varphi \) is false at some world \( w \) in a universal model \( \mathfrak{M} \) . Let \( \Gamma \) contain \( \varphi \) as well as all of its sub-formulas; clearly \( \Gamm... | Yes |
Theorem 51.17. S5 is decidable. | Proof. Let \( \varphi \) be given, and suppose the propositional variables occurring in \( \varphi \) are among \( {p}_{1},\ldots ,{p}_{k} \) . Since for each \( n \) there are only finitely many models with \( n \) worlds assigning a value to \( {p}_{1},\ldots ,{p}_{k} \), we can enumerate, in parallel, all the theore... | Yes |
1. Suppose \( {R}^{ * }\left\lbrack u\right\rbrack \left\lbrack v\right\rbrack \) if and only if \( {C}_{1}\left( {u, v}\right) \land {C}_{2}\left( {u, v}\right) \) . Then \( {R}^{ * } \) is symmetric, and \( {\mathfrak{M}}^{ * } = \left\langle {{W}^{ * },{R}^{ * },{V}^{ * }}\right\rangle \) is a filtration if \( \math... | 1. It’s immediate that \( {R}^{ * } \) is symmetric, since \( {C}_{1}\left( {u, v}\right) \Leftrightarrow {C}_{2}\left( {v, u}\right) \) and \( {C}_{2}\left( {u, v}\right) \Leftrightarrow {C}_{1}\left( {v, u}\right) \) . So it’s left to show that if \( \mathfrak{M} \) is symmetric then \( {\mathfrak{M}}^{ * } \) is a f... | No |
1. If \( \mathfrak{M} \) is symmetric, so is \( {\mathfrak{M}}^{ * } \) . | 1. If \( {\mathfrak{M}}^{ * } \) is a coarsest filtration, then by definition \( {R}^{ * }\left\lbrack u\right\rbrack \left\lbrack v\right\rbrack \) holds if and only if \( {C}_{1}\left( {u, v}\right) \) . For transitivity, suppose \( {C}_{1}\left( {u, v}\right) \) and \( {C}_{1}\left( {v, w}\right) \) ; we have to sho... | No |
We give a closed tableau that shows \( \vdash \diamond \left( {\varphi \vee \psi }\right) \rightarrow \left( {\diamond \varphi \vee \diamond \psi }\right) \) |  | No |
Proposition 52.5. If \( \Gamma \) contains both \( \sigma \mathbb{T}\varphi \) and \( \sigma \mathbb{F}\varphi \), for some formula \( \varphi \) and prefix \( \sigma \), then \( \Gamma \) is unsatisfiable. | Proof. There cannot be a model \( \mathfrak{M} \) and interpretation \( f \) of \( P\left( \Gamma \right) \) such that both \( \mathfrak{M}, f\left( \sigma \right) \Vdash \varphi \) and \( \mathfrak{M}, f\left( \sigma \right) \nVdash \varphi \) . | Yes |
Corollary 52.7. If \( \Gamma \vdash \varphi \) then \( \Gamma \vDash \varphi \) . | Proof. If \( \Gamma \vdash \varphi \) then for some \( {\psi }_{1},\ldots ,{\psi }_{n} \in \Gamma ,\Delta = \left\{ {1\mathbb{F}\varphi ,1\mathbb{T}{\psi }_{1},\ldots ,1\mathbb{T}{\psi }_{n}}\right\} \) has a closed tableau. We want to show that \( \Gamma \vDash \varphi \) . Suppose not, so for some \( \mathfrak{M} \) ... | Yes |
We give a closed tableau that shows \( \mathbf{S}\mathbf{5} \vdash 5 \), i.e., \( ▱\varphi \rightarrow ▱\diamond \varphi \) . | <table><tr><td>1.</td><td>\( 1\mathbb{F}\;▱\varphi \rightarrow ▱\Diamond \varphi \)</td><td>Assumption</td></tr><tr><td>2.</td><td>\( 1 \) T \( ▱\varphi \)</td><td>\( \rightarrow \mathbb{F}1 \)</td></tr><tr><td>3.</td><td>\( 1\mathbb{F}\;▱\Diamond \varphi \)</td><td>\( \rightarrow \mathbb{F}1 \)</td></tr><tr><td>4.</td... | Yes |
Proposition 52.10. T \( ▱ \) and T \( \diamond \) are sound for reflexive models. | Proof. 1. The branch is expanded by applying \( \mathrm{T}▱ \) to \( \sigma \mathbb{T}▱\psi \in \Gamma \) : This results in a new signed formula \( \sigma \mathbb{T}\psi \) on the branch. Suppose \( \mathfrak{M}, f \Vdash \Gamma \), in particular, \( \mathfrak{M}, f\left( \sigma \right) \Vdash ▱\psi \) . Since \( R \) ... | No |
Proposition 52.11. \( \mathrm{D}▱ \) and \( \mathrm{D}\diamond \) are sound for serial models. | Proof. 1. The branch is expanded by applying \( \mathrm{D}▱ \) to \( \sigma \mathbb{T}▱\psi \in \Gamma \) : This results in a new signed formula \( \sigma \mathbb{T}\diamond \psi \) on the branch. Suppose \( \mathfrak{M}, f \Vdash \) \( \Gamma \), in particular, \( \mathfrak{M}, f\left( \sigma \right) \Vdash ▱\psi \) .... | Yes |
Proposition 52.12. \( \mathrm{B}▱ \) and \( \mathrm{B}\diamond \) are sound for symmetric models. | Proof. 1. The branch is expanded by applying \( \mathrm{B}▱ \) to \( \sigma .n\mathbb{T}▱\psi \in \Gamma \) : This results in a new signed formula \( \sigma \mathbb{T}\psi \) on the branch. Suppose \( \mathfrak{M}, f \Vdash \Gamma \) , in particular, \( \mathfrak{M}, f\left( {\sigma .n}\right) \Vdash ▱\psi \) . Since \... | No |
Proposition 52.13. \( 4▱ \) and \( 4\diamond \) are sound for transitive models. | Proof. 1. The branch is expanded by applying \( 4▱ \) to \( \sigma \mathbb{T}▱\psi \in \Gamma \) : This results in a new signed formula \( \sigma .n\mathbb{T}▱\psi \) on the branch. Suppose \( \mathfrak{M}, f \Vdash \) \( \Gamma \), in particular, \( \mathfrak{M}, f\left( \sigma \right) \Vdash ▱\psi \) . Since \( f \) ... | No |
Proposition 52.14. \( 4\mathrm{r}▱ \) and \( 4\mathrm{r}\diamond \) are sound for euclidean models. | Proof. 1. The branch is expanded by applying \( 4\mathrm{r}▱ \) to \( \sigma .n\mathbb{T}▱\psi \in \Gamma \) : This results in a new signed formula \( \sigma \mathbb{T}▱\psi \) on the branch. Suppose \( \mathfrak{M}, f \Vdash \Gamma \) , in particular, \( \mathfrak{M}, f\left( {\sigma .n}\right) \Vdash ▱\psi \) . Since... | No |
Example 52.16. We give a simplified closed tableau that shows \( \mathrm{S}5 \vdash 5 \), i.e., \( \diamond \varphi \rightarrow ▱\diamond \varphi \) . | <table><tr><td>1.</td><td>1 \( \mathbb{F} \land \varphi \rightarrow ▱ \land \varphi \)</td><td>Assumption</td></tr><tr><td>2.</td><td>1 T \( \Diamond \varphi \)</td><td>\( \rightarrow \mathbb{F}1 \)</td></tr><tr><td>3.</td><td>\( 1\mathbb{F}▱\Diamond \varphi \)</td><td>\( \rightarrow \mathbb{F}1 \)</td></tr><tr><td>4.<... | Yes |
Proposition 52.18. Every finite \( \Gamma \) has a tableau in which every branch is complete. | Proof. Consider an open branch in a tableau for \( \Gamma \) . There are finitely many prefixed formulas in the branch to which a rule could be applied. In some fixed order (say, top to bottom), for each of these prefixed formulas for which the conditions (1)-(4) do not already hold, apply the rules that can be applied... | No |
We know that \( \nvdash ▱\left( {p \vee q}\right) \rightarrow \left( {▱p \vee ▱q}\right) \) . | <table><tr><td>1.</td><td>\( 1\;F\;\square \left( {p \vee q}\right) \rightarrow \left( {\square p\vee \square q}\right) \;\checkmark \)</td><td>Assumption</td></tr><tr><td>2.</td><td>1 \( \mathbb{T}\;▱\left( {p \vee q}\right) \)</td><td>\( \rightarrow \mathbb{F}1 \)</td></tr><tr><td>3.</td><td>\( 1\;\mathbb{F}\;▱p \vee... | Yes |
Theorem 53.1. There are irrational numbers \( a \) and \( b \) such that \( {a}^{b} \) is rational. | Proof. Consider \( {\sqrt{2}}^{\sqrt{2}} \) . If this is rational, we are done: we can let \( a = b = \sqrt{2} \) . Otherwise, it is irrational. Then we have\n\n\[ \n{\left( {\sqrt{2}}^{\sqrt{2}}\right) }^{\sqrt{2}} = {\sqrt{2}}^{\sqrt{2} \cdot \sqrt{2}} = {\sqrt{2}}^{2} = 2, \n\]\n\nwhich is rational. So, in this case... | Yes |
Example 53.3. Take \( \neg \bot \) for example. A construction of it is a function which, given any construction of \( \bot \) as input, provides a construction of \( \bot \) as output. | Obviously, the identity function Id is such a construction: given a construction \( M \) of \( \bot ,\operatorname{Id}\left( M\right) = M \) yields a construction of \( \bot \) . | Yes |
Let us prove \( \varphi \rightarrow \neg \neg \varphi \) for any proposition \( \varphi \), which is \( \varphi \rightarrow \) \( \left( {\left( {\varphi \rightarrow \bot }\right) \rightarrow \bot }\right) \) . | The construction should be a function \( f \) that, given a construction \( M \) of \( \varphi \), returns a construction \( f\left( M\right) \) of \( \left( {\varphi \rightarrow \bot }\right) \rightarrow \bot \) . Here is how \( f \) constructs the construction of \( \left( {\varphi \rightarrow \bot }\right) \rightarr... | Yes |
Let us give a construction for \( \neg \left( {\varphi \land \neg \varphi }\right) \), i.e., \( \left( {\varphi \land \left( {\varphi \rightarrow \bot }\right) }\right) \rightarrow \) \( \bot \) . | This is a function \( f \) which, given as input a construction \( M \) of \( \varphi \land \left( {\varphi \rightarrow \bot }\right) \) , yields a construction of \( \bot \) . A construction of a conjunction \( {\psi }_{1} \land {\psi }_{2} \) is a pair \( \left\langle {{N}_{1},{N}_{2}}\right\rangle \) where \( {N}_{1... | Yes |
Let us give a construction of \( \left( {\left( {\varphi \land \psi }\right) \rightarrow \chi }\right) \rightarrow \left( {\varphi \rightarrow \left( {\psi \rightarrow \chi }\right) }\right) \) , i.e., a function \( f \) which turns a construction \( g \) of \( \left( {\varphi \land \psi }\right) \rightarrow \chi \) in... | The construction \( g \) is itself a function (from constructions of \( \varphi \land \psi \) to constructions of \( C \) ). And the output \( f\left( g\right) \) is a function \( {h}_{g} \) from constructions of \( \varphi \) to functions from constructions of \( \psi \) to constructions of \( \chi \) . \( \mathrm{{Ok... | Yes |
To prove \( \neg \neg \left( {\varphi \vee \neg \varphi }\right) \) | we need a function \( f \) that transforms a construction of \( \neg \left( {\varphi \vee \neg \varphi }\right) \), i.e., of \( \left( {\varphi \vee \left( {\varphi \rightarrow \bot }\right) }\right) \rightarrow \bot \), into a construction of \( \bot \) . In other words, we need a function \( f \) such that \( f\left(... | Yes |
Proposition 53.8. If \( \Gamma \vdash \varphi \) in intuitionistic logic, \( \Gamma \vdash \varphi \) in classical logic. In particular, if \( \varphi \) is an intuitionistic theorem, it is also a classical theorem. | Proof. Every natural deduction rule is also a rule in classical natural deduction, so every derivation in intuitionistic logic is also a derivation in classical logic. | Yes |
Proposition 53.13. If \( \Gamma \vdash \varphi \) in intuitionistic logic, \( \Gamma \vdash \varphi \) in classical logic. In particular, if \( \varphi \) is an intuitionistic theorem, it is also a classical theorem. | Proof. Every intuitionistic axiom is also a classical axiom, so every derivation in intuitionistic logic is also a derivation in classical logic. | Yes |
Proposition 54.3. Truth at worlds is monotonic with respect to \( R \), i.e., if \( \mathfrak{M}, w \Vdash \varphi \) and Rww \( {}^{\prime } \), then \( \mathfrak{M},{w}^{\prime } \Vdash \varphi \) . | Proof. Exercise. | No |
Proposition 54.5. 1. If \( \mathfrak{M}, w \Vdash \Gamma \) and \( \Gamma \vDash \varphi \), then \( \mathfrak{M}, w \Vdash \varphi \) . | Proof. 1. Suppose \( \mathfrak{M} \Vdash \Gamma \) . Since \( \Gamma \vDash \varphi \), we know that if \( \mathfrak{M}, w \Vdash \Gamma \), then \( \mathfrak{M}, w \Vdash \varphi \) . Since \( \mathfrak{M}, u \Vdash \Gamma \) for all every \( u \in W,\mathfrak{M}, w \Vdash \Gamma \) . Hence \( \mathfrak{M}, w \Vdash \... | Yes |
Theorem 55.1 (Soundness). If \( \Gamma \vdash \varphi \), then \( \Gamma \vDash \varphi \) . | Proof. We prove that if \( \Gamma \vdash \varphi \), then \( \Gamma \vDash \varphi \) . The proof is by induction on the number \( n \) of formulas in the derivation of \( \varphi \) from \( \Gamma \) . We show that if \( {\varphi }_{1} \) , \( \ldots ,{\varphi }_{n} = \varphi \) is a derivation from \( \Gamma \), then... | Yes |
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