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Proposition 42.15. Every general recursive function is \( \lambda \) -definable.
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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.
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Yes
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Theorem 42.17. If a partial function \( f \) is \( \lambda \) -definable, it is partial recursive.
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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 lambda term as argument, and which returns the Gödel number of the normal form if it has one, or is undefined otherwise. Then define two partial recursive functions toChurch and fromChurch that maps natural numbers to and from the Gödel numbers of the corresponding Church numeral.\n\nUsing these recursive functions, we can define the function \( f \) as a partial recursive function. There is a lambda term \( F \) that \( \lambda \) -defines \( f \) . To compute \( f\left( {{n}_{1},\ldots ,{n}_{k}}\right) \), first obtain the Gödel numbers of the corresponding Church numerals using toChurch \( \left( {n}_{i}\right) \), append these to * \( {F}^{\# } \) to obtain the Gödel number of the term \( F{n}_{1}\ldots \overline{{n}_{k}} \) . Now use normalize on this Gödel number. If \( f\left( {{n}_{1},\ldots ,{n}_{k}}\right) \) is defined, \( F\overline{{n}_{1}}\ldots \overline{{n}_{k}} \) has a normal form (which must be a Church numeral), and otherwise it has no normal form (and so\n\n\[ \text{normalize}\left( {{}^{\# }F\overline{{n}_{1}}\ldots {\overline{{n}_{k}}}^{\# }}\right) \]\n\nis undefined). Finally, use fromChurch on the Gödel number of the normalized term.
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No
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Proposition 43.11. 1. \( \varphi \) is a tautology if and only if \( \varnothing \vDash \varphi \) ;
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Proof. Exercise.
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No
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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. \( {\widetilde{\neg }}_{\mathbf{L}}\left( x\right) = {\widetilde{\neg }}_{\mathbf{C}}\left( x\right) \) if \( x = \mathbb{T} \) or \( x = \mathbb{F} \) ;\n\n2. \( {\widetilde{ \land }}_{\mathbf{L}}\left( {x, y}\right) = {\widetilde{ \land }}_{\mathbf{C}}\left( {x, y}\right) \) ,\n\n3. \( {\widetilde{ \vee }}_{\mathbf{L}}\left( {x, y}\right) = {\widetilde{ \vee }}_{\mathbf{C}}\left( {x, y}\right) \) ,\n\n\[ \text{4.}{\widetilde{ \rightarrow }}_{\mathbf{L}}\left( {x, y}\right) = {\widetilde{ \rightarrow }}_{\mathbf{C}}\left( {x, y}\right) \text{, if}x, y \in \{ \mathbb{T},\mathbb{F}\} \text{.} \]\n\nThen, for any valuation \( \mathfrak{v} \) into \( V \) such that \( \mathfrak{v}\left( p\right) \in \{ \mathbb{T},\mathbb{F}\} ,{\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \varphi \right) = {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \varphi \right) \) .
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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\[ {\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \varphi \right) = {\widetilde{\neg }}_{\mathbf{L}}\left( {{\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \psi \right) }\right) \]\nby Definition 43.8\n\n\[ = {\widetilde{\neg }}_{\mathbf{L}}\left( {{\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \psi \right) }\right) \]\nby inductive hypothesis\n\n\[ = {\widetilde{\neg }}_{\mathbf{C}}\left( {{\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \psi \right) }\right) \]\nby assumption (1),\n\nsince \( {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \psi \right) \in \{ \mathbb{T},\mathbb{F}\} \)\n\n\( = {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \varphi \right) \; \) by Definition 43.8.\n\n3. If \( \varphi \equiv \left( {\psi \land \chi }\right) \), we have\n\n\[ {\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \varphi \right) = {\widetilde{ \land }}_{\mathbf{L}}\left( {{\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \psi \right) ,{\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \chi \right) }\right) \;\text{by Definition 43.8} \]\n\n\[ = {\widetilde{ \land }}_{\mathbf{L}}\left( {{\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \psi \right) ,{\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \chi \right) }\right) \;\text{by inductive hypothesis} \]\n\n\[ = {\widetilde{ \land }}_{\mathbf{C}}\left( {{\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \psi \right) ,{\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \chi \right) }\right) \;\text{by assumption (2),} \]\n\nsince \( {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \psi \right) ,{\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \chi \right) \in \{ \mathbb{T},\mathbb{F}\} \)\n\n\[ = {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \varphi \right) \]\n\nThe cases where \( \varphi \equiv \left( {\psi \vee \chi }\right) \) and \( \varphi \equiv \left( {\psi \rightarrow \chi }\right) \) are similar.
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Yes
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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 \( \Gamma { \vDash }_{\mathbf{C}}\psi \) . In particular, every tautology of \( \mathbf{L} \) is also a classical tautology.
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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 \( \varphi \in \Gamma \) and \( {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \psi \right) = \mathbb{F} \) . Since \( \mathbb{T},\mathbb{F} \in V \), the valuation \( \mathfrak{v} \) is also a valuation for \( \mathbf{L} \) . By Proposition 43.12, \( {\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \varphi \right) = \mathbb{T} \) for all \( \varphi \in \Gamma \) and \( {\overline{\mathfrak{v}}}_{\mathbf{L}}\left( \psi \right) = \mathbb{F} \) . Since \( \mathbb{T} \in {V}^{ + } \) and \( \mathbb{F} \notin {V}^{ + } \) that means \( \mathfrak{v}{ \vDash }_{\mathbf{L}}\Gamma \) and \( \mathfrak{v}{ \mathrel{\text{\vDash \not{} }} }_{\mathbf{L}}\psi \), i.e., \( \Gamma { \mathrel{\text{\vDash \not{} }} }_{\mathbf{L}}\psi \) .
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Yes
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Proposition 44.2. If \( \mathfrak{v}\left( p\right) \in \{ \mathbb{T},\mathbb{F}\} \) for all \( p \) in \( \varphi \), then \( {\overline{\mathfrak{v}}}_{{Ł}_{3}}\left( \varphi \right) = {\overline{\mathfrak{v}}}_{\mathbf{C}}\left( \varphi \right) \) .
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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
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No
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Proposition 44.5. Ks and Kw have no tautologies.
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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\[ \widetilde{\neg }\left( \mathbb{U}\right) = \widetilde{ \vee }\left( {\mathbb{U},\mathbb{U}}\right) = \widetilde{ \land }\left( {\mathbb{U},\mathbb{U}}\right) = \widetilde{ \rightarrow }\left( {\mathbb{U},\mathbb{U}}\right) = \mathbb{U} \]\n\nin both logics. As \( \mathbb{U} \notin {V}^{ + } \) for either \( \mathbf{{Ks}} \) or \( \mathbf{{Kw}} \), on this valuation, \( \varphi \) will not be designated.
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Yes
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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.
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Proof. Exercise.
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No
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Problem 44.6. Which of the following relations hold in (a) strong and (b) weak Kleene logic? Give a truth table for each.
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\[ \text{1.}p, p \rightarrow q \vDash q \]
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No
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Problem 44.10. Which of the following relations hold in Gödel logic? Give a truth table for each.
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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. \( p \vDash p \vee q \)
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No
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Problem 44.13. Which of the following relations hold in (a) LP and in (b) Hal? Give a truth table for each.
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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. \( p \rightarrow q, q \rightarrow r \vDash p \rightarrow r \)
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No
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Proposition 45.2. The logic \( {L}_{3} \) defined by Definition 44.1 is the same as \( {L}_{3} \) defined by Definition 45.1.
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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:  
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Yes
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Proposition 45.3. If \( \Gamma { \vDash }_{{L}_{\infty }}\psi \) then \( \Gamma { \vDash }_{{L}_{m}}\psi \) for all \( m \geq 2 \) .
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Proof. Exercise.
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No
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Proposition 45.5. The logic \( {\mathbf{G}}_{3} \) defined by Definition 44.6 is the same as \( {\mathbf{G}}_{3} \) defined by Definition 45.4.
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Proof. This can be seen by comparing the truth tables for the connectives given in Definition 44.6 with the truth tables determined by the equations in Definition 45.4: 
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Yes
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Proposition 45.6. If \( \Gamma { \vDash }_{{\mathbf{G}}_{\infty }}\psi \) then \( \Gamma { \vDash }_{{\mathbf{G}}_{m}}\psi \) for all \( m \geq 2 \) .
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Proof. Exercise.
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No
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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
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\[ \diamond \left( {{p}_{2} \rightarrow {p}_{3}}\right) \rightarrow ▱\left( {\diamond \left( {{p}_{2} \rightarrow {p}_{3}}\right) \land \neg ▱{p}_{1}}\right) \]
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Yes
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Proposition 47.8. 1. \( \mathfrak{M}, w \Vdash ▱\varphi \) iff \( \mathfrak{M}, w \Vdash \neg \Diamond \neg \varphi \) .
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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 \( {Rw}{w}^{\prime },\mathfrak{M},{w}^{\prime } \Vdash \neg \varphi \) . Hence, \( \mathfrak{M}, w \nVdash \diamond \neg \varphi \) iff for all \( {w}^{\prime } \) with \( {Rw}{w}^{\prime },\mathfrak{M},{w}^{\prime } \nVdash \neg \varphi \) . We also have \( \mathfrak{M},{w}^{\prime } \nVdash \neg \varphi \) iff \( \mathfrak{M},{w}^{\prime } \Vdash \varphi \) . Together we have \( \mathfrak{M}, w \Vdash \neg \diamond \neg \varphi \) iff for all \( {w}^{\prime } \) with \( {Rw}{w}^{\prime },\mathfrak{M},{w}^{\prime } \Vdash \varphi \) . Again by definition of \( \mathfrak{M}, w \Vdash \), that is the case iff \( \mathfrak{M}, w \Vdash ▱\varphi \) .
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Yes
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1. If \( \mathfrak{M} \Vdash \varphi \) then \( \mathfrak{M} \nVdash \neg \varphi \), but not vice-versa.
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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 \( \mathfrak{M} \nVdash \neg \varphi \) then \( \varphi \) is true at some world \( w \in W \) . It does not follow that \( \mathfrak{M}, w \Vdash \varphi \) for every \( w \in W \) . For instance, in the model of Figure 47.1, \( \mathfrak{M} \nVdash \neg p \), and also \( \mathfrak{M} \nVdash p \) .
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Yes
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Proposition 47.13. If \( \varphi \) is valid, then so is \( ▱\varphi \) .
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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 \( \mathfrak{M} \) and \( w \) were arbitrary, \( \vDash ▱\varphi \) .
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Yes
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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)
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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 \( \mathfrak{M}, w \Vdash ▱\left( {\varphi \rightarrow \psi }\right) \) and \( \mathfrak{M}, w \Vdash ▱\varphi \) . We need to show \( \mathfrak{M} \Vdash ▱\psi \) . So let \( {w}^{\prime } \) be arbitrary such that \( {Rw}{w}^{\prime } \) . Then by the first assumption \( \mathfrak{M},{w}^{\prime } \Vdash \varphi \rightarrow \psi \) and by the second assumption \( \mathfrak{M},{w}^{\prime } \Vdash \varphi \) . It follows that \( \mathfrak{M},{w}^{\prime } \Vdash \psi \) . Since \( {w}^{\prime } \) was arbitrary, \( \mathfrak{M}, w \Vdash ▱\psi \) .
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Yes
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Proposition 47.20. The following schema DUAL is valid\n\n\[ \diamond \varphi \leftrightarrow \neg ▱\neg \varphi \]
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Proof. Exercise.
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No
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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.
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Proof. The \
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No
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To show \( p \rightarrow \diamond p \vDash ▱\neg p \rightarrow \neg p \)
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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 \( \mathfrak{M}, w \nVdash \neg p \) . Since \( \mathfrak{M}, w \nVdash \neg p,\mathfrak{M}, w \Vdash p \) . By assumption, \( \mathfrak{M}, w \Vdash p \rightarrow \diamond p \), hence \( \mathfrak{M}, w \Vdash \diamond p \) . By definition of \( \mathfrak{M}, w \Vdash \diamond p \), there is some \( {w}^{\prime } \) with \( {Rw}{w}^{\prime } \) such that \( \mathfrak{M},{w}^{\prime } \Vdash p \) . Since also \( \mathfrak{M}, w \Vdash ▱\neg p,\mathfrak{M},{w}^{\prime } \Vdash \neg p \), a contradiction.
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Yes
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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 \( \mathfrak{M} \) .
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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 at \( w \), in order to show that \( ▱\diamond \varphi \) is true at \( w \) . So we need to show that \( \diamond \varphi \) is true at all \( {w}^{\prime } \) accessible from \( w \) . Now, for any \( {w}^{\prime } \) such that \( {Rw}{w}^{\prime } \) we have, using the hypothesis of symmetry, that also \( R{w}^{\prime }w \) (see Figure 48.1). Since \( \mathfrak{M}, w \Vdash \varphi \), we have \( \mathfrak{M},{w}^{\prime } \Vdash \diamond \varphi \) . Since \( {w}^{\prime } \) was an arbitrary world such that \( {Rw}{w}^{\prime } \), we have \( \mathfrak{M}, w \Vdash ▱\diamond \varphi \) .
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Yes
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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 \( \varphi : \mathfrak{M}, u \Vdash \varphi \) if and only if \( \mathfrak{M}, v \Vdash \varphi \) (use induction on \( \varphi \) ).\n\n2. Every instance of \( \mathrm{T} \) is true in \( \mathfrak{M} \) .
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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).
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No
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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.
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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 \( {Rwv} \) . Suppose not: then both \( \mathfrak{M} \Vdash ▱\varphi \) and \( \mathfrak{M}, w \nVdash \diamond \varphi \) for any \( \varphi \), including \( p \) . But then \( \mathfrak{M}, w \nVdash ▱p \rightarrow \diamond p \), contradicting the assumption that \( \mathfrak{F} \vDash ▱p \rightarrow \diamond p \) .
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Yes
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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.
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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 right is valid in \( \mathfrak{F},\mathfrak{F} \) has the property on the left.
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Yes
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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.
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Proof. Exercise.
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No
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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 \( \mathbf{{S5}} \) .
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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 \( w \) in a model \( \mathfrak{M} = \langle W, R, V\rangle \) based on a frame \( \langle W, R\rangle \), where \( R \) is an equivalence on \( W \) . So \( \mathfrak{M}, w \nVdash \psi \) . Define a model \( {\mathfrak{M}}^{\prime } = \left\langle {{W}^{\prime },{R}^{\prime },{V}^{\prime }}\right\rangle \) as follows:\n\n\[ \n\text{1.}{W}^{\prime } = \left\lbrack w\right\rbrack \text{;} \n\]\n\n2. \( {R}^{\prime } \) is universal on \( {W}^{\prime } \) ;\n\n3. \( {V}^{\prime }\left( p\right) = V\left( p\right) \cap {W}^{\prime } \) .\n\n(So the set \( {W}^{\prime } \) of worlds in \( {\mathfrak{M}}^{\prime } \) is represented by the shaded area in Figure 48.2.) It is easy to see that \( R \) and \( {R}^{\prime } \) agree on \( {W}^{\prime } \) . Then one can show by induction on formulas that for all \( {w}^{\prime } \in {W}^{\prime } : {\mathfrak{M}}^{\prime },{w}^{\prime } \Vdash \varphi \) if and only if \( \mathfrak{M},{w}^{\prime } \Vdash \varphi \) for each \( \varphi \) (this makes sense since \( {W}^{\prime } \subseteq W \) ). In particular, \( {\mathfrak{M}}^{\prime }, w \nVdash \psi \), and \( \psi \) fails in a model based on a universal frame.
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Yes
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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 \( s\left( x\right) = w \) . Then\n\n\[ \mathfrak{M}, w \Vdash \varphi \text{iff}{\mathfrak{M}}^{\prime }, s \vDash {\mathrm{{ST}}}_{x}\left( \varphi \right) \]
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Proof. By induction on \( \varphi \) .
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No
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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 the second-order formula\n\n\[ \forall {X}_{1}\ldots \forall {X}_{n}\forall x{\mathrm{{ST}}}_{x}\left( \varphi \right) \left\lbrack {{X}_{1}/{P}_{1},\ldots ,{X}_{n}/{P}_{n}}\right\rbrack \]\n\nwhere \( {P}_{1},\ldots ,{P}_{n} \) are all one-place predicate symbols in \( {\mathrm{{ST}}}_{x}\left( \varphi \right) \) . Then\n\n\[ \mathfrak{F} \vDash \varphi \text{iff}{\mathfrak{F}}^{\prime } \vDash {\varphi }^{\prime } \]
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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 \( s\left( x\right) \in W,{\mathfrak{M}}^{\prime }, s \vDash {\mathrm{{ST}}}_{x}\left( \varphi \right) \) . By Proposition 48.16, that is the case iff for all models \( \mathfrak{M} \) based on \( \mathfrak{F} \) and every world \( w \in W,\mathfrak{M}, w \Vdash \varphi \), i.e., \( \mathfrak{F} \vDash \varphi \) .
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Yes
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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.
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Proof. The monadic second-order sentence \( {\varphi }^{\prime } \) of the preceding proof has the required property.
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No
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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.
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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.
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No
|
Proposition 49.6. Every normal modal logic is closed under rule RK,
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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\[{\varphi }_{1} \rightarrow \left( {{\varphi }_{2} \rightarrow \cdots \left( {{\varphi }_{n - 1} \rightarrow {\varphi }_{n}}\right) \cdots }\right) \in \sum\]\n\nBy the induction hypothesis, we have\n\n\[▱{\varphi }_{1} \rightarrow \left( {▱{\varphi }_{2} \rightarrow \cdots ▱\left( {{\varphi }_{n - 1} \rightarrow {\varphi }_{n}}\right) \cdots }\right) \in \sum\]\n\nSince \( \sum \) is a normal modal logic, it contains all instances of \( \mathrm{K} \), in particular\n\n\[▱\left( {{\varphi }_{n - 1} \rightarrow {\varphi }_{n}}\right) \rightarrow \left( {▱{\varphi }_{n - 1} \rightarrow ▱{\varphi }_{n}}\right) \in \sum\]\n\nUsing modus ponens and suitable tautological instances we get\n\n\[▱{\varphi }_{1} \rightarrow \left( {▱{\varphi }_{2} \rightarrow \cdots \left( {▱{\varphi }_{n - 1} \rightarrow ▱{\varphi }_{n}}\right) \cdots }\right) \in \sum .\]
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Yes
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Proposition 49.8. Let \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) be formulas. Then there is a smallest modal logic \( \sum \) containing all instances of \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) .
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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 modal logic.
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Yes
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Proposition 49.12. \( \mathbf{K} \vdash ▱\varphi \rightarrow ▱\left( {\psi \rightarrow \varphi }\right) \)
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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. \( ▱\left( {\varphi \rightarrow \left( {\psi \rightarrow \varphi }\right) }\right) \rightarrow \left( {▱\varphi \rightarrow ▱\left( {\psi \rightarrow \varphi }\right) }\right) \)\n\n4. \( ▱\varphi \rightarrow ▱\left( {\psi \rightarrow \varphi }\right) \; \) MP,2,3
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Yes
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Proposition 49.13. \( \mathbf{K} \vdash ▱\left( {\varphi \land \psi }\right) \rightarrow \left( {▱\varphi \land ▱\psi }\right) \)
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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. \( ▱\left( {\left( {\varphi \land \psi }\right) \rightarrow \varphi }\right) \rightarrow \left( {▱\left( {\varphi \land \psi }\right) \rightarrow ▱\varphi }\right) \;\mathrm{K} \)\n\n4. \( ▱\left( {\varphi \land \psi }\right) \rightarrow ▱\varphi \; \) MP,2,3\n\n5. \( \left( {\varphi \land \psi }\right) \rightarrow \psi \; \) TAUT\n\n6. \( ▱\left( {\left( {\varphi \land \psi }\right) \rightarrow \psi }\right) \; \) NEC\n\n7. \( ▱\left( {\left( {\varphi \land \psi }\right) \rightarrow \psi }\right) \rightarrow \left( {▱\left( {\varphi \land \psi }\right) \rightarrow ▱\psi }\right) \)\n\n8. \( ▱\left( {\varphi \land \psi }\right) \rightarrow ▱\psi \; \) MP,6,7\n\n9. \( \left( {▱\left( {\varphi \land \psi }\right) \rightarrow ▱\varphi }\right) \rightarrow \)\n\n\( (\left( {▱\left( {\varphi \land \psi }\right) \rightarrow ▱\psi }\right) \rightarrow \)\n\n\( \left( {▱\left( {\varphi \land \psi }\right) \rightarrow \left( {▱\varphi \land ▱\psi }\right) }\right) )\; \) TAUT\n\n10. \( \left( {▱\left( {\varphi \land \psi }\right) \rightarrow ▱\psi }\right) \rightarrow \)\n\n\( \left( {▱\left( {\varphi \land \psi }\right) \rightarrow \left( {▱\varphi \land ▱\psi }\right) }\right) \; \) MP, \( 4,9 \)\n\n11. \( ▱\left( {\varphi \land \psi }\right) \rightarrow \left( {▱\varphi \land ▱\psi }\right) \; \) MP,8,10.
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Yes
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Proposition 49.14. \( \mathbf{K} \vdash \left( {▱\varphi \land ▱\psi }\right) \rightarrow ▱\left( {\varphi \land \psi }\right) \)
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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. \( ▱\left( {\varphi \rightarrow \left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) }\right) \rightarrow \left( {▱\varphi \rightarrow ▱\left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) }\right) \)\n\n\( \mathrm{K} \)\n\n4. \( ▱\varphi \rightarrow ▱\left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) \)\n\nMP, 2, 3\n\n5. \( ▱\left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) \rightarrow \left( {▱\psi \rightarrow ▱\left( {\varphi \land \psi }\right) }\right) \)\n\n\( \mathrm{K} \)\n\n6. \( \;\left( {▱\varphi \rightarrow ▱\left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) }\right) \rightarrow \)\n\n\[ \left( {▱\left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) \rightarrow \left( {▱\psi \rightarrow ▱\left( {\varphi \land \psi }\right) }\right) }\right) \rightarrow \)\n\n\( \left( {\square \varphi \rightarrow \left( {\square \psi \rightarrow \square \left( {\varphi \land \psi }\right) }\right) }\right) ) \)\n\nTAUT\n\n7. \( \;\left( {▱\left( {\psi \rightarrow \left( {\varphi \land \psi }\right) }\right) \rightarrow \left( {▱\psi \rightarrow ▱\left( {\varphi \land \psi }\right) }\right) }\right) \rightarrow \)\n\n\( \left( {▱\varphi \rightarrow \left( {▱\psi \rightarrow ▱\left( {\varphi \land \psi }\right) }\right) }\right) \)\n\nMP, 4, 6\n\n8. \( ▱\varphi \rightarrow \left( {▱\psi \rightarrow ▱\left( {\varphi \land \psi }\right) }\right) ) \)\n\nMP, 5, 7\n\n9. \( \left( {▱\varphi \rightarrow \left( {▱\psi \rightarrow ▱\left( {\varphi \land \psi }\right) }\right) }\right) ) \rightarrow \)\n\n\[ \left( {\left( {▱\varphi \land ▱\psi }\right) \rightarrow ▱\left( {\varphi \land \psi }\right) }\right) \]\n\nTAUT\n\n10. \( \left( {▱\varphi \land ▱\psi }\right) \rightarrow ▱\left( {\varphi \land \psi }\right) \)\n\nMP, 8, 9\n\nThe formulas on lines 6 and 9 are instances of the tautologies\n\n\[ \left( {p \rightarrow q}\right) \rightarrow \left( {\left( {q \rightarrow r}\right) \rightarrow \left( {p \rightarrow r}\right) }\right) \]\n\n\[ \left( {p \rightarrow \left( {q \rightarrow r}\right) }\right) \rightarrow \left( {\left( {p \land q}\right) \rightarrow r}\right) \]
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Yes
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Proposition 49.15. \( \mathbf{K} \vdash \neg ▱p \rightarrow \Diamond \neg p \)
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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. \( \neg \neg p \rightarrow p\; \) TAUT\n\n5. \( ▱\left( {\neg \neg p \rightarrow p}\right) \; \) NEC,4\n\n6. \( ▱\left( {\neg \neg p \rightarrow p}\right) \rightarrow \left( {▱\neg \neg p \rightarrow ▱p}\right) \;\mathrm{K} \)\n\n7. \( \left( {▱\neg \neg p \rightarrow ▱p}\right) \;\mathrm{{MP}},5,6 \)\n\n8. \( \left( {▱\neg \neg p \rightarrow ▱p}\right) \rightarrow \left( {\neg ▱p \rightarrow \neg ▱\neg \neg p}\right) \; \) TAUT\n\n9. \( \neg ▱p \rightarrow \neg ▱\neg \neg p\;\mathrm{{MP}},7,8 \)\n\n10. \( \left( {\neg ▱p \rightarrow \neg ▱\neg \neg p}\right) \rightarrow \)\n\n\( \left( {\left( {\neg \square \neg \neg p \rightarrow \Diamond \neg p}\right) \rightarrow \left( {\neg \square p \rightarrow \Diamond \neg p}\right) }\right) \; \) TAUT\n\n11. \( \;\left( {\neg ▱\neg \neg p \rightarrow \Diamond \neg p}\right) \rightarrow \left( {\neg ▱p \rightarrow \Diamond \neg p}\right) \; \) MP,9,10\n\n12. \( \neg ▱p \rightarrow \diamond \neg p\;\mathrm{{MP}},3,{11} \)
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Yes
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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 \( \mathbf{K} \vdash \psi \) .
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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 derivation of \( \psi \) .
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Yes
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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 \( \mathbf{K} \vdash ▱{\varphi }_{1} \rightarrow \left( {▱{\varphi }_{2} \rightarrow \cdots \left( {▱{\varphi }_{n - 1} \rightarrow ▱{\varphi }_{n}}\right) \ldots }\right) \) .
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Proof. By induction on \( n \), just as in the proof of Proposition 49.6.
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No
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Proposition 49.18. \( \mathbf{K} \vdash \left( {▱\varphi \land ▱\psi }\right) \rightarrow ▱\left( {\varphi \land \psi }\right) \)
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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. \( \mathbf{K} \vdash \left( {▱\varphi \land ▱\psi }\right) \rightarrow ▱\left( {\varphi \land \psi }\right) \; \) PL,2
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Yes
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Proposition 49.19. If \( \mathbf{K} \vdash \varphi \leftrightarrow \psi \) and \( \mathbf{K} \vdash \chi \left\lbrack {\varphi /q}\right\rbrack \) then \( \mathbf{K} \vdash \chi \left\lbrack {B/q}\right\rbrack \)
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Proof. Exercise.
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No
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Proposition 49.20. \( \mathbf{K} \vdash \neg ▱p \rightarrow \Diamond \neg p \)
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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 \
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Yes
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Proposition 49.21. If \( \varphi \) is a substitution instance of \( \psi \) and \( \mathbf{K} \vdash \psi \), then \( \mathbf{K} \vdash \varphi \) .
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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 of instances of DUAL and \( \mathrm{K} \) are themselves instances of DUAL and \( \mathrm{K} \), and applications of MP and NEC remain correct when substituting formulas for propositional variables in both premise(s) and conclusion.
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No
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Proposition 49.22. \( \mathbf{K} \vdash ▱\left( {\varphi \rightarrow \psi }\right) \rightarrow \left( {\Diamond \varphi \rightarrow \Diamond \psi }\right) \)
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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. \( \mathbf{K} \vdash \left( {▱\neg \psi \rightarrow ▱\neg \varphi }\right) \rightarrow \left( {\neg ▱\neg \varphi \rightarrow \neg ▱\neg \psi }\right) \; \) TAUT\n\n4. \( \;\mathbf{K} \vdash \left( {\square \neg \psi \rightarrow \square \neg \varphi }\right) \rightarrow \left( {\neg \square \neg \varphi \rightarrow \neg \square \neg \psi }\right) \; \) PL,2,3\n\n5. \( \;\mathbf{K} \vdash ▱\left( {\varphi \rightarrow \psi }\right) \rightarrow \left( {\diamond \varphi \rightarrow \diamond \psi }\right) \;\diamond \) for \( \neg ▱\neg \)
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Yes
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Proposition 49.23. \( \mathbf{K} \vdash ▱\varphi \rightarrow \left( {\Diamond \left( {\varphi \rightarrow \psi }\right) \rightarrow \Diamond \psi }\right) \)
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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. \( \mathbf{K} \vdash ▱\varphi \rightarrow \left( {\neg ▱\neg \left( {\varphi \rightarrow \psi }\right) \rightarrow \neg ▱\neg \psi }\right) \; \) PL,2\n\n4. \( \mathbf{K} \vdash ▱\varphi \rightarrow \left( {\diamond \left( {\varphi \rightarrow \psi }\right) \rightarrow \diamond \psi }\right) \;\diamond \) for \( \neg ▱\neg \) .
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Yes
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Proposition 49.24. \( \mathbf{K} \vdash \left( {\Diamond \varphi \vee \Diamond \psi }\right) \rightarrow \Diamond \left( {\varphi \vee \psi }\right) \)
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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. \( \mathbf{K} \vdash \neg ▱\neg \varphi \rightarrow \neg ▱\neg \left( {\varphi \vee \psi }\right) \; \) PL,2\n\n4. \( \mathbf{K} \vdash \Diamond \varphi \rightarrow \Diamond \left( {\varphi \vee \psi }\right) \;\Diamond \) for \( \neg ▱\neg \)\n\n5. \( \mathbf{K} \vdash \diamond \psi \rightarrow \diamond \left( {\varphi \vee \psi }\right) \; \) similarly\n\n6. \( \mathbf{K} \vdash \left( {\Diamond \varphi \vee \Diamond \psi }\right) \rightarrow \Diamond \left( {\varphi \vee \psi }\right) \; \) PL,4,5.
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Yes
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Proposition 49.25. \( \mathbf{K} \vdash \diamond \left( {\varphi \vee \psi }\right) \rightarrow \left( {\diamond \varphi \vee \diamond \psi }\right) \)
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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. \( \;\mathbf{K} \vdash ▱\neg \varphi \rightarrow \left( {\neg ▱\neg \left( {\varphi \vee \psi }\right) \rightarrow \neg ▱\neg \psi }\right) )\; \) PL,2\n\n4. \( \mathbf{K} \vdash \neg ▱\neg \left( {\varphi \vee \psi }\right) \rightarrow \left( {▱\neg \varphi \rightarrow \neg ▱\neg \psi }\right) \)\n\n5. \( \mathbf{K} \vdash \neg ▱\neg \left( {\varphi \vee \psi }\right) \rightarrow \left( {\neg \neg ▱\neg \psi \rightarrow \neg ▱\neg \varphi }\right) \)\n\n6. \( \mathbf{K} \vdash \diamond \left( {\varphi \vee \psi }\right) \rightarrow \left( {\neg \diamond \psi \rightarrow \diamond \varphi }\right) \;\diamond \) for \( \neg ▱\neg \)\n\n7. \( \;\mathbf{K} \vdash \diamond \left( {\varphi \vee \psi }\right) \rightarrow \left( {\diamond \psi \vee \diamond \varphi }\right) \; \) PL,6.
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Yes
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Proposition 49.27. The following provability results obtain:\n\n1. \( \mathrm{KT}5 \vdash \mathrm{B} \) ;
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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.
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Yes
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Proposition 49.29. KTB4 \( = \) KT5 \( = \) KDB4 \( = \) KDB5.
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Proof. Exercise.
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No
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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 models \( {\mathcal{C}}_{1} \cap \cdots \cap {\mathcal{C}}_{n} \) .
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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 \( {\varphi }_{1},\ldots ,{\varphi }_{n} \). In the first case, \( \psi \) is valid in \( \mathcal{C} \), since tautological instance are valid in any class of models, by Proposition 47.16. Similarly in the second case, by Proposition 47.19 and Proposition 47.20. Finally in the third case, since \( \psi \) is valid in \( {\mathcal{C}}_{i} \) and \( \mathcal{C} \subseteq {\mathcal{C}}_{i} \), we have that \( \psi \) is valid in \( \mathcal{C} \) as well.\n\n2. Inductive step: Suppose \( \psi \) has a proof of length \( k > 1 \). If \( \psi \) is a tautological instance or an instance of one of \( {\varphi }_{1},\ldots ,{\varphi }_{n} \), we proceed as in the previous step. So suppose \( \psi \) is obtained by MP from previous formulas \( \chi \rightarrow \psi \) and \( \chi \). Then \( \chi \rightarrow \psi \) and \( \chi \) have proofs of length \( < k \), and by inductive hypothesis they are valid in \( \mathcal{C} \). By Proposition 47.21, \( \psi \) is valid in \( \mathcal{C} \). Finally suppose \( \psi \) is obtained by NEC from \( \chi \) (so that \( \psi = ▱\chi \)). By inductive hypothesis, \( \chi \) is valid in \( \mathcal{C} \), and by Proposition 47.13 so is \( \psi \).
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Yes
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Theorem 49.34. \( \mathrm{{KD}}5 \neq \mathrm{{KT}}4 = \mathrm{S}4 \) .
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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 \( \mathfrak{M},{w}_{1} \nVdash ▱p \rightarrow ▱▱p \) .
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Yes
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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 \( \Gamma { \vdash }_{\sum }\varphi \) ;\n\n3. Cut: If \( \Gamma { \vdash }_{\sum }\varphi \) and \( \Delta \cup \{ \varphi \} { \vdash }_{\sum }\psi \) then \( \Gamma \cup \Delta { \vdash }_{\sum }\psi \) ;\n\n4. Deduction theorem: \( \Gamma \cup \{ \psi \} { \vdash }_{\sum }\varphi \) if and only if \( \Gamma { \vdash }_{\sum }\psi \rightarrow \varphi \) ;\n\n5. \( \Gamma { \vdash }_{\sum }{\varphi }_{1} \) and \( \ldots \) and \( \Gamma { \vdash }_{\sum }{\varphi }_{n} \) and \( {\varphi }_{1} \rightarrow \left( {{\varphi }_{2} \rightarrow \cdots \left( {{\varphi }_{n} \rightarrow \psi }\right) \cdots }\right) \) is a tautological instance, then \( \Gamma { \vdash }_{\sum }\psi \) .
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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 of \( \Gamma \cup \{ \varphi \} { \vdash }_{\sum }\psi \) .
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No
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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 \( \Gamma { \nvdash }_{\sum }\varphi \) .
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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 \( \Gamma ,\neg \varphi { \vdash }_{\sum } \bot \) . By the deduction theorem \( \Gamma { \vdash }_{\sum }\varphi \rightarrow \bot \) and \( \Gamma { \vdash }_{\sum }\neg \varphi \rightarrow \bot \) . But \( \left( {\varphi \rightarrow \bot }\right) \rightarrow ((\neg \varphi \rightarrow \) \( \bot ) \rightarrow \bot ) \) is a tautological instance, hence by Proposition \( {49.36}\left( 5\right) ,\Gamma { \vdash }_{\sum } \bot \) .
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Yes
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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. \( \varphi \land \psi \in \Gamma \) iff \( \varphi \in \Gamma \) and \( \psi \in \Gamma \)\n\n6. \( \varphi \vee \psi \in \Gamma \) iff \( \varphi \in \Gamma \) or \( \psi \in \Gamma \)\n\n7. \( \varphi \rightarrow \psi \in \Gamma \) iff \( \varphi \notin \Gamma \) or \( \psi \in \Gamma \)
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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 \( \varphi \in \sum \) then \( \Gamma { \vdash }_{\sum }\varphi \), and \( \varphi \in \Gamma \) by deductive closure, i.e., case (1).\n\n3. If \( \bot \in \Gamma \), then \( \Gamma { \vdash }_{\sum } \bot \), so \( \Gamma \) would be \( \sum \) -inconsistent.\n\n4. If \( \neg \varphi \in \Gamma \), then by consistency \( \varphi \notin \Gamma \) ; and if \( \varphi \notin \Gamma \) then \( \varphi \in \Gamma \) since \( \Gamma \) is complete \( \sum \) -consistent.\n\n5. Exercise.\n\n6. Suppose \( \varphi \vee \psi \in \Gamma \), and \( \varphi \notin \Gamma \) and \( \psi \notin \Gamma \) . Since \( \Gamma \) is complete \( \sum \) - consistent, \( \neg \varphi \in \Gamma \) and \( \neg \psi \in \Gamma \) . Then \( \neg \left( {\varphi \vee \psi }\right) \in \Gamma \) since \( \neg \varphi \rightarrow \) \( \left( {\neg \psi \rightarrow \neg \left( {\varphi \vee \psi }\right) }\right) \) is a tautological instance. This would mean that \( \Gamma \) is \( \sum \) -inconsistent, a contradiction.\n\n7. Exercise.
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No
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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 \( \sum \) .)
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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 \( \Delta { \vdash }_{\sum }\neg \varphi \) (by reflexivity), and so \( \Delta \) is inconsistent. Conversely if \( \Gamma { \nvdash }_{\sum }\varphi \), then \( \Gamma \cup \{ \neg \varphi \} \) is \( \sum \) -consistent, and by Lindenbaum’s Lemma there is a complete consistent set \( \Delta \) extending \( \Gamma \cup \{ \neg \varphi \} \) . By consistency, \( \varphi \notin \Delta \) .
|
Yes
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Lemma 50.7. If \( {▱}^{-1}\Gamma { \vdash }_{\sum }\varphi \) then \( \Gamma { \vdash }_{\sum }▱\varphi \) .
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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 \
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No
|
Lemma 50.9. Suppose \( \Gamma \) and \( \Delta \) are complete \( \sum \) -consistent. Then: \( {▱}^{-1}\Gamma \subseteq \Delta \) if and only if \( \Diamond \Delta \subseteq \Gamma \) .
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Proof. \
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No
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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 \( \varphi \in \Delta \) .
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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 Proposition 50.8, \( ▱\neg \varphi \notin \Gamma \) iff, for some complete \( \sum \) -consistent \( \Delta \) with \( {▱}^{-1}\Gamma \subseteq \Delta ,\neg \varphi \notin \Delta \) . Now consider any such \( \Delta \) . By Lemma 50.9, \( {▱}^{-1}\Gamma \subseteq \Delta \) iff \( \langle \Delta \subseteq \Gamma \) . Also, \( \neg \varphi \notin \Delta \) iff \( \varphi \in \Delta \) by Proposition 50.2(4). So \( \diamond \varphi \in \Gamma \) iff, for some complete \( \sum \) -consistent \( \Delta \) with \( \diamond \Delta \subseteq \Gamma \) , \( \varphi \in \Delta \) .
|
Yes
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Proposition 50.12 (Truth Lemma). For every formula \( \varphi ,{\mathfrak{M}}^{\sum },\Delta \Vdash \varphi \) if and only if \( \varphi \in \Delta \) .
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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 47.7. Also, \( \Delta \in {V}^{\sum }\left( p\right) \) iff \( p \in \Delta \) by definition of \( {V}^{\sum } \).\n\n3. \( \varphi \equiv \neg \psi : {\mathfrak{M}}^{\sum },\Delta \Vdash \neg \psi \) iff \( {\mathfrak{M}}^{\sum },\Delta \nVdash \psi \) (Definition 47.7) iff \( \psi \notin \Delta \) (by inductive hypothesis) iff \( \neg \psi \in \Delta \) (by Proposition 50.2(4)).\n\n4. \( \varphi \equiv \psi \land \chi \) : Exercise.\n\n5. \( \varphi \equiv \psi \vee \chi : {\mathfrak{M}}^{\sum },\Delta \Vdash \psi \vee \chi \) iff \( {\mathfrak{M}}^{\sum },\Delta \Vdash \psi \) or \( {\mathfrak{M}}^{\sum },\Delta \Vdash \chi \) (by Definition 47.7) iff \( \psi \in \Delta \) or \( \chi \in \Delta \) (by inductive hypothesis) iff \( \psi \vee \chi \in \Delta \) (by Proposition 50.2(6)).\n\n6. \( \varphi \equiv \psi \rightarrow \chi \) : Exercise.\n\n7. \( \varphi \equiv ▱\psi \) : First suppose that \( {\mathfrak{M}}^{\sum },\Delta \Vdash ▱\psi \) . By Definition 47.7, for every \( {\Delta }^{\prime } \) such that \( {R}^{\sum }\Delta {\Delta }^{\prime },{\mathfrak{M}}^{\sum },{\Delta }^{\prime } \Vdash \psi \) . By inductive hypothesis, for every \( {\Delta }^{\prime } \) such that \( {R}^{\sum }\Delta {\Delta }^{\prime },\psi \in {\Delta }^{\prime } \) . By definition of \( {R}^{\sum } \), for every \( {\Delta }^{\prime } \) such that \( {▱}^{-1}\Delta \subseteq {\Delta }^{\prime },\psi \in {\Delta }^{\prime } \) . By Proposition 50.8, \( ▱\psi \in \Delta \) .\n\nNow assume \( ▱\psi \in \Delta \) . Let \( {\Delta }^{\prime } \in {W}^{\sum } \) be such that \( {R}^{\sum }\Delta {\Delta }^{\prime } \), i.e., \( {▱}^{-1}\Delta \subseteq \) \( {\Delta }^{\prime } \) . Since \( ▱\psi \in \Delta ,\psi \in {▱}^{-1}\Delta \) . Consequently, \( \psi \in {\Delta }^{\prime } \) . By inductive hypothesis, \( {\mathfrak{M}}^{\sum },{\Delta }^{\prime } \Vdash \psi \) . Since \( {\Delta }^{\prime } \) is arbitrary with \( {R}^{\sum }\Delta {\Delta }^{\prime } \), for all \( {\Delta }^{\prime } \in {W}^{\sum } \) such that \( {R}^{\sum }\Delta {\Delta }^{\prime },{\mathfrak{M}}^{\sum },{\Delta }^{\prime } \Vdash \psi \) . By Definition 47.7, \( {\mathfrak{M}}^{\sum },\Delta \Vdash ▱\psi \) .\n\n8. \( \varphi \equiv \diamond \psi \) : Exercise.
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No
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Theorem 50.14 (Determination). \( {\mathfrak{M}}^{\sum } \Vdash \varphi \) if and only if \( \sum \vdash \varphi \) .
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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 \( \Gamma = \varnothing \) ), \( \sum \vdash \varphi \) .\n\nConversely, if \( \sum \vdash \varphi \) then by Proposition 50.2(1), every complete \( \sum \) -consistent \( \Delta \) contains \( \varphi \), and hence by the Truth Lemma, \( {\mathfrak{M}}^{\sum },\Delta \Vdash \varphi \) for every \( \Delta \in {W}^{\sum } \) , i.e., \( {\mathfrak{M}}^{\sum } \Vdash \varphi \) .
|
Yes
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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 \( \mathbf{K} \vdash \varphi \) .
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Proof. Contrapositively, if \( \mathbf{K} \nvdash \varphi \) then by Determination \( {\mathfrak{M}}^{\mathbf{K}} \nVdash \varphi \) and hence \( \varphi \) is not valid.
|
Yes
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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.
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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 then by Lindenbaum’s Lemma, there is a complete \( \sum \) -consistent set \( {\Delta }^{\prime } \supseteq {▱}^{-1}\Delta \), and by definition of \( {R}^{\sum } \) we have \( {R}^{\sum }\Delta {\Delta }^{\prime } \) . So, suppose for contradiction that \( {▱}^{-1}\Delta \) is not \( \sum \) -consistent, i.e., \( {▱}^{-1}\Delta { \vdash }_{\sum } \bot \) . By Lemma 50.7, \( \Delta { \vdash }_{\sum }▱ \bot \), and since \( \sum \)\n\n<table><thead><tr><th>If \( \sum \) contains …</th><th>. . . the canonical model for \( \sum \) is:</th></tr></thead><tr><td>D： \( ▱\varphi \rightarrow \Diamond \varphi \)</td><td>serial;</td></tr><tr><td>T:\( ▱\varphi \rightarrow \varphi \)</td><td>reflexive;</td></tr><tr><td>B:\( \varphi \rightarrow ▱\Diamond \varphi \)</td><td>symmetric;</td></tr><tr><td>4:\( ▱\varphi \rightarrow ▱▱\varphi \)</td><td>transitive;</td></tr><tr><td>5:\( \Diamond \varphi \rightarrow ▱\Diamond \varphi \)</td><td>euclidean.</td></tr></table>\n\nTable 50.1: Basic correspondence facts.\n\ncontains D, also \( \Delta { \vdash }_{\sum }\Diamond \bot \) . But \( \sum \) is normal, so \( \sum \vdash \neg \Diamond \bot \) (Proposition 49.7), whence also \( \Delta { \vdash }_{\sum }\neg \diamond \bot \), against the consistency of \( \Delta \) .
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No
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1. If \( \sum \) contains the schema \( \diamond \varphi \rightarrow ▱\varphi \) then the canonical model for \( \sum \) is partially functional.
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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 \( {R}^{\sum }{\Delta }_{1}{\Delta }_{3} \) then \( {\Delta }_{2} = {\Delta }_{3} \) . Since \( {R}^{\sum }{\Delta }_{1}{\Delta }_{2} \) we have \( {▱}^{-1}{\Delta }_{1} \subseteq \) \( {\Delta }_{2} \) and since \( {R}^{\sum }{\Delta }_{1}{\Delta }_{3} \) also \( {▱}^{-1}{\Delta }_{1} \subseteq {\Delta }_{3} \) . The identity \( {\Delta }_{2} = {\Delta }_{3} \) will follow if we can establish the two inclusions \( {\Delta }_{2} \subseteq {\Delta }_{3} \) and \( {\Delta }_{3} \subseteq {\Delta }_{2} \) . For the first inclusion, let \( \varphi \in {\Delta }_{2} \) ; then \( \diamond \varphi \in {\Delta }_{1} \), and by the schema and deductive closure of \( {\Delta }_{1} \) also \( ▱\varphi \in {\Delta }_{1} \), whence by the hypothesis that \( {R}^{\sum }{\Delta }_{1}{\Delta }_{3},\varphi \in {\Delta }_{3} \) . The second inclusion is similar.
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Yes
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\[ {\varphi }_{1},\ldots ,{\varphi }_{n}{ \vdash }_{\sum }\diamond \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \rightarrow \bot \]
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\[ {\varphi }_{1},\ldots ,{\varphi }_{n}{ \vdash }_{\sum }\neg \diamond \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \]
\[ \text{by PL} \]
\[ {\varphi }_{1},\ldots ,{\varphi }_{n}{ \vdash }_{\sum }▱\neg \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \]
\[ ▱\neg \text{for}\neg \diamond \]
\[ ▱{\varphi }_{1},\ldots ,▱{\varphi }_{n}{ \vdash }_{\sum }▱▱\neg \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \]
by Lemma 50.6
\[ ▱{\varphi }_{1},\ldots ,▱{\varphi }_{n}{ \vdash }_{\sum }▱\neg \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \]
\[ \text{by schema}▱▱\varphi \rightarrow ▱\varphi \]
\[ {\Delta }_{1}{ \vdash }_{\sum }▱\neg \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \]
by monotony, Proposition 49.36(1)
\[ ▱\neg \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \in {\Delta }_{1} \]
by deductive closure;
\[ \neg \left( {{\psi }_{1} \land \cdots \land {\psi }_{m}}\right) \in {\Delta }_{2} \]
\[ \text{since}{R}^{\sum }{\Delta }_{1}{\Delta }_{2}\text{.} \]
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Yes
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Proposition 51.3. Given \( \mathfrak{M} \) and \( \Gamma \) , \( \equiv \) as defined above is an equivalence relation, i.e., it is reflexive, symmetric, and transitive.
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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 formulas from \( \Gamma \) true as \( v \), and \( v \) as \( w \), then \( u \) makes the same formulas from \( \Gamma \) true as \( w \) .
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Yes
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Proposition 51.8. The finest filtration \( {\mathfrak{M}}^{ * } \) is indeed a filtration.
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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 (2a) is satisfied.\n\nFor (2b), suppose \( ▱\varphi \in \Gamma ,{R}^{ * }\left\lbrack u\right\rbrack \left\lbrack v\right\rbrack \), and \( \mathfrak{M}, u \Vdash ▱\varphi \) . By definition of \( {R}^{ * } \) , there are \( {u}^{\prime } \equiv u \) and \( {v}^{\prime } \equiv v \) such that \( R{u}^{\prime }{v}^{\prime } \) . Since \( u \) and \( {u}^{\prime } \) agree on \( \Gamma \), also \( \mathfrak{M},{u}^{\prime } \Vdash ▱\varphi \), so that \( \mathfrak{M},{v}^{\prime } \Vdash \varphi \) . By closure of \( \Gamma \) under sub-formulas, \( v \) and \( {v}^{\prime } \) agree on \( \varphi \), so \( \mathfrak{M}, v \Vdash \varphi \), as desired.\n\nWe leave the verification of (2c) as an exercise.
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No
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Proposition 51.10. The coarsest filtration \( {\mathfrak{M}}^{ * } \) is indeed a filtration.
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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 \( \mathfrak{M}, v \Vdash \varphi \), and (1) is satisfied. Suppose \( \diamond \varphi \in \Gamma \) and \( \mathfrak{M}, v \Vdash \varphi \) . Then \( \mathfrak{M}, u \Vdash \diamond \varphi \) since \( {Ruv} \), and (2) is satisfied.
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No
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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 only the even numbers.
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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 makes \( ▱p \) true and \( p \) and \( ▱p \rightarrow p \) false; every even number makes \( p \) and \( ▱p \rightarrow p \) true, but \( ▱p \) false. So \( {W}^{ * } = \{ \left\lbrack 1\right\rbrack ,\left\lbrack 2\right\rbrack \} \), where \( \left\lbrack 1\right\rbrack = \) \( \{ 1,3,5,\ldots \} \) and \( \left\lbrack 2\right\rbrack = \{ 2,4,6,\ldots \} \) . Since \( 2 \in V\left( p\right) ,\left\lbrack 2\right\rbrack \in {V}^{ * }\left( p\right) \) ; since \( 1 \notin \) \( V\left( p\right) ,\left\lbrack 1\right\rbrack \notin {V}^{ * }\left( p\right) \) . So \( {V}^{ * }\left( p\right) = \{ \left\lbrack 2\right\rbrack \} \) . Any filtration based on \( {W}^{ * } \) must have an accessibility relation that includes \( \langle \left\lbrack 1\right\rbrack ,\left\lbrack 2\right\rbrack \rangle ,\langle \left\lbrack 2\right\rbrack ,\left\lbrack 1\right\rbrack \rangle \) : since \( {R12} \), we must have \( {R}^{ * }\left\lbrack 1\right\rbrack \left\lbrack 2\right\rbrack \) by Definition 51.4(2a), and since \( {R23} \) we must have \( {R}^{ * }\left\lbrack 2\right\rbrack \left\lbrack 3\right\rbrack \), and \( \left\lbrack 3\right\rbrack = \left\lbrack 1\right\rbrack \) . It cannot include \( \langle \left\lbrack 1\right\rbrack ,\left\lbrack 1\right\rbrack \rangle \) : if it did, we’d have \( {R}^{ * }\left\lbrack 1\right\rbrack \left\lbrack 1\right\rbrack ,\mathfrak{M},1 \Vdash ▱p \) but \( \mathfrak{M},1 \Vdash p \), contradicting (2b). Nothing requires or rules out that \( {R}^{ * }\left\lbrack 2\right\rbrack \left\lbrack 2\right\rbrack \) . So, there are two possible filtrations of \( \mathfrak{M} \), corresponding to the two accessibility relations \[ \{ \langle \left\lbrack 1\right\rbrack ,\left\lbrack 2\right\rbrack \rangle ,\langle \left\lbrack 2\right\rbrack ,\left\lbrack 1\right\rbrack \rangle \} \text{and}\{ \langle \left\lbrack 1\right\rbrack ,\left\lbrack 2\right\rbrack \rangle ,\langle \left\lbrack 2\right\rbrack ,\left\lbrack 1\right\rbrack \rangle ,\langle \left\lbrack 2\right\rbrack ,\left\lbrack 2\right\rbrack \rangle \} \text{.} \] In either case, \( p \) and \( ▱p \rightarrow p \) are false and \( ▱p \) is true at \( \left\lbrack 1\right\rbrack ;p \) and \( ▱p \rightarrow p \) are true and \( ▱p \) is false at \( \left\lbrack 2\right\rbrack \) .
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Yes
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Proposition 51.12. If \( \Gamma \) is finite then any filtration \( {\mathfrak{M}}^{ * } \) of a model \( \mathfrak{M} \) through \( \Gamma \) is also finite.
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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 \( \Gamma ,\varphi \in \Gamma \) either \( \varphi \) is true at both \( u \) and \( v \), or at neither. So each class \( \left\lbrack w\right\rbrack \) corresponds to subset of \( \Gamma \), namely the set of all \( \varphi \in \Gamma \) such that \( \varphi \) is true at the worlds in \( \left\lbrack w\right\rbrack \) . No two different classes \( \left\lbrack u\right\rbrack \) and \( \left\lbrack v\right\rbrack \) correspond to the same subset of \( \Gamma \) . For if the set of formulas true at \( u \) and that of formulas true at \( v \) are the same, then \( u \) and \( v \) agree on all formulas in \( \Gamma \), i.e., \( u \equiv v \) . But then \( \left\lbrack u\right\rbrack = \left\lbrack v\right\rbrack \) . So, there is an injective function from \( {W}^{ * } \) to \( \wp \left( \Gamma \right) \), and hence \( \left| {W}^{ * }\right| \leq \left| {\wp \left( \Gamma \right) }\right| \) . Hence if \( \Gamma \) contains \( n \) sentences, the cardinality of \( {W}^{ * } \) is no greater than \( {2}^{n} \) .
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Yes
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Proposition 51.14. K has the finite model property.
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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 formula only has finitely many sub-formulas, so \( \Gamma \) is finite. By Proposition 51.12, \( \left| {W}^{ * }\right| \leq {2}^{n} \), where \( n \) is the number of formulas in \( \Gamma \) . And since \( \mathbf{K} \) imposes no restriction on models, \( {\mathfrak{M}}^{ * } \) is a \( \mathbf{K} \) -model.
|
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 \( {\mathcal{U}}_{\text{Fin }} \) .
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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 \( \Gamma \) is finite. Take a filtration \( {\mathfrak{M}}^{ * } \) of \( \mathfrak{M} \) ; then \( {\mathfrak{M}}^{ * } \) is finite by Proposition 51.12, and by Theorem 51.5, \( \varphi \) is false at \( \left\lbrack w\right\rbrack \) in \( {\mathfrak{M}}^{ * } \) . It remains to observe that \( {\mathfrak{M}}^{ * } \) is also universal: given \( u \) and \( v \), by hypothesis \( {Ruv} \) and by Definition 51.4(2), also \( {R}^{ * }\left\lbrack u\right\rbrack \left\lbrack v\right\rbrack \) .
|
Yes
|
Theorem 51.17. S5 is decidable.
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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 theorems of S5 by generating proofs in some systematic way; and all the models containing \( 1,2,\ldots \) worlds and checking whether \( \varphi \) fails at a world in some such model. Eventually one of the two parallel processes will give an answer, as by Theorem 50.17 and Corollary 51.16, either \( \varphi \) is derivable or it fails in a finite universal model.
|
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 \( \mathfrak{M} \) is symmetric.
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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 filtration through \( \Gamma \) . Condition \( {C}_{1}\left( {u, v}\right) \) guarantees that (2b) and (2c) of Definition 51.4 are satisfied. So we just have to verify Definition 51.4(2a), i.e., that \( {Ruv} \) implies \( {R}^{ * }\left\lbrack u\right\rbrack \left\lbrack v\right\rbrack \) .\n\nSo suppose \( {Ruv} \) . To show \( {R}^{ * }\left\lbrack u\right\rbrack \left\lbrack v\right\rbrack \) we need to establish that \( {C}_{1}\left( {u, v}\right) \) and \( {C}_{2}\left( {u, v}\right)
|
No
|
1. If \( \mathfrak{M} \) is symmetric, so is \( {\mathfrak{M}}^{ * } \) .
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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 show \( {C}_{1}\left( {u, w}\right) \) . Suppose \( \mathfrak{M}, u \Vdash ▱\varphi \) ; then \( \mathfrak{M}, u \Vdash ▱▱\varphi \) since 4 is valid in all transitive models; since \( ▱▱\varphi \in \Gamma \) by closure, also by\n\n\( {C}_{1}\left( {u, v}\right) ,\mathfrak{M}, v \Vdash ▱\varphi \) and by \( {C}_{1}\left( {v, w}\right) \), also \( \mathfrak{M}, w \Vdash \varphi \) . Suppose \( \mathfrak{M}, w \Vdash \varphi \) ; then \( \mathfrak{M}, v \Vdash \diamond \varphi \) by \( {C}_{1}\left( {v, w}\right) \), since \( \diamond \varphi \in \Gamma \) by modal closure. By \( {C}_{1}\left( {u, v}\right) \) , we get \( \mathfrak{M}, u \Vdash \diamond \diamond \varphi \) since \( \diamond \diamond \varphi \in \Gamma \) by modal closure. Since \( {4}_{\diamond } \) is valid in all transitive models, \( \mathfrak{M}, u \Vdash \diamond \varphi \) .
|
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 \( \Gamma \vDash \varphi \) .
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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 \( \mathfrak{M} \) and \( w,\mathfrak{M}, w \Vdash {\psi }_{i} \) for \( i = 1,\ldots, n \), but \( \mathfrak{M}, w \nVdash \varphi \) . Let \( f\left( 1\right) = w \) ; then \( f \) is an interpretation of \( P\left( \Delta \right) \) into \( \mathfrak{M} \), and \( \mathfrak{M} \) satisfies \( \Delta \) with respect to \( f \) . But by Theorem 52.6, \( \Delta \) is unsatisfiable since it has a closed tableau, a contradiction. So we must have \( \Gamma \vdash \varphi \) after all.
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Yes
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We give a closed tableau that shows \( \mathbf{S}\mathbf{5} \vdash 5 \), i.e., \( ▱\varphi \rightarrow ▱\diamond \varphi \) .
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<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><td>\( {1.1} \) F \( \land \varphi \)</td><td>\( ▱ \) F 3</td></tr><tr><td>5.</td><td>\( 1\mathbb{F}\;\Diamond \varphi \)</td><td>4r \( \diamond \) 4</td></tr><tr><td>6.</td><td>\( {1.1}\mathrm{\;F}\varphi \)</td><td>◇ IF 5</td></tr><tr><td>7.</td><td>\( {1.1} \) T \( \varphi \)</td><td>\( ▱ \) T2</td></tr><tr><td></td><td>\( \otimes \)</td><td></td></tr></table>
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Yes
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Proposition 52.10. T \( ▱ \) and T \( \diamond \) are sound for reflexive models.
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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 \( R \) is reflexive, we know that \( {Rf}\left( \sigma \right) f\left( \sigma \right) \) . Hence, \( \mathfrak{M}, f\left( \sigma \right) \Vdash \psi \), i.e., \( \mathfrak{M}, f \) satisfies \( \sigma \mathbb{T}\psi \) .\n\n2. The branch is expanded by applying \( \mathrm{T}\diamond \) to \( \sigma \mathbb{F}\diamond \psi \in \Gamma \) : Exercise.
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No
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Proposition 52.11. \( \mathrm{D}▱ \) and \( \mathrm{D}\diamond \) are sound for serial models.
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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, \( \mathfrak{M}, f\left( \sigma \right) \Vdash ▱\psi \) . Since \( R \) is serial, there is a \( w \in W \) such\n\nthat \( {Rf}\left( \sigma \right) w \) . Then \( \mathfrak{M}, w \Vdash \psi \), and hence \( \mathfrak{M}, f\left( \sigma \right) \Vdash \diamond \psi \) . So, \( \mathfrak{M}, f \) satisfies \( \sigma \mathbb{T}\diamond \psi \
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Yes
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Proposition 52.12. \( \mathrm{B}▱ \) and \( \mathrm{B}\diamond \) are sound for symmetric models.
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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 \( f \) is an interpretation of prefixes on the branch into \( \mathfrak{M} \), we know that \( {Rf}\left( \sigma \right) f\left( {\sigma .n}\right) \) . Since \( R \) is symmetric, \( {Rf}\left( {\sigma .n}\right) f\left( \sigma \right) \) . Since \( \mathfrak{M}, f\left( {\sigma .n}\right) \Vdash ▱\psi ,\mathfrak{M}, f\left( \sigma \right) \Vdash \psi \) . Hence, \( \mathfrak{M}, f \) satisfies \( \sigma \mathbb{T}\psi \) .\n\n2. The branch is expanded by applying B \( \diamond \) to \( \sigma .n\mathbb{F}\diamond \psi \in \Gamma \) : Exercise.
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No
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Proposition 52.13. \( 4▱ \) and \( 4\diamond \) are sound for transitive models.
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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 \( f \) is an interpretation of prefixes on the branch into \( \mathfrak{M} \) and \( \sigma .n \) must be used, we know that \( R\widetilde{f}\left( \sigma \right) f\left( {\sigma .n}\right) \) . Now let \( w \) be any world such that \( {Rf}\left( {\sigma .n}\right) w \) . Since \( R \) is transitive, \( {Rf}\left( \sigma \right) w \) . Since \( \mathfrak{M}, f\left( \sigma \right) \Vdash ▱\psi ,\mathfrak{M}, w \Vdash \psi \) . Hence, \( \mathfrak{M}, f\left( {\sigma .n}\right) \Vdash ▱\psi \) , and \( \mathfrak{M}, f \) satisfies \( \sigma .n\mathbb{T}▱\psi \) .\n\n2. The branch is expanded by applying \( 4\diamond \) to \( \sigma \mathbb{F}\diamond \psi \in \Gamma \) : Exercise.
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No
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Proposition 52.14. \( 4\mathrm{r}▱ \) and \( 4\mathrm{r}\diamond \) are sound for euclidean models.
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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 \( f \) is an interpretation of prefixes on the branch into \( \mathfrak{M} \), we know that \( {Rf}\left( \sigma \right) f\left( {\sigma .n}\right) \) . Now let \( w \) be any world such that \( {Rf}\left( \sigma \right) w \) . Since \( R \) is euclidean, \( {Rf}\left( {\sigma .n}\right) w \) . Since \( \mathfrak{M}, f\left( \sigma \right) .n \Vdash \) \( ▱\psi ,\mathfrak{M}, w \Vdash \psi \) . Hence, \( \mathfrak{M}, f\left( \sigma \right) \Vdash ▱\psi \), and \( \mathfrak{M}, f \) satisfies \( \sigma \mathbb{T}▱\psi \) .\n\n2. The branch is expanded by applying \( 4\mathrm{r}\diamond \) to \( \sigma .n\mathbb{F}\diamond \psi \in \Gamma \) : Exercise.
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No
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Example 52.16. We give a simplified closed tableau that shows \( \mathrm{S}5 \vdash 5 \), i.e., \( \diamond \varphi \rightarrow ▱\diamond \varphi \) .
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<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.</td><td>2 \( \mathbb{F}\;\Diamond \varphi \)</td><td>\( ▱ \) F3</td></tr><tr><td>5.</td><td>3 T \( \varphi \)</td><td>\( \Diamond \mathbb{T} \) 2</td></tr><tr><td>6.</td><td>3 IF \( \varphi \)</td><td>\( \Diamond \mathbb{F} \) 4</td></tr><tr><td></td><td>\( \otimes \)</td><td></td></tr></table>
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Yes
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Proposition 52.18. Every finite \( \Gamma \) has a tableau in which every branch is complete.
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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 to it to extend the branch. In some cases this will result in branching; apply the rule at the tip of each resulting branch for all remaining prefixed formulas. Since the number of prefixed formulas is finite, and the number of used prefixes on the branch is finite, this procedure eventually results in (possibly many) branches extending the original branch. Apply the procedure to each, and repeat. But by construction, every branch is closed.
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No
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We know that \( \nvdash ▱\left( {p \vee q}\right) \rightarrow \left( {▱p \vee ▱q}\right) \) .
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<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>\( 1\;\mathbb{F}\;▱p \vee ▱q\;\checkmark \)</td><td>\( \rightarrow \mathbb{F}1 \)</td></tr><tr><td>4.</td><td>\( 1\;\mathbb{F}\;▱p\;\checkmark \)</td><td>VIF 3</td></tr><tr><td>5.</td><td>\( 1\;\mathbb{F}\;▱q\;\checkmark \)</td><td>VIF 3</td></tr><tr><td>6.</td><td>\( {1.1} \) F \( p\checkmark \)</td><td>\( ▱ \) IF 4</td></tr><tr><td>7.</td><td>\( {1.2} \) F \( q\checkmark \)</td><td>\( ▱ \) IF 5</td></tr></table>
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Yes
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Theorem 53.1. There are irrational numbers \( a \) and \( b \) such that \( {a}^{b} \) is rational.
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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, let \( a \) be \( {\sqrt{2}}^{\sqrt{2}} \), and let \( b \) be \( \sqrt{2} \) .
|
Yes
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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.
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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
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Let us prove \( \varphi \rightarrow \neg \neg \varphi \) for any proposition \( \varphi \), which is \( \varphi \rightarrow \) \( \left( {\left( {\varphi \rightarrow \bot }\right) \rightarrow \bot }\right) \) .
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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 \( \left( {\varphi \rightarrow \bot }\right) \rightarrow \bot \) : We have to define a function \( g \) which, when given a construction \( h \) of \( \varphi \rightarrow \bot \) as input, outputs a construction of \( \bot \) . We can define \( g \) as follows: apply the input \( h \) to the construction \( M \) of \( \varphi \) (that we received earlier). Since the output \( h\left( M\right) \) of \( h \) is a construction of \( \bot \) , \( f\left( M\right) \left( h\right) = h\left( M\right) \) is a construction of \( \bot \) if \( M \) is a construction of \( \varphi \) .
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Yes
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Let us give a construction for \( \neg \left( {\varphi \land \neg \varphi }\right) \), i.e., \( \left( {\varphi \land \left( {\varphi \rightarrow \bot }\right) }\right) \rightarrow \) \( \bot \) .
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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 \( {N}_{1} \) is a construction of \( {\psi }_{1} \) and \( {N}_{2} \) is a construction of \( {\psi }_{2} \) . We can define functions \( {p}_{1} \) and \( {p}_{2} \) which recover from a construction of \( {\psi }_{1} \land {\psi }_{2} \) the constructions of \( {\psi }_{1} \) and \( {\psi }_{2} \), respectively: \[ {p}_{1}\left( \left\langle {{N}_{1},{N}_{2}}\right\rangle \right) = {N}_{1} \] \[ {p}_{2}\left( \left\langle {{N}_{1},{N}_{2}}\right\rangle \right) = {N}_{2} \] Here is what \( f \) does: First it applies \( {p}_{1} \) to its input \( M \) . That yields a construction of \( \varphi \) . Then it applies \( {p}_{2} \) to \( M \), yielding a construction of \( \varphi \rightarrow \bot \) . Such a construction, in turn, is a function \( {p}_{2}\left( M\right) \) which, if given as input a construction of \( \varphi \), yields a construction of \( \bot \) . In other words, if we apply \( {p}_{2}\left( M\right) \) to \( {p}_{1}\left( M\right) \), we get a construction of \( \bot \) . Thus, we can define \( f\left( M\right) = \) \( {p}_{2}\left( M\right) \left( {{p}_{1}\left( M\right) }\right) \) .
|
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 \( \left( {\varphi \land \psi }\right) \rightarrow \chi \) into a construction of \( \left( {\varphi \rightarrow \left( {\psi \rightarrow \chi }\right) }\right) \) .
|
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 \( \chi \) . \( \mathrm{{Ok}} \), this is confusing. We have to construct a certain function \( {h}_{g} \), which will be the output of \( f \) for input \( g \) . The input of \( {h}_{g} \) is a construction \( M \) of \( \varphi \) . The output of \( {h}_{g}\left( M\right) \) should be a function \( {k}_{M} \) from constructions \( N \) of \( \psi \) to constructions of \( \chi \) . Let \( {k}_{g, M}\left( N\right) = g\left( {\langle M, N\rangle }\right) \) . Remember that \( \langle M, N\rangle \) is a construction of \( \varphi \land \psi \) . So \( {k}_{g, M} \) is a construction of \( \psi \rightarrow \chi \) : it maps constructions \( N \) of \( \psi \) to constructions of \( \chi \) . Now let \( {h}_{g}\left( M\right) = {k}_{g, M} \) . That’s a function that maps constructions \( M \) of \( \varphi \) to constructions \( {k}_{g, M} \) of \( \psi \rightarrow \chi \) . Now let \( f\left( g\right) = {h}_{g} \) . That’s a function that maps constructions \( g \) of \( \left( {\varphi \land \psi }\right) \rightarrow \chi \) to constructions of \( \varphi \rightarrow \left( {\psi \rightarrow \chi }\right) \) . Whew!
|
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 \( f\left( g\right) \) is a construction of \( \bot \) if \( g \) is a construction of \( \neg \left( {\varphi \vee \neg \varphi }\right) \) .\n\nSuppose \( g \) is a construction of \( \neg \left( {\varphi \vee \neg \varphi }\right) \), i.e., a function that transforms a construction of \( \varphi \vee \neg \varphi \) into a construction of \( \bot \) . A construction of \( \varphi \vee \neg \varphi \) is a pair \( \langle s, M\rangle \) where either \( s = 1 \) and \( M \) is a construction of \( \varphi \), or \( s = 2 \) and \( M \) is a construction of \( \neg \varphi \) . Let \( {h}_{1} \) be the function mapping a construction \( {M}_{1} \) of \( \varphi \) to a construction of \( \varphi \vee \neg \varphi \) : it maps \( {M}_{1} \) to \( \left\langle {1,{M}_{2}}\right\rangle \) . And let \( {h}_{2} \) be the function mapping a construction \( {M}_{2} \) of \( \neg \varphi \) to a construction of \( \varphi \vee \neg \varphi \) : it maps \( {M}_{2} \) to \( \left\langle {2,{M}_{2}}\right\rangle \) .\n\nLet \( k \) be \( g \circ {h}_{1} \) : it is a function which, if given a construction of \( \varphi \), returns a construction of \( \bot \), i.e., it is a construction of \( \varphi \rightarrow \bot \) or \( \neg \varphi \) . Now let \( l \) be \( g \circ {h}_{2} \) . It is a function which, given a construction of \( \neg \varphi \), provides a construction of \( \bot \) . Since \( k \) is a construction of \( \neg \varphi, l\left( k\right) \) is a construction of \( \bot \) .\n\nTogether, what we’ve done is describe how we can turn a construction \( g \) of \( \neg \left( {\varphi \vee \neg \varphi }\right) \) into a construction of \( \bot \), i.e., the function \( f \) mapping a construction \( g \) of \( \neg \left( {\varphi \vee \neg \varphi }\right) \) to the construction \( l\left( k\right) \) of \( \bot \) is a construction of \( \neg \neg \left( {\varphi \vee \neg \varphi }\right) \) .
|
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.
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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.
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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 \( \mathfrak{M},{w}^{\prime } \Vdash \varphi \) .
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Proof. Exercise.
|
No
|
Proposition 54.5. 1. If \( \mathfrak{M}, w \Vdash \Gamma \) and \( \Gamma \vDash \varphi \), then \( \mathfrak{M}, w \Vdash \varphi \) .
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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 \( \mathfrak{M}, w \Vdash \varphi \) .
|
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 \( \Gamma \vDash {\varphi }_{n} \) . Note that if \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) is a derivation, so is \( {\varphi }_{1},\ldots ,{\varphi }_{k} \) for any \( k < n \) .\n\nThere are no derivations of length 0, so for \( n = 0 \) the claim holds vacuously. So the claim holds for all derivations of length \( < n \) . We distinguish cases according to the justification of \( {\varphi }_{n} \) .\n\n1. \( {\varphi }_{n} \) is an axiom. All axioms are valid, so \( \Gamma \vDash {\varphi }_{n} \) for any \( \Gamma \) .\n\n2. \( {\varphi }_{n} \in \Gamma \) . Then for any \( \mathfrak{M} \) and \( w \), if \( \mathfrak{M}, w \Vdash \Gamma \), obviously \( \mathfrak{M} \Vdash \Gamma {\varphi }_{n}\left\lbrack w\right\rbrack \), i.e., \( \Gamma \vDash \varphi \) .\n\n3. \( {\varphi }_{n} \) follows by MP from \( {\varphi }_{i} \) and \( {\varphi }_{j} \equiv {\varphi }_{i} \rightarrow {\varphi }_{n}.{\varphi }_{1},\ldots ,{\varphi }_{i} \) and \( {\varphi }_{1},\ldots ,{\varphi }_{j} \) are derivations from \( \Gamma \), so by inductive hypothesis, \( \Gamma \vDash {\varphi }_{i} \) and \( \Gamma \vDash {\varphi }_{i} \rightarrow {\varphi }_{n} \) .\n\nSuppose \( \mathfrak{M}, w \Vdash \Gamma \) . Since \( \mathfrak{M}, w \Vdash \Gamma \) and \( \Gamma \vDash {\varphi }_{i} \rightarrow {\varphi }_{n},\mathfrak{M}, w \Vdash {\varphi }_{i} \rightarrow {\varphi }_{n} \) . By definition, this means that for all \( {w}^{\prime } \) such that \( {Rw}{w}^{\prime } \), if \( \mathfrak{M},{w}^{\prime } \Vdash {\varphi }_{i} \) then \( \mathfrak{M},{w}^{\prime } \Vdash {\varphi }_{n} \) . Since \( R \) is reflexive, \( w \) is among the \( {w}^{\prime } \) such that \( {Rw}{w}^{\prime } \) , i.e., we have that if \( \mathfrak{M}, w \Vdash {\varphi }_{i} \) then \( \mathfrak{M}, w \Vdash {\varphi }_{n} \) . Since \( \Gamma \vDash {\varphi }_{i},\mathfrak{M}, w \Vdash {\varphi }_{i} \) . So, \( \mathfrak{M}, w \Vdash {\varphi }_{n} \), as we wanted to show.
|
Yes
|
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