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Lecture 8 Modal Logic 1 Review of Greek Lower-Case Letters ✤ Recall: we define formula as follows: ✤ ✤ ✤ We use lower-case greek letters for arbitrary formulas: ✤ ϕ or φ, phi: pronounced “phee” ψ, psi: pronounced “sai” χ, chi; pronouced “kai” α, alpha β, beta γ, gamma δ, delta ρ, rho, pronounced “row” θ, theta the basic propositional letters p,q,r, . . . . are formulas if ϕ is a formula, so is ¬ϕ, □ϕ, ⬦ϕ ✤ if ϕ, ψ, are formulas, then so are (ϕ ∧ ψ), (ϕ ∨ ψ), (ϕ → ψ), and (ϕ ↔ ψ). ✤ Nothing else is a formula. 2 Outline of Today’s Lecture ✤ I. Hilbert-Style Deductive System for Propositional Logic ✤ II. Hilbert-Style Deductive System for Modal Propositional Logic 3 Outline of Today’s Lecture ✤ I. Hilbert-Style Deductive System for Propositional Logic ✤ II. Hilbert-Style Deductive System for Modal Propositional Logic 4 Two Kinds of Deductive Systems for Propositional Logic ✤ One basic kind of deductive system for propositional logic is a natural deduction system. In this kind of system, there is an “introduction” rule for each connective and an “elimination” rule for each connective. For instance, the introduction rule for “and” might say: if you can deduce ϕ and if you can deduce ψ, then you can deduce ϕ ∧ ψ. 5 ✤ While natural deduction systems are easy to learn, they have lots of rules, and there is no real analogue for modal logic. ✤ Another kind of deductive system is a Hilbert-style deductive system. It has very few rules and compensates by having more axioms. A Hilbert-Style Deductive System for Propositional Logic ✤ The three parts of the system: ✤ I. A set of definitions and tautologies, which serve as our axioms. ✤ II. A set of rules, which say things like “if such and such is a theorem, then this related thing is as well.” ✤ ✤ III. A specification of what counts as a proof in the system. 6 Let’s begin with the Part III. Because it’s the simplest. A proof of ϕ, or a demonstration that ⊢ϕ, consists of a numbered list of well-formed formulas such that ✤ the list ends with ϕ ✤ each formula on the list is an axiom (from Part I) or an earlier theorem, or follows from some of the earlier lines by the rules (from Part II). Description of Part I ✤ We assume, when we are setting up the deductive system, that all the connectives besides arrow and negation are defined in terms of arrow and negation as follows (cf. section 2.1 of Sider): ✤ ϕ ∧ ψ is defined as ~(ϕ➝~ψ) ✤ ϕ ∨ ψ is defined as ~ϕ➝ψ ✤ ✤ ϕ ↔ ψ is defined as (ϕ ➝ ψ) ∧ (ψ ➝ ϕ) 7 Practically, this means that we can just state a small set of tautologies concerning arrow and negation, and not have to worry about stating specific axioms concerning “or” and “and” and “biconditional.” Description of Part I, Continued ✤ ✤ ✤ ✤ The tautologies of the deductive system are the following list from section 2.6 of Sider. ✤ ϕ➝(ψ➝ϕ) ✤ ϕ ➝ ( ψ ➝ χ)) ➝ ((ϕ ➝ ψ ) ➝ (ϕ ➝ χ)) PL2 You can draw a truth-table to check that each of these is a validity. ✤ (~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ) The “PL” stands for “propositional logic.” ✤ The way to think about PL1 is: if you assume ϕ, then whatever else ψ you assume, you still have ϕ. The way to view PL2 is to think of the “ϕ➝” as distributing through the conditional ψ➝χ. ✤ The way to think about PL3 is to think of ~ψ➝~ϕ as ψ∨~ϕ, and to think of ~ψ➝ϕ as ψ∨ϕ. Hence, an equivalent formulation of PL3 would be: (ψ∨~ϕ)➝((ψ∨ϕ) ➝ ψ) 8 PL1 PL3 Description of Part II Part II of the deductive system is a set of rules, which say things like “if such and such has a proof, then this related thing has a proof as well.” ✤ ✤ This means: if you have ϕ➝ψ at line ℓ𝓁, and you have ϕ at line ℓ𝓁’, then you may write ψ at any subsequent line ℓ𝓁’’. ✤ ℓ𝓁. ϕ➝ψ ✤ ℓ𝓁’. ϕ ✤ ℓ𝓁’’. ψ The first rule is: ✤ ϕ➝ψ ϕ ψ MP The MP stands for “modus ponens”. 9 MP, ℓ𝓁, ℓ𝓁’. Description of Part II ✤ The second rule is that you can substitute defined notions. ✤ For instance, you can do: ✤ ✤ ℓ𝓁. (ϕ➝ψ) ∧ (ψ➝ϕ) ✤ ✤ ℓ𝓁’. (ϕ ↔ ψ) ✤ (Defn of ↔, ℓ𝓁). ✤ ✤ You don’t need to substitute everywhere either. 10 For instance, suppose you had a proof that started out like: ℓ𝓁. (~ϕ➝ψ) ∧ (~ϕ➝ψ) ℓ𝓁’. (ϕ ∨ ψ)∧ (~ϕ➝ψ) (Defn, ℓ𝓁) In moving from ℓ𝓁 to ℓ𝓁’, we substitute the defined notion ϕ∨ψ in the first conjunct but not in the second conjunct of line ℓ𝓁. Theorem 1 + Discussion ✤ Let’s look at an example derivation of ϕ➝(ψ ∨ ϕ). ✤ 1. ϕ➝(~ψ➝ϕ) 2. ϕ➝(ψ ∨ ϕ) ✤ ✤ ✤ (PL1) (Defn 1) Note that line 1 is an instance of PL1 since we substitute in ~ψ for ψ. Of course, in general, you should not confuse ψ and ~ψ. 11 The idea of the axioms is that they hold for all formulas, and so PL1 ϕ➝(ψ➝ϕ) has all the following instances: ✤ ϕ➝(~ψ➝ϕ) ✤ (~ϕ ∧ ψ) ➝(~ψ➝ (~ϕ ∧ ψ)) ✤ ϕ➝(ϕ➝ϕ) ✤ ~ϕ➝(ψ➝~ϕ) Discussion, Continued ✤ But there’s a general warning that should be given here: ✤ ✤ ✤ In doing this kind of substitution, you have to substitute uniformly. For instance, PL1 ϕ➝(ψ➝ϕ) does not have the following substitution instance: (~ϕ ∧ ψ) ➝(ψ➝ϕ) For, you substituted in (~ϕ ∧ ψ) for ϕ only in the first instance. 12 The way to remember this is as follows: ✤ When doing substitution instances of axioms (or theorems), you must substitute uniformly. ✤ When doing substitution instances of definitions, you may substitute selectively. Example of Finding Substitution Instance of an Axiom ✤ Consider PL3: (~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ) ✤ What do you get if you substitute in ϕ ∧ ψ for ψ? ✤ A. (~ψ ➝ ~(ϕ ∧ ψ))➝((~ψ ➝ (ϕ ∧ ψ)) ➝ ψ) ✤ B. (~(ϕ ∧ ψ) ➝ ~ϕ)➝((~(ϕ ∧ ψ) ➝ ϕ) ➝ (ϕ ∧ ψ)) ✤ C. (~ ϕ ∧ ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ) ✤ D. (~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ϕ ∧ ψ) 13 Theorem 2 ✤ Prove: ϕ ➝ ϕ ✤ Proof: ✤ 1. ϕ ➝ (ϕ ➝ ϕ) PL1 ✤ 2. ϕ ➝ ((ϕ ➝ϕ) ➝ ϕ) PL1 ✤ 3. [ϕ ➝ ((ϕ ➝ϕ) ➝ ϕ)] ➝ [(ϕ ➝ (ϕ ➝ϕ)) ➝ (ϕ ➝ ϕ) ] PL2 ✤ 4. (ϕ ➝ (ϕ ➝ϕ)) ➝ (ϕ ➝ ϕ) MP, 2, 3 ✤ 5. ϕ ➝ ϕ MP, 1, 4 14 Theorem 3 ✤ Let’s look at an example derivation of ~~ϕ ∨ ~ϕ. ✤ 1. ~ϕ➝~ϕ 2. ~~ϕ ∨ ~ϕ ✤ (prev.) (Defn, 1) 15 In the first line, we appealed to the previous result Theorem 2. Of course, as we stated it there, the conclusion of Theorem 2 read: ϕ ➝ ϕ But again, the idea is that if you have a result, you can substitute uniformly. Completeness Theorem ✤ The important result about the Hilbert-style deductive calculus itself is the completeness theorem: ✤ A formula ϕ of propositional logic is has a proof if and only if ϕ is a tautology. ✤ Hence if you can use a truthtable to test that a formula is a tautology, then it has a proof in the Hilbert-style system. 16 ✤ It’s beyond the scope of the methods we’ve developed to actually prove the completeness theorem. ✤ But we can use this result to justify that anything we know to be a tautology of ordinary propositional logic has a proof in the Hilbert-style system. ✤ The following slides contain a big long list of tautologies. A Big List of Tautologies ✤ #1. ϕ➝(ψ➝ϕ) PL1 ✤ #2. ϕ ➝ ( ψ ➝ χ)) ➝ ((ϕ ➝ ψ ) ➝ (ϕ ➝ χ)) PL2 ✤ #3. (~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ) PL3 ✤ #4. (ϕ ∧ ψ) ➝ ϕ (analogue of “and” elim.) ✤ #5. (ϕ ∧ ψ) ➝ ψ (analogue of “and” elim.) ✤ #6. (ϕ ➝ (ψ ➝ (ϕ ∧ ψ)) (analogue of “and” intro.) ✤ #7. (ϕ ➝ ψ) ➝ ((ϕ ➝ χ) ➝ (ϕ ➝ (ψ ∧ χ) )) (analogue of “and” intro, with hypothesis ϕ) ✤ #8. (ϕ ➝ (ψ ➝ χ)) ➝ ( (ϕ ∧ ψ) ➝χ ) (analogue of “and” elim) 17 A Big List of Tautologies, Continued ✤ #9. ϕ ➝ (ϕ ∨ ψ) (analogue of “or” intro.) ✤ #10. ψ ➝ (ϕ ∨ ψ) ✤ #11. (ϕ ➝ χ) ➝ ( (ψ➝ χ) ➝ ((ϕ ∨ ψ) ➝ χ)) ) (analogue of “or” elimination) ✤ #12. (ϕ ∧ ψ) ↔ (ψ ∧ ϕ) (commutativity of conjunction) ✤ #13. (ϕ ∨ ψ) ↔ (ψ ∨ ϕ) (commutative of disjunction) ✤ #14. ( ϕ ∧ ( ψ ∧ χ )) ↔ ( (ϕ ∧ ψ) ∧ χ ) (associativity of conjunction) ✤ #15. ( ϕ ∨ ( ψ ∨ χ )) ↔ ( (ϕ ∨ ψ) ∨ χ ) (associativity of disjunction) (analogue of “or” intro.) 18 A Big List of Tautologies, Continued ✤ #16. ϕ ↔ ~~ ϕ (Double negation Law 1) ✤ #17. ϕ ∨ ~ ϕ (Double negation Law 2) ✤ #18. (ϕ ∨ ψ) ↔ ~(~ ϕ ∧ ~ ψ) (DeMorgan Law 1) ✤ #19. (ϕ ∧ ψ) ↔ ~(~ ϕ ∨ ~ψ) (DeMorgan Law 2) ✤ #20. (ϕ ∧ (ψ ∨ χ)) ↔ (ϕ ∧ ψ ) ∨ (ϕ ∧ χ)) (Distribution Law 1) ✤ #21. (ϕ ∨ (ψ ∧ χ)) ↔ (ϕ ∨ ψ ) ∧ (ϕ ∨ χ)) (Distribution Law 2) 19 A Helpful Derived Rule ✤ Here’s a helpful derived rule: If ϕ is a theorem, and if ψ is a theorem, then ϕ ∧ ψ is a theorem. ✤ Why should we trust this rule? ✤ Well, suppose that we had (1) ϕ assumption (2) ψ assumption ✤ From our list (#6) we have (3) (ϕ ➝ (ψ ➝ (ϕ ∧ ψ)) (PL) 20 ✤ Then from MP we obtain: (4) ψ ➝ (ϕ ∧ ψ) MP 1, 3 ✤ And from MP we obtain (5) ϕ ∧ ψ MP 2, 4 Another Helpful Derived Rule ✤ Here’s a helpful derived rule: If ϕ ➝ ψ is a theorem, and if ψ ➝ χ is a theorem, then ϕ ➝ χ is a theorem. ✤ Why should we trust this rule? ✤ Well, suppose that we had: (1) ϕ ➝ ψ assumption (2) ψ ➝ χ assumption ✤ From our derived rule, we have (3) (ϕ ➝ ψ) ∧ (ψ ➝ χ) DR 1,2 21 ✤ From truth-tables, we know that the following is a tautology: ((ϕ ➝ ψ) ∧ (ψ ➝ χ)) ➝ (ϕ ➝χ) Hence by the completeness theorems, we can put it on a line and justify it by ‘prop logic’: (4) ((ϕ ➝ ψ) ∧ (ψ ➝ χ)) ➝ (ϕ ➝χ) PL ✤ Then we can argue by MP: (5) ϕ ➝χ MP 3, 4 Summary of Axioms, Rules, Derived Rules, and Theorems of Hilbert-Style System Axioms: ✤ Derived Rules: ϕ ✤ ϕ➝(ψ➝ϕ) PL1 ϕ ➝ ( ψ ➝ χ)) ➝ ((ϕ ➝ ψ ) ➝ (ϕ ➝ χ)) PL2 (~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ) PL3 ϕ∧ψ ϕ ∧ ψ is defined as ~(ϕ➝~ψ) ϕ ∨ ψ is defined as ~ϕ➝ψ ϕ ↔ ψ is defined as (ϕ ➝ ψ) ∧ (ψ ➝ ϕ) ψ➝χ ϕ ➝χ Abbreviated in proofs as ‘DR’, simply because context allows us to figure out which one is intended. MP ψ Abbreviated in proofs as ‘MP’ ϕ➝ψ Rules: Modus Ponens: ϕ➝ψ ϕ ✤ ψ 22 Outline of Today’s Lecture ✤ I. Hilbert-Style Deductive System for Propositional Logic ✤ II. Hilbert-Style Deductive System for Modal Propositional Logic 23 The K-Deductive System ✤ Today we describe the so-called Kdeductive system for Modal Propositional Logic. ✤ The three parts of the system: ✤ I. A set of definitions and tautologies, which serve as our axioms. ✤ II. A set of rules, which say things like “if such and such is a theorem, then this related thing is as well.” ✤ III. A specification of what counts as a proof in the system. 24 ✤ Part I expands the list of tautologies by the K-axiom; and by a definition of diamond in terms of box. ✤ Part II expands the list of rules by a single rule, known as the necessitation rule. ✤ Part III is exactly like the deductive system for ordinary propositional logic. The K-axiom ✤ Our first axiom is the K-axiom, where the ‘K’ stands for Kripke. ✤ It say: □( ϕ → ψ ) → (□ϕ → □ψ) ✤ ✤ Intuitively, it’s saying that whatever necessarily follows from a necessary truth is itself necessary. ✤ Or, in a word, necessary truth is closed under logical consequence. That is, it says: if it’s necessary that ϕ implies ψ, then necessarily ϕ implies necessarily ψ. ✤ 25 Or, more colloquially, if ϕ→ψ had to be the case, and if ϕ had to be the case, then also ψ had to be the case. Example of Finding Substitution Instance of an Axiom ✤ Consider the K-axiom: □( ϕ → ψ ) → (□ϕ → □ψ) ✤ What do you get if you substitute in ϕ ∨ ψ for ψ? ✤ A. □( ϕ → ψ ) → (□ϕ → □(ϕ ∨ ψ)) ✤ B. □( (ϕ ∨ ψ) → ψ ) → (□(ϕ ∨ ψ) → □ψ) ✤ C. □( ϕ → (ϕ ∨ ψ) ) → (□ϕ → □ψ) ✤ D. □( ϕ → (ϕ ∨ ψ) ) → (□ϕ → □(ϕ ∨ ψ)) 26 The Necessitation Rule ✤ ✤ The necessitation rule says that if ϕ is a theorem, then also □ϕ is a theorem. ✤ So what does this rule say? ✤ It says that if ϕ is proven, then so is □ϕ. ✤ The necessitation rule is not saying that this is a tautology: Visually, it looks like this: ϕ □ϕ NEC (*) ϕ ➝ □ϕ Note: (*) is usually false: just because Bill went to beach doesn’t mean he had to. 27 A Helpful Derived Rule ✤ It’s helpful to note that we have the following derived rule: if ϕ ➝ ψ is a theorem, then so is □ϕ ➝ □ψ. ✤ Then by the Necessity Rule NEC, we know that the following is a theorem: (2) □(ϕ ➝ ψ) NEC, 1 ✤ Why should we trust this rule? ✤ ✤ Well, suppose that know that the following is proven: (1) ϕ➝ψ assumption Now, one instance of the K-axiom is the following: (3) □(ϕ ➝ ψ) ➝ (□ϕ ➝ □ψ) K ✤ So by modus ponens rule from (2) and (3), the following is a tautology: (4) (□ϕ ➝ □ψ) MP 2,3 28 First Theorem ✤ Our first theorem is: □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ) ✤ Now recall this tautology of propositional logic (#7 on list): (ϕ ➝ ψ) ➝ ((ϕ ➝ χ) ➝ (ϕ ➝ (ψ ∧ χ) )) ✤ Here’s the proof. ✤ First, we have that the following are theorems of prop. logic: (1) (ϕ ∧ ψ) ➝ ϕ PL (2) (ϕ ∧ ψ) ➝ ψ PL ✤ Hence, we have the following: (5) (□(ϕ∧ψ) ➝ □ϕ) ➝ ((□(ϕ∧ψ) ➝ □ψ) ➝ (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) ) PL ✤ Then by derived rule, we have the following are theorems: (3) □(ϕ∧ψ) ➝ □ϕ DR, 1 (4) □(ϕ∧ψ) ➝ □ψ DR, 2 29 Then by two applications of MP: (6) ((□(ϕ∧ψ) ➝ □ψ) ➝ (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) )) MP 3,6 (7) (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) )) MP 4,7 First Theorem ✤ On the previous slide, we had the commentary mixed in with the proof. The proof itself would just look like the following: ✤ Prove in the K-deductive system: □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ). ✤ Proof: ✤ (1) (ϕ ∧ ψ) ➝ ϕ PL ✤ (2) (ϕ ∧ ψ) ➝ ψ PL ✤ (3) □(ϕ∧ψ) ➝ □ϕ DR, 1 ✤ (4) □(ϕ∧ψ) ➝ □ψ DR, 2 ✤ (5) (□(ϕ∧ψ) ➝ □ϕ) ➝ ((□(ϕ∧ψ) ➝ □q) ➝ (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) )) PL ✤ (6) ((□(ϕ∧ψ) ➝ □ψ) ➝ (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) )) MP 3,5 ✤ (7) (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) )) MP 4,6 30 Second Theorem ✤ Our second theorem: (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) ✤ Recall the following analogue of the introduction rule for ‘and’ (#6 on list): (1) ϕ ➝ (ψ ➝ (ϕ ∧ ψ)) ✤ ✤ By applying derived rule to it, (2) □ϕ ➝ □(ψ➝ (ϕ ∧ ψ)) DR 1 ✤ Then by our derived rule on p. 19 of the slides today, we get: (4) □ϕ ➝ (□ψ➝ □(ϕ ∧ ψ)) DR 2,3 ✤ The following is an instance of a tautology from the list (#8): (5) (□ϕ ➝ (□ψ ➝ □(ϕ ∧ ψ))) ➝ ( (□ϕ ∧ □ψ) ➝□(ϕ ∧ ψ) ) PL ✤ One instance of K is (3) □(ψ➝ (ϕ ∧ ψ)) ➝ (□ψ➝ □(ϕ ∧ ψ)) K 31 (6) Then by MP, we get: (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) MP 4,5 Second Theorem ✤ Prove in the K-deductive system: (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) ✤ Proof: ✤ (1) ϕ ➝ (ψ ➝ (ϕ ∧ ψ)) PL ✤ (2) □ϕ ➝ □(ψ➝ (ϕ ∧ ψ)) DR 1 ✤ (3) □(ψ➝ (ϕ ∧ ψ)) ➝ (□ψ➝ □(ϕ ∧ ψ)) K ✤ (4) □ϕ ➝ (□ψ➝ □(ϕ ∧ ψ)) DR 2,3 ✤ (5) (□ϕ ➝ (□ψ ➝ □(ϕ ∧ ψ))) ➝ ( (□ϕ ∧ □ψ) ➝□(ϕ ∧ ψ) ) PL ✤ (6) (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) MP 4,5 32 Third theorem. Putting the first two together. ✤ So we just proved: □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ) (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) ✤ Now let’s prove: □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ) ✤ First, we list our earlier results, and just justify them as previous result: (1) □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ) prev. (2) (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) prev. 33 ✤ Then we use the derived rule from p. 18 of the slides to obtain the conjunction of the two previous lines: (3) ((□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)) ∧ ((□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)) DR 1,2 ✤ Then we recall from p. 6 of the slides that p↔q is defined as (p➝q) ∧ (q➝p). So we can write: (4) □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ) Defn, 3 Third Theorem ✤ Prove in the K-deductive system: □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ) ✤ Proof: ✤ (1) □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ) prev. ✤ (2) (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) prev. ✤ (3) ((□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)) ∧ ((□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)) DR 1,2 ✤ (4) □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ) Defn, 3 34 A fourth theorem ✤ Our final theorem is: (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ) ✤ We have the following tautologies of propositional logic: (1) ϕ ➝ (ϕ ∨ ψ) PL (2) ψ ➝ (ϕ ∨ ψ) PL ✤ Applying our derived rule to these, we get (3) □ϕ ➝ □(ϕ ∨ ψ) DR 1 (4) □ψ ➝ □(ϕ ∨ ψ) DR 2 ✤ Then we look at our big list, and we find (#11) the following: (ϕ ➝ χ) ➝ ( (ψ➝ χ) ➝ ((ϕ ∨ ψ) ➝ χ)) ) A substitution instance of this is the following: (5) (□ϕ ➝ □(ϕ ∨ ψ)) ➝ [ (□ψ ➝ □(ϕ ∨ ψ)) ➝ ((□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ))] PL ✤ 35 Then by modus ponens, we get: (6) (□ψ ➝ □(ϕ ∨ ψ) ) ➝ ((□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ)) MP 3,5 (7) (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ) MP 4,6 Fourth Theorem ✤ Prove in the K-deductive system: (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ) ✤ Proof: ✤ (1) ϕ ➝ (ϕ ∨ ψ) PL ✤ (2) ψ ➝ (ϕ ∨ ψ) PL ✤ (3) □ϕ ➝ □(ϕ ∨ ψ) DR 1 ✤ (4) □ψ ➝ □(ϕ ∨ ψ) DR 2 ✤ (5) (□ϕ ➝ □(ϕ ∨ ψ)) ➝ [ (□ψ ➝ □(ϕ ∨ ψ)) ➝ ((□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ))] ✤ (6) (□ψ ➝ □(ϕ ∨ ψ) ) ➝ ((□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ)) MP 3,5 ✤ (7) (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ) MP 4,6 36 PL Summary of Axioms, Rules, Derived Rules, and Theorems of System K K-axiom: □( ϕ → ψ ) → (□ϕ → □ψ) Abbreviation in proofs: ‘K’ ✤ □ϕ □ϕ→□ψ Abbreviation in proofs: DR ϕ ϕ→ψ Necessitation Rule: ✤ Derived Rule: ✤ NEC ✤ Abbreviation in proofs: ‘NEC’ 37 Theorems: 1. □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ) 2. (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) 3. □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ) 4. (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ) Costs and Benefits of the Deductive System ✤ One huge cost of the deductive system is that you have to be able to do clever substitutions on really long formulas, and you have to remember what you proved earlier, so that you don’t have to recreate proofs on the fly. 38 ✤ One benefit of the deductive system is that you don’t have to mention possible worlds at all. Hence, we have a route to the notions of necessity and possibility which does not depend on the notion of possible world. ✤ Compare: proving things about geometry using a deductive system vs. using intuitions about space. What do I need to know about the deductive system (for the purposes of this class)? ✤ You need to know the following things: ✤ 1. What the K-axiom and the Necessitation rule say. ✤ 2. The additional axioms that we will go over next time, which are specific to the systems D, T, B, S4, S5. ✤ ✤ 3. How to read proofs in the deductive system. 39 You don’t need to know the following things: ✤ 1. How to create/write proofs in the deductive system. ✤ 2. How to tell automatically whether any of the big long formulas we went over today are tautologies of propositional logic. Ω 40