Propositional Logic: Gentzen System, G

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1 CS402, Spring 2017

2 Quiz on Thursday, 6th April: 15 minutes, two questions.

3 Sequent Calculus in G In Natural Deduction, each line in the proof consists of exactly one proposition. That is, A 1, A 2,..., A n B. In Sequent calculus, each line in the proof consists of zero or more propositions. That is, A 1, A 2,..., A n B 1, B 2,..., B k. The standard semantic is, whenever every A i is true, at least one B j will also be true.

4 Axioms in G Definition 1 (3.2, Ben-Ari) An axiom of G is a set of literals U containing a complementary pair. Note that sets in G are implicitly disjunctive. For example, { p, q, p} is an axiom, i.e. p, q, p in G.

5 Inference Rules in G Definition 2 (3.2, Ben-Ari) There are two types of inference rules, defined with reference to tables below: Let {α 1, α 2} U 1 and let U 1 = U 1 {α 1, α 2}. Then U = U 1 {α} can be inferred. Let {β 1} U 1, {β 2} U 2 and let U 1 = U 1 {β 1}, U 2 = U 2 {β 2}. Then U = U 1 U 2 {β} can be inferred.

6 Inference Rules in G U 1 {α 1, α 2 } U 1 {α} α U 1 {β 1} U 2 {β 2} U 1 U 2 {β} β α α 1 α 2 β β 1 β 2 α (α 1 α 2 ) α 1 α 2 α 1 α 2 α 1 α 2 α 1 α 2 α 1 α 2 α 1 α 2 α 1 α 2 (α 1 α 2 ) α 1 α 2 α (α 1 α 2 ) (α 1 α 2 ) (α 2 α 1 ) α 1 α 2 (α 1 α 2 ) (α 2 α 1 ) That is, α rules build up disjunctions. β 1 β 2 β 1 β 2 (β 1 β 2 ) β 1 β 2 (β 1 β 2 ) β 1 β 2 (β 1 β 2 ) β 1 β 2 β 1 β 2 β 1 β 2 β 1 β 2 β 1 β 2 β 2 β 1 (β 1 β 2 ) β 1 β 2 β 2 β 1 That is, β rules build up conjuntions (consider (a b) (c d) = a c (b d)).

7 Example Proof Prove that p (q r) (p q) (p r) in G. 1. p, p, q Axiom 2. p, (p q) α, 1 3. p, p, r Axiom 4. p, (p r) α, 3 5. p, (p q) (p r) β, 2, 4 6. q, r, p, q Axiom 7. q, r, (p q) α, 6 8. q, r, p, r Axiom 9. q, r, (p r) α, q, r, (p q) (p r) β, 7, (q r), (p q) (p r) α, (p (q r)), (p q) (p r) β, 5, p (q r) (p q) (p r) α, 12

8 Wait... How do you magically come up with the axioms { p, p, q}, { p, p, r}, { q, r, p, q}, and { q, r, p, r}? Haven t we seen something like this before?

9 (p q) (q p) ((p q) (q p)) Proof in G (p q), (q p) p, q, p q, q, p (p q), q, p (p q), q, p p, q, p q, q, p (p q), (q p) (p q) (q p) UNSAT UNSAT Semantic Tableau (Sets are conjunctive)

10 G and Semantic Tableau Theorem 1 (3.6, Ben-Ari) Let A be a formula in propositional logic. Then A in G if and only if there is a closed semantic tableaux for A. Theorem 2 (3.7, Ben-Ari) Let U be a set of formulas and let Ū be the set of complements of formulas in U. Then, U in G if and only if there is a closed semantic tableau for Ū.

11 G and Semantic Tableau We prove that, if there exists a closed semantic tableau for Ū, then U in G. The opposite direction is left for you. Proof. Let T be a closed semantic tableau for Ū. We prove U by induction on h, the height of T. If h = 0, then T consists of a single node labeled by Ū. By assumption, T is closed, so it contains a complementary pair of literals {p, p}, that is, Ū = Ū {p, p}. Obviously, U = U { p, p} is an axiom in G, hence U.

12 G and Semantic Tableau Proof. Cont. If h > 0, then some tableau rule was used on an α- or β-formula at the root of T on a formula φ Ū, that is, Ū = Ū φ. The proof proceeds by cases, where you must be careful to distinguish between applications of the tableau rules and applications of the Gentzen rules of the same name. Case 1: φ is an α-formula (such as) (A 1 A 2 ). The tableau rule created a child node labeled by the set of formulas Ū { A 1, A 2 }. By assumption, the subtree rooted at this node is a closed tableau, so by the inductive hypothesis, U {A 1, A 2 }. Using the appropriate rule of inference from G, we obtain U {A 1 A 2 }, that is, U {φ}, which is U.

13 G and Semantic Tableau Proof. If h > 0, then some tableau rule was used on an α- or β-formula at the root of T on a formula φ Ū, that is, Ū = Ū φ. The proof proceeds by cases, where you must be careful to distinguish between applications of the tableau rules and applications of the Gentzen rules of the same name. Case 2: φ is a β-formula (such as) (B 1 B 2 ). The tableau rule created two child nodes labeled by the sets of formulas Ū { B 1 } and Ū { B 2 }. By assumption, the subtrees rooted at this node are closed, so by the inductive hypothesis U {B 1 } and U {B 2 }. Using the appropriate rule of inference from G, we obtain U {B 1 B 2 }, that is, U {φ}, which is U.

14 Why G and not natural deduction? Taste. Or, more appropriately, aesthetics. Natural deduction feels more, umm, natural. It is also more simplistic; having multiple disjunct on the right hand side, in G, is clearly cumbersome and adds complexity. G shows the symmetric nature of negation more vividly. A 1,..., A n B 1,..., B k (A 1 A n ) (B 1 B k ) A 1 A 2 A n B 1 B 2 B k (A 1 A 2 A n B 1 B 2 B k )

15 Soundness and Completeness of G Theorem 3 (3.8 in Ben-Ari) = A if and only if A in G. Proof. A is valid iff A is unsatisfiable iff there is a closed semanti tableau for A iff there is a proof of A in G.

16 Exercises Prove the following in G: (A B) ( B A) (A B) (( A B) B) ((A B) A) A

Computational Logic. Davide Martinenghi. Spring Free University of Bozen-Bolzano. Computational Logic Davide Martinenghi (1/30)

Computational Logic. Davide Martinenghi. Spring Free University of Bozen-Bolzano. Computational Logic Davide Martinenghi (1/30) Computational Logic Davide Martinenghi Free University of Bozen-Bolzano Spring 2010 Computational Logic Davide Martinenghi (1/30) Propositional Logic - sequent calculus To overcome the problems of natural