CS402, Spring 2017
Quiz on Thursday, 6th April: 15 minutes, two questions.
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.
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.
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.
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)).
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, 8 10. q, r, (p q) (p r) β, 7, 9 11. (q r), (p q) (p r) α, 10 12. (p (q r)), (p q) (p r) β, 5, 11 13. p (q r) (p q) (p r) α, 12
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?
(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)
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.
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.
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.
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.
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 )
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.
Exercises Prove the following in G: (A B) ( B A) (A B) (( A B) B) ((A B) A) A