Automated Reasoning Jacques Fleuriot September 14, 2013 1 / 18 Lecture 3 Natural Deduction in Propositional Logic (II)1 Jacques Fleuriot 1With contributions by Paul Jackson 2 / 18 Recap I Last time we introduced natural deduction. I We looked at the introduction rule conjI PQ conjI P ^ Q I Now for the other rules of our formal deductive system .... 3 / 18 Conjunction Elimination I Elimination rules work in the opposite direction to introduction rules P ^ Q P ^ Q conjunct1 conjunct2 P Q : I An alternative conjunction elimination rule is: [P][Q] . P and Q are assumptions in a P ^ Q R conjE R proof of R. This can be useful for mechanized proofs. 4 / 18 Disjunction Elimination Likewise, there are introduction and elimination rules for disjunction: Q P disjI 1 disjI 2 P _ Q P _ Q [P] [Q] To show R given P _ Q, it . suffices to show R given P and P _ Q R R R given Q. disjE R The [φ] notation indicates that the rule discharges assumption φ. 5 / 18 Natural Deduction as Calculus of Derivations Introduce sequent notation for derivations Γ ` φ Conclusion φ is derivable from assumptions Γ Elimination rules can then be formulated using these sequents. For instance: [P] [Q] . P _ Q R R disjE R can be re-expressed as Γ ` P _ Q Γ; P ` R Γ; Q ` R disjE Γ ` R 6 / 18 Disjunction Rules in Action Prove that B _ A is true given that the assumption A _ B holds. Rules: P disjI 1 P _ Q [A]1 [B]1 Q disjI 2 disjI 1 disjI 2 A _ B B _ A B _ A P _ Q disjE1 B _ A [P] [Q] . P _ Q R R disjE R Subscript 1 indicates which rule discharges which hypotheses 7 / 18 Rules for Implication [P] . impI Forward: If on the assumption that P is true, Q . can be shown to hold, then we can conclude P −! Q. Q impI Backward: To prove P −! Q, assume P is true impI and prove that Q follows. P −! Q P −! QP mp The familiar modus ponens (mp) rule. Q [Q] . Another possible implication rule is this one. Note: this . is not necessarily a standard ND rule but may be useful . in mechanized proofs. P −! QP R impE R In general derivation assumptions may occur multiple times, and only a subset of the occurrences need be discharged. 8 / 18 Other Rules Negation: [P] . False :PP notI notE :P R : If and only if ( !): [Q] [P] . P Q Q ! PQ P ! QQ iffI iffD1 iffD2 P ! Q P P 9 / 18 Proof Recall the logic problems from lecture 2: we can now prove (Sunny _ Rainy) ^ :Sunny −! Rainy as follows [(S _ R) ^ :S]1 conjunct2 [(S _ R) ^ :S]1 [S]2 :S conjunct1 notE (S _ R) R [R]2 disjE2 R impI1 (S _ R) ^ :S −! R 10 / 18 Problematic lemma I Can every valid statement be proved using only the inference rules we have encountered so far? I Consider Peirce's Law: ((A −! B) −! A) −! A. I Our inference rules cannot prove this! I Classical logic: We can prove it with the law of excluded middle: A _:A I Intuitionistic logic: It cannot generally be proved! Evidence must be given for every statement, so we would need either evidence of A or a refutation of A to assert A _:A. 11 / 18 Classical Rules in Isabelle/HOL I The logic we will be using in Isabelle (HOL) is a classical logic and adds the rule: True or False P = True _ P = False I From this, we can derive the rules [:P] . False excluded middle ccontr :P _ P P 12 / 18 Summary of Rules PQ . PQ P ^ Q P ^ Q . ^ conjI conjunct1 conjunct2 . P ^ Q P Q P ^ Q R conjE R Q P. P Q . _ disjI 1 disjI 2 . P _ Q P _ Q P _ Q R R disjE R P [Q] . ! . Q P −! QP R impI impE P −! Q R Q . P. Q ! PQ P ! QQ $ . iffD1 iffD2 P Q P P iffI P ! Q [P] :P . :PP . : . excluded middle . notE . A _:A False R False notI ccontr :P P 13 / 18 Summary I Natural deduction in propositional logic. I Concept of a sequent calculus Next time: Propositional reasoning in Isabelle 14 / 18.