10. Propositional Logic Soundness The Lecture Soundness ! Soundness of natural deduction means that deductions respect truth in the following sense: If A can be derived from the assumptions B1,…,Bn, and v(B1)=…=v(Bn)=1, then also v(A)=1. Last Jouko Väänänen: Propositional logic viewed We show: If A has a natural deduction from B1,…,Bn, and v(B1)=v(Bn)=1, then v(A)=1. Last Jouko Väänänen: Propositional logic viewed We show: If A has a natural deduction from B1,…,Bn, and v(B1)=v(Bn)=1, then v(A)=1. ! The proof is “by induction” on the structure of a natural deduction. Last Jouko Väänänen: Propositional logic viewed We show: If A has a natural deduction from B1,…,Bn, and v(B1)=v(Bn)=1, then v(A)=1. ! The proof is “by induction” on the structure of a natural deduction. ! We proceed from simpler deductions to more complex ones. Last Jouko Väänänen: Propositional logic viewed Conjunction introduction rule Last Jouko Väänänen: Propositional logic viewed Conjunction introduction rule ! We assume v(A)=v(B) =1. Last Jouko Väänänen: Propositional logic viewed Conjunction introduction rule ! We assume v(A)=v(B) =1. ! We show v(A∧B)=1. Last Jouko Väänänen: Propositional logic viewed Conjunction introduction rule ! We assume v(A)=v(B) =1. ! We show v(A∧B)=1. ! But this is trivial! Last Jouko Väänänen: Propositional logic viewed Conjunction elimination rule Last Jouko Väänänen: Propositional logic viewed Conjunction elimination rule ! We assume v(A∧B)=1. Last Jouko Väänänen: Propositional logic viewed Conjunction elimination rule ! We assume v(A∧B)=1. ! We show v(A)=v(B)=1. Last Jouko Väänänen: Propositional logic viewed Conjunction elimination rule ! We assume v(A∧B)=1. ! We show v(A)=v(B)=1. ! But this is again trivial! Last Jouko Väänänen: Propositional logic viewed Disjunction introduction rule Last Jouko Väänänen: Propositional logic viewed Disjunction introduction rule ! We assume v(A)=1. ! We show v(AvB)=1. ! But this is trivial! Last Jouko Väänänen: Propositional logic viewed Disjunction introduction rule ! We assume v(A)=1. ! We show v(AvB)=1. ! But this is trivial! ! We assume v(B)=1. ! We show v(AvB)=1. ! Again, this is trivial! Last Jouko Väänänen: Propositional logic viewed Disjunction elimination rule Last Jouko Väänänen: Propositional logic viewed Disjunction elimination rule ! We assume v(AvB)=1. Last Jouko Väänänen: Propositional logic viewed Disjunction elimination rule ! We assume v(AvB)=1. ! We also assume that the derivation of C from A, as well as the derivation of C from B, are sound i.e. if v(A)=1, then v(C)=1, and if v(B)=1, then v(C)=1. Last Jouko Väänänen: Propositional logic viewed Disjunction elimination rule ! We assume v(AvB)=1. ! We also assume that the derivation of C from A, as well as the derivation of C from B, are sound i.e. if v(A)=1, then v(C)=1, and if v(B)=1, then v(C)=1. ! We show v(C)=1. Last Jouko Väänänen: Propositional logic viewed Disjunction elimination rule ! We assume v(AvB)=1. ! We also assume that the derivation of C from A, as well as the derivation of C from B, are sound i.e. if v(A)=1, then v(C)=1, and if v(B)=1, then v(C)=1. ! We show v(C)=1. ! But v(AvB)=1 implies v(A)=1 or v(B)=1. Last Jouko Väänänen: Propositional logic viewed Disjunction elimination rule ! We assume v(AvB)=1. ! We also assume that the derivation of C from A, as well as the derivation of C from B, are sound i.e. if v(A)=1, then v(C)=1, and if v(B)=1, then v(C)=1. ! We show v(C)=1. ! But v(AvB)=1 implies v(A)=1 or v(B)=1. ! In either case we have v(C)=1. Last Jouko Väänänen: Propositional logic viewed Implication introduction rule Last Jouko Väänänen: Propositional logic viewed Implication introduction rule ! We assume that the derivation of B from A is sound, i.e. if v(A)=1, then v(B)=1. Last Jouko Väänänen: Propositional logic viewed Implication introduction rule ! We assume that the derivation of B from A is sound, i.e. if v(A)=1, then v(B)=1. ! We prove v(A→B)=1. Last Jouko Väänänen: Propositional logic viewed Implication introduction rule ! We assume that the derivation of B from A is sound, i.e. if v(A)=1, then v(B)=1. ! We prove v(A→B)=1. ! Case 1: v(A)=0. Clear. Last Jouko Väänänen: Propositional logic viewed Implication introduction rule ! We assume that the derivation of B from A is sound, i.e. if v(A)=1, then v(B)=1. ! We prove v(A→B)=1. ! Case 1: v(A)=0. Clear. ! Case 2: v(A)=1. By assumption, in this case v(B)=1, so v(A→B)=1. Last Jouko Väänänen: Propositional logic viewed Implication elimination rule Last Jouko Väänänen: Propositional logic viewed Implication elimination rule ! We assume v(A→B) =v(A)=1. Last Jouko Väänänen: Propositional logic viewed Implication elimination rule ! We assume v(A→B) =v(A)=1. ! We show v(B)=1. Last Jouko Väänänen: Propositional logic viewed Implication elimination rule ! We assume v(A→B) =v(A)=1. ! We show v(B)=1. ! This is trivial! Last Jouko Väänänen: Propositional logic viewed Equivalence introduction rule Last Jouko Väänänen: Propositional logic viewed Equivalence introduction rule ! We leave both the formulation of the claim, and the details of the proof as an exercise. Last Jouko Väänänen: Propositional logic viewed Equivalence elimination rule ! We leave both the formulation of the claim, and the details of the proof as an exercise. Last Jouko Väänänen: Propositional logic viewed Negation introduction rule Last Jouko Väänänen: Propositional logic viewed Negation introduction rule ! We assume that the inference of B∧¬B from A is sound i.e. if v(A)=1, then v(B∧¬B)=1. Last Jouko Väänänen: Propositional logic viewed Negation introduction rule ! We assume that the inference of B∧¬B from A is sound i.e. if v(A)=1, then v(B∧¬B)=1. ! But v(B∧¬B)=0 always. Last Jouko Väänänen: Propositional logic viewed Negation introduction rule ! We assume that the inference of B∧¬B from A is sound i.e. if v(A)=1, then v(B∧¬B)=1. ! But v(B∧¬B)=0 always. ! So v(A)=0. Last Jouko Väänänen: Propositional logic viewed Negation elimination rule Last Jouko Väänänen: Propositional logic viewed Negation elimination rule ! We assume v(¬¬A)=1. Last Jouko Väänänen: Propositional logic viewed Negation elimination rule ! We assume v(¬¬A)=1. ! We show v(A)=1. Last Jouko Väänänen: Propositional logic viewed Negation elimination rule ! We assume v(¬¬A)=1. ! We show v(A)=1. ! Clear! Last Jouko Väänänen: Propositional logic viewed Soundness Theorem Last Jouko Väänänen: Propositional logic viewed Soundness Theorem ! If a propositional formula has a natural deduction, then it is a tautology. Last Jouko Väänänen: Propositional logic viewed Soundness Theorem ! If a propositional formula has a natural deduction, then it is a tautology. ! If a propositional formula A has a natural deduction from assumptions which have truth value 1 in a valuation v, then also v(A)=1. Last Jouko Väänänen: Propositional logic viewed Applications of Soundness ! We can show that a formula B is not derivable by natural deduction from a formula A by finding a valuation v such that v(A)=1 and v(B)=0. Last Jouko Väänänen: Propositional logic viewed Applications of Soundness ! We can show that a formula B is not derivable by natural deduction from a formula A by finding a valuation v such that v(A)=1 and v(B)=0. ! Example: We show that p0∨(p1∧p2) is not derivable from (p0∨p2)→p1. Last Jouko Väänänen: Propositional logic viewed Applications of Soundness ! We can show that a formula B is not derivable by natural deduction from a formula A by finding a valuation v such that v(A)=1 and v(B)=0. ! Example: We show that p0∨(p1∧p2) is not derivable from (p0∨p2)→p1. ! Solution: Let v(p0)=v(p1)=v(p2)=0. Last Jouko Väänänen: Propositional logic viewed.