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Symbolic Logic II, Lecture 3 1 in Order to Apply the Axiomatic Method, We Need the Following

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Symbolic Logic II, Lecture 3 1 in Order to Apply the Axiomatic Method, We Need the Following Propositional Logic Axiomatic proofs in propositional logic We wil now consider a different approach to proof theory, using axiomatic (or \Hilbert-style") systems. Axiomatic proofs are harder to construct than sequent proofs because we cannot use conditional proofs or reductio ad absurdum. An axiomatic proof is a series of formulas, the last of which is the conclusion of the proof. Each line in the proof must be justified in one of two ways: it may be inferred by a rule of inference from earlier lines in the proof, or it may be an axiom. Brandon C. Look: Symbolic Logic II, Lecture 3 1 In order to apply the axiomatic method, we need the following: I a set of rules, and I a set of axioms. One of our rules is going to be modus ponens: (φ ! ) φ MP Brandon C. Look: Symbolic Logic II, Lecture 3 2 Definition of axiomatic proof from a set: Where Γ is a set of wffs and φ is a wff, an axiomatic proof from Γ is a finite sequence of wffs whose last line is φ, in which each line either (i) is an axiom or (ii) is a member of Γ, or (iii) follows from earlier wffs in the sequence of a rule. Definition of an axiomatic proof: An axiomatic proof of φ is an axiomatic proof of φ from ; (i.e., a finite sequence of wffs whose last line is φ, in which each line either (i) is an axiom, or (ii) follows from earlier wffs in the sequence via a rule). Brandon C. Look: Symbolic Logic II, Lecture 3 3 We will write “Γ ` φ" when we mean that φ is provable from the set of wffs Γ. (For “Γ ` φ" just think “Γ proves φ." Also, “Γ ` φ" means \φ is a syntactic consequence of Γ”, whereas “Γ j= φ" means \φ is a semantic consequence of Γ”.) Brandon C. Look: Symbolic Logic II, Lecture 3 4 Axiomatic System for PL: I Rule: MP I Axioms: The result of substituting wffs for φ, , and χ in any of the following schemas is an axiom: pφ ! ( ! φ)q (PL1) p(φ ! ( ! χ)) ! ((φ ! ) ! (φ ! χ))q (PL2) p(∼ →∼ φ) ! ((∼ ! φ) ! )q (PL3) Brandon C. Look: Symbolic Logic II, Lecture 3 5 Propositional Logic Soundness of Propositional Logic Two very important results of metalogic: 1. Soundness of PL: Every PL-theorem is PL-valid. 2. Completeness of PL: Every PL-valid wff is a PL-theorem. Brandon C. Look: Symbolic Logic II, Lecture 3 6 Why is this important? Consider, first, Soundness. Here the claim is often represented as \if ` φ then j= φ". And this means that whatever is provable by propositional logic is valid (or true given true premises). So, you can't go from true premises, use our rules of inference and so on, and arrive at something that is false. Next, Completeness. Here the claim is represented as \if j= φ then ` φ". And this means that everything that is true or valid is provable in propositional logic. Brandon C. Look: Symbolic Logic II, Lecture 3 7 So, there is no valid consequence that is not derivable from propositional logic. And our logical syntax matches up with our semantics (for every semantic consequence in PL is a syntactic consequence in PL and every syntactic consequence in PL is a semantic consequence in PL). That's a big deal. Brandon C. Look: Symbolic Logic II, Lecture 3 8 Propositional Logic Aside: method of induction In order to prove both the soundness and completeness of PL, we have to discuss an important method of proof: the method of induction. Basically, the idea is simple: if we can prove a particular property of a first object in some group and show that whenever one object in the group has the property the next one must have the property, then we can conclude that all objects have that property. In other words, we need to proceed in two steps: I the base case (b) I the inductive step (i) Brandon C. Look: Symbolic Logic II, Lecture 3 9 So, when want to establish that every wff has a certain property, p, we have to establish the following two claims: (b) every atomic wff has property p, and (i) for any wffs φ and , if both φ and have property p, then the wffs ∼ φ and φ ! have property p. Brandon C. Look: Symbolic Logic II, Lecture 3 10 Propositional Logic The proof of soundness The basic idea behind this proof is this. We are going to take our axioms as the starting points or base case(s) and show that they are PL-valid. Then we are going to show that our inference rule, modus ponens, preserves validity. If we do this, we can claim that every theorem in PL is valid. Brandon C. Look: Symbolic Logic II, Lecture 3 11 So, here's Sider's proof: Base case: We need to show that every PL-axiom is valid. Start with PL1, pφ ! ( ! φ)q. (PL2 and PL3 will be exercises.) Suppose for a reductio that some instance of PL1 is invalid; that means for some PL-interpretation I , VI (φ ! ( ! φ)) = 0. Thus, VI (φ) = 1 and VI ( ! φ) = 0. Given the latter, VI (φ) = 0. Since this leads to a contradiction, PL1 must be valid. Brandon C. Look: Symbolic Logic II, Lecture 3 12 Induction step: We assume that every line in a proof up to a certain point is valid, and then we show that, if one adds another line that follows from the earlier lines by modus ponens, then that line must be valid, too. So, this means that we are claiming that some formula can be inferred from two earlier lines containing the formulas φ ! and φ. We must show that is a valid formula | that VI ( ) = 1 for every PL-interpretation I . By the inductive hypothesis, all earlier lines in the proof are valid, and hence both φ ! and φ are valid. Thus, VI (φ) = 1 and VI (φ ! ) = 1. But if VI (φ) = 1, then VI ( ) cannot be 0, for if it were, then VI (φ ! ) would be 0, and it is not. Thus, VI ( ) = 1. Brandon C. Look: Symbolic Logic II, Lecture 3 13 Another proof of Soundness: Suppose Γ ` φ. Now, one metatherem of PL is that Γ ` φ iff there is a finite subset ∆ of Γ such that ∆ ` φ. (i) If ∆ is empty, then ` φ and therefore φ is a theorem and Γ ` φ. Brandon C. Look: Symbolic Logic II, Lecture 3 14 (ii) If ∆ is not empty, let φ1; : : : ; φn be the members of ∆. Then, φ1; : : : ; φn ` φ. By the Deduction Theorem (which we will discuss next time), applied as many times as necessary, ` φ1 ! (φ2 ! (::: (φn ! φ))). A consequence of this is that j= φ1 ! (φ2 ! (::: (φn ! φ))). And, by the definition of true for an interpretation of PL, there is no model of fφ1; : : : ; φng, that is, of ∆, that is not also a model of φ. Therefore, since ∆ is a subset of Γ, there is no model of Γ that is not also a model of φ; that is, either Γ has no model or every model of Γ is also a model of φ. That is, Γ j= φ. So, if Γ ` φ, then Γ j= φ. Brandon C. Look: Symbolic Logic II, Lecture 3 15.
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