# Logic. Propositional Logic: Syntax

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1 Logic Propositional Logic: Syntax Logic is a tool for formalizing reasoning. There are lots of different logics: probabilistic logic: for reasoning about probability temporal logic: for reasoning about time (and programs) epistemic logic: for reasoning about knowledge The simplest logic (on which all the rest are based) is propositional logic. It is intended to capture features of arguments such as the following: Borogroves are mimsy whenever it is brillig. It is now brillig and this thing is a borogrove. Hence this thing is mimsy. Propositional logic is good for reasoning about conjunction, negation, implication ( if... then... ) Amazingly enough, it is also useful for circuit design program verification 1 To formalize the reasoning process, we need to restrict the kinds of things we can say. Propositional logic is particularly restrictive. The syntax of propositional logic tells us what are legitimate formulas. We start with primitive propositions. Think of these as statements like It is now brillig This thing is mimsy It s raining in San Francisco n is even We can then form more complicated compound propositions using connectives like: : not : and : or : implies : equivalent (if and only if) 2 Examples: P : it is not the case that P P Q: P and Q P Q: P or Q P Q: P implies Q (if P then Q) Typical formula: P ( P (Q (R P ))) Wffs Formally, we define well-formed formulas (wffs or just formulas) inductively (remember Chapter 2!): The wffs consist of the least set of strings such that: 1. Every primitive proposition P, Q, R,... is a wff 2. If A is a wff, so is A 3. If A and B are wffs, so are A B, A B, A B, and A B 3 4

2 Disambiguating Wffs Translating English to Wffs We use parentheses to disambiguate wffs: P Q R can be either (P Q) R or P (Q R) Mathematicians are lazy, so there are standard rules to avoid putting in parentheses. In arithmetic expressions, binds more tightly than +, so means 3 + (2 5) In wffs, here is the precedence order: Thus, P Q R is P (Q R); P Q R is P (( Q) R) P Q R is (P ( Q)) R With two or more instances of the same binary connective, evaluate left to right: P Q R is (P Q) R To analyze reasoning, we have to be able to translate English to wffs. Consider the following sentences: 1. Bob doesn t love Alice 2. Bob loves Alice and loves Ann 3. Bob loves Alice or Ann 4. Bob loves Alice but doesn t love Ann 5. If Bob loves Alice then he doesn t love Ann First find appropriate primitive propositions: P : Bob loves Alice Q: Bob loves Ann Then translate: 1. P 2. P Q 3. P Q 4. P Q (note: but becomes and ) 5. P Q 5 6 Evaluating Formulas Truth Tables Given a formula, we want to decide if it is true or false. How do we deal with a complicated formula like: P ( P (Q (R P ))) The truth or falsity of such a formula depends on the truth or falsity of the primitive propositions that appear in it. We use truth tables to describe how the basic connectives (,,,, ) work. For : P P For : P Q P Q T T F F For : P Q P Q T T F F This means is inclusive or, not exclusive or. 7 8

3 Exclusive Or Truth Table for Implication What s the truth table for exclusive or? P Q P Q T T T F F F P Q is equivalent to (P Q) ( P Q) P Q P Q P Q Q P (P Q) ( P Q) T F F F F T T F F F For : P Q P Q T T F F Why is this right? What should the truth value of P Q be when P is false? This choice is mathematically convenient As long as Q is true when P is true, then P Q will be true no matter what. For : P Q P Q T T T F F F 9 10 How many possible truth tables are there with two primitive propositions? P Q? T T F F By the product rule, there are 16. We ve defined connectives corresponding to 4 of them:,,,. Why didn t we bother with the rest? They re definable! Other Equivalences It s not hard to see that P Q is also equivalent to (P Q) Thus, P Q is equivalent to (P Q), which is equivalent to ((P Q) ( P Q)) Thus, we don t need either. We also don t need : P Q is equivalent to P Q We also don t need : P Q is equivalent to ( P Q) Each of the sixteen possible connectives can be expressed using and (or ) 11 12

4 Tautologies Arguments A truth assignment is an assignment of T or F to every proposition. How hard is it to check if a formula is true under a given truth assignment? Easy: just plug it in and evaluate. Time linear in the length of the formula A tautology (or theorem) is a formula that evaluates to T for every truth assignment. Examples: (P Q) ( P Q) P Q ( P Q) (P Q) (Q P ) It s necessarily true that if elephants are pink then the moon is made of green cheese or if the moon is made of green cheese, then elephants are pink. How hard is it to check if a formula is a tautology? How many truth assignments do we have to try? Definition: An argument has the form A 1 A 2. A n B A 1,..., A n are called the premises of the argument; B is called the conclusion. An argument is valid if, whenever the premises are true, then the conclusion is true Logical Implication A formula A logically implies B if A B is a tautology. Theorem: An argument is valid iff the conjunction of its premises logically implies the conclusion. Proof: Suppose the argument is valid. We want to show (A 1... A n ) B is a tautology. Do we have to try all 2 k truth assignments (where k = #primitive propositions in A 1,..., A n, B). It s not that bad. Because of the way we defined, A 1... A n B is guaranteed to be true if A 1... A n is false. But if A 1... A n is true, B is true, since the argument is valid. Thus, (A 1... A n ) B is a tautology. For the converse, suppose (A 1... A n ) B is a tautology. If A 1,..., A n are true, then B must be true. Hence the argument is valid. Remember: Borogroves are mimsy whenever it is brillig. It is now brillig and this thing is a borogrove. Hence this thing is mimsy. Suppose P : It is now brillig Q: This thing is a borogrove R: This thing is mimsy This becomes: P (Q R) P Q R This argument is valid if is a tautology. [(P (Q R)) (P Q)] R 15 16

5 Natural Deduction Formal Deductive Systems Are there better ways of telling if a formula is a tautology than trying all possible truth assignments. In the worst case, it appears not. The problem is co-np-complete. The satisfiability problem deciding if at least one truth assignment makes the formula true is NPcomplete. Nevertheless, it often seems that the reasoning is staightforward: Why is this true: ((P Q) (Q R)) (P R) We want to show that if P Q and Q R is true, then P R is true. So assume that P Q and Q R are both true. To show that P R, assume that P is true. Since P Q is true, Q must be true. Since Q R is true, R must be true. Hence, P R is true. We want to codify such reasoning. A formal deductive system (also known as an axiom system) consists of axioms (special formulas) rules of inference: ways of getting new formulas from other formulas. These have the form A 1 A 2. A n B Read this as from A 1,..., A n, infer B. Sometimes written A 1,..., A n B Think of the axioms as tautologies, while the rules of inference give you a way to derive new tautologies from old ones Derivations Typical Axioms A derivation (or proof ) in an axiom system AX is a sequence of formulas C 1,..., C N ; each formula C k is either an axiom in AX or follows from previous formulas using an inference rule in AX: i.e., there is an inference rule A 1,..., A n B such that A i = C ji for some j i < N and B = C N. This is said to be a derivation or proof of C N. A derivation is a syntactic object: it s just a sequence of formulas that satisfy certain constraints. Whether a formula is derivable depends on the axiom system Different axioms different formulas derivable Derivation has nothing to do with truth! How can we connect derivability and truth? P P P (Q P ) What makes an axiom acceptable? it s a tautology 19 20

6 Typical Rules of Inference Sound and Complete Axiomatizations Modus Ponens A B A B Modus Tollens A B B A What makes a rule of inference acceptable? It preserves validity: if the antecedents are valid, so is the conclusion Both modus ponens and modus tollens are acceptable Standard question in logic: Can we come up with a nice sound and complete axiomatization: a (small, natural) collection of axioms and inference rules from which it is possible to derive all and only the tautologies? Soundness says that only tautologies are derivable Completeness says you can derive all tautologies If all the axioms are valid and all rules of inference preserve validity, then all formulas that are derivable must be valid. Proof: by induction on the length of the derivation It s not so easy to find a complete axiomatization A Sound and Complete Axiomatization for Propositional Logic Consider the following axiom schemes: A1. A (B A) A2. (A (B C)) ((A B) (A C)) A3. ((A B) (A B)) A These are axioms schemes; each one encodes an infinite set of axioms: P (Q P ), (P R) (Q (P R)) are instances of A1. Theorem: A1, A2, A3 + modus ponens give a sound and complete axiomatization for formulas in propositional logic involving only and. Recall: can define and using and P Q is equivalent to P Q P Q is equivalent to (P Q) Derivation of P P : A Sample Proof 1. P ((P P ) P ) [instance of A1: take A = P, B = P P ] 2. (P ((P P ) P )) ((P (P P )) (P P )) [instance of A2: take A = C = P, B = P P ] 3. (P (P P )) (P P ) [applying modus ponens to 1, 2] 4. P (P P ) [instance of A1: take A = B = P ] 5. P P [applying modus ponens to 3, 4] Try deriving P P from these axioms it s hard! 23 24

7 Algorithm Verification Algorithm Verification: Example This is (yet another) hot area of computer science. How do you prove that your program is correct? You could test it on a bunch of instances. That runs the risk of not exercising all the features of the program. In general, this is an intractable problem. For small program fragments, formal verification using logic is useful It also leads to insights into program design. Consider the following algorithm for multiplication: Input x [Integer 0] y [Integer] Algorithm Mult prod 0 u 0 repeat u = x prod prod + y u u + 1 end repeat How do we prove this is correct? Idea (due to Floyd and Hoare): annotate program with assertions that are true of the line of code immediately following them. An assertion just before a loop is true each time the loop is entered. This is a loop invariant. An assertion at the end of a program is true after running the program Input x [Integer 0] y [Integer] Algorithm Mult prod 0 u 0 {prod = uy} [Loop invariant] repeat u = x prod prod + y u u + 1 end repeat {prod = uy u = x} Thus, we must show prod = uy is true each time we enter the loop. Proof is by induction (big surprise) It follows that prod = uy u = x holds after exiting the program, since we exit after trying the loop (so prod = uy) and discovering u = x. It follows that prod = xy at termination. But how do we know the program terminates? We prove (by induction!) that after the kth iteration of the loop, u = k. Since x 0, eventually u = x, and we terminate the loop (and program) We won t be covering Boolean algebra (it s done in CS 314), although you should read Section 7.5! 27 28

8 Predicate Calculus There are lots of things that can t be expressed by propositional formulas. In first-order logic, we can: Talk about individuals and the properties they have: Bob and Alice are both American American(Bob) American(Alice) Talk about the relations between individuals Bob loves Alice but Bob doesn t love Anne Loves(Bob, Alice) Loves(Bob, Anne). Quantify: Everybody loves somebody x yloves(x, y) First-order logic lets us capture arguments like: All men are mortal Socrates is a man Therefore Socrates is mortal All prime numbers are integers 7 is a prime number Therefore 7 is an integer 29

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