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Classical Logic Lecture Notes for SWE 623 by Duminda Wijesekera SWE 623Classical Logic Duminda Wijesekera 1 Propositional and Predicate Logic • Propositional Logic – The study of statements and their connectivity structure. • Predicate Logic – The study of individuals and their properties. • Study syntax and semantics for both. • Propositional logic more abstract and hence less detailed than predicate logic. SWE 623Classical Logic Duminda Wijesekera 2 Propositional Logic: Syntax • A collection of atomic propositional symbols. – Say A = { ai : 0 < i }. A special atom _|_ for contradiction • A collection of logical connectives. – (and) ^, (or) v, ( not ) , (implies) => • Inductively define propositions as: If X,Y are propositions, so are :– X ^ Y, X v Y, X => Y, X. • Examples: – a1^a2, (a =>a2)v(a3^( a4)) are propositions. SWE 623Classical Logic Duminda Wijesekera 3 Propositional Logic: Semantics • A model M of a propositional language consists of – a collection of atoms, say B = { bi : 0 < i }, where _|_ is excluded from B, and – a partial mapping M from A = { ai : 0 < i } to B = { bi : 0 < i }. – If M(ai) e B, we say that ai is true in M. We write “ai is true in M” as M |= ai. (Read M satisfies ai). – |= is referred to as the satisfaction relation. SWE 623Classical Logic Duminda Wijesekera 4 Propositional Semantics: Continued • Extend M, and therefore the satisfaction relation to all propositions using the following inductive definition: – M |= X ^ Y iff M |= X and M |= Y. – M |= X v Y iff M |= X or M |= Y. – M |= X => Y if M |= X then M |= Y. – M |= X, if it is not the case that M |= X. • Notice usage of truth tables SWE 623Classical Logic Duminda Wijesekera 5 Propositional Logic: Example • B = { a1, a3} where M given as M(a1) = a1 and M(a2) = a2 has the following properties. – M |= a1 – M |= a1 ^ a3 – M |= a2 – M |= a2 => a4 • M does not satisfy the following propositions. – M |= a4 – M |= a1 => a4 SWE 623Classical Logic Duminda Wijesekera 6 Propositional Logic: Proofs • What formulas hold in all models ? • I.e. can we check if a given proposition is true in all models without going through all possible models? • Need proofs to answer this question. • We use Natural Deduction proofs. • Recommended: Read Ch 2 of Logic and Computation by L.C. Paulson. SWE 623Classical Logic Duminda Wijesekera 7 Natural Deduction for Prop. Logic • Proofs are trees of formulae made by applying inference rules. • An inference rule is of the form: A1 …… An B • Here A1 ….. An are said to be premises (or antecedents) of the rule, and B is said to be the conclusion (consequent) of the rule. SWE 623Classical Logic Duminda Wijesekera 8 Natural Deduction for Prop. Logic • Hence a proofs is a trees whose – Root is the theorem to be proved, – Branches are rules, and – Leaves are the assumptions (axioms) of the proof. • Example – A1 A2 A3 C1 C2 Assumptions B1 B2 Applications of rules D Theorem being proved • There are introduction and elimination rules for each connective in Natural Deduction proof systems. SWE 623Classical Logic Duminda Wijesekera 9 Rules for Conjunction • Introduction A B A^B • Elimination A^B A^B A B SWE 623Classical Logic Duminda Wijesekera 10 Rules for Disjunction • Introduction A B AvB AvB • Elimination [A] [B] AvB C C C • [X] denotes discharged assumption X. SWE 623Classical Logic Duminda Wijesekera 11 Rules for Implication • Introduction [A] B A => B • Elimination (Modus Ponens) A => B A B SWE 623Classical Logic Duminda Wijesekera 12 Rules for Negation • B interpreted as ( B => _|_). Hence we get the following rules from those of implication. • Introduction Elimination [B] B B _|_ _|_ ________ B • Special Contradiction Rule: B _|_ __________ B SWE 623Classical Logic Duminda Wijesekera 13 Propositional Proofs: Examples • Prove: ( A ^ B ) => (A v B) • Notice: – The outermost connective is =>. Hence, the last step of the proof must be an implication introduction. – That means, we must assume ( A ^ B ) and prove (A v B), and then discharge the assumption by using => introduction rule. – In order to prove (A v B) from ( A ^ B ), we must use v –introduction, and hence must prove either A or B from ( A ^ B ). – This plan forms a skeleton of a proof. SWE 623Classical Logic Duminda Wijesekera 14 Prop. Proof: Example Continued • Prove: ( A ^ B ) => (A v B) [A ^ B ] A ^ elimination AvB v introduction ( A ^ B ) => (A v B) => introduction • Proofs are analyzed backwards, I.e. start unraveling the logical structure of the conclusion and work backwards to the assumptions. Draw out a plan based on your analysis and write down the formal proof. SWE 623Classical Logic Duminda Wijesekera 15 Derived Rules • These are rules derived from other rules. • Example: A^B B^A • Here is the derivation: A^B B^A B A ^ elimination B^ A ^ introduction SWE 623Classical Logic Duminda Wijesekera 16 Soundness and Completeness • A rule A1 …… An is said to be sound if for every B model in which all of A1 …… An are true, then so is B. I.e. if M |= A1 , …… , M |= An, then M |= B. • A collection of rules are sound if all rules in the collection is sound. • A collection of rules is complete if M |= A for all models M, then A is provable. I.e. there is a proof of A using the given set of rules. (Denoted |R-- A ) where R is the set of rules. SWE 623Classical Logic Duminda Wijesekera 17 Predicate Logic • Language to describe properties of individuals. • Thus, syntax is able to describe individuals, their properties (relationships) and functions. – These are to be thought of as names of individuals, properties (relationships) and functions. – Models are “incarnations” of these individuals, properties (relationships) and functions. • More detailed than propositional logic. SWE 623Classical Logic Duminda Wijesekera 18 Predicate Logic: Syntax • A collection of constants– say { ci : i >= 0 }. – Constants are names for individuals. E.g.: 0, 1. – Note: not all individuals in a model have names. • A collection of variables– say { xi : i >= 0 }. – Needed to generically refer to individuals. – Think of them as standing in place of pronouns like it, she. • A collection of function symbols- say { fi : i >= 0 }. – May be of different arities, and may be typed. E.g.: +(x,y) • A collection of predicate symbols- say { pi : i >= 0 }. – May be of different arities. – Encodes properties of individuals. E.g.: prime(x). SWE 623Classical Logic Duminda Wijesekera 19 Predicate Logic Recursive Definition of Terms • Every variable is a term. • Every constant is a term. • If fi is an n-ary function symbol and t1, .., tn are terms, then fi(t1, .., tn) is a term. • We use {ti : i <=0 } for the collection of terms. • Examples: – f(x, g(2, y)) is a term, where f, g are function symbols and x, y are variables. – +( x, *(3,y)) is a term in arithmetic usually written as x + (3*y) SWE 623Classical Logic Duminda Wijesekera 20 Recursive Definition of Formulas • If pi is an n-ary predicate symbol and t1, .., tn are terms, then pi(t1, .., tn ) is an atomic formula. • If A and B are formulas, then so are: – A ^ B, A v B, A, A => B. – xi A(xi), xi A(xi), where xi is a variable. • , are referred to as the universal and existential quantifier, respectively. • A formula that does not have either quantifier is said to be a quantifier free. SWE 623Classical Logic Duminda Wijesekera 21 Free and bound Variables • In x A(x), the variable x is said to be bound; meaning the name x plays no significant role. (compare with he, she, it) • A variable x occurs bound in a formula if x or x is a part of it. More precisely, x occurs bound in: – y A(y) or y A(y) if x and y are the same variable. – A if x occurs bound in A. – A ^ B, A v B, A => B if x occurs bound in either A or B. SWE 623Classical Logic Duminda Wijesekera 22 Substitutions • If A is a formula, t is a term and x is a variable, then A[t/x] is the formula obtained by substituting t for x in A. – A[t1/x1, … tn/xn] is the formula resulting in simultaneously substituting x1, …xn by t1, …tn. – Note: Simultaneous substitution Q(x,y)[x/y,y/x] yields Q(y,x) but iterated substitution Q(x,y)[x/y][y/x] yields Q(y,y). SWE 623Classical Logic Duminda Wijesekera 23 Substituting Terms for Variables • In A[t/x], the free variables of t stand the danger of becoming bound in A. Hence, need a precise definition. – If x is y then y A(y) [x/y] is y A(y). If not let z be a fresh variable (I.e. not in t, x) then (y A(y) )[t/x] is z (A(z/y) [t/x]). – Similar definition for y A(y). • Examples: – y (y = 1) [y/y] is y (y = 1). Here x is y and t is x. – y (y+1 > x) [2y+x/x] is z ((z+1>x)[2y+x/x] I.e. z (z+1>2y+x). Here t is (2y+x). SWE 623Classical Logic Duminda Wijesekera 24 Substituting Terms Continued • ( A )[t/x] is (A [t/x]) • (A ^ B) [t/x] is (A[t/x] ^ B[t/x]) • (A v B) [t/x] is (A[t/x] v B[t/x]) • (A => B) [t/x] is (A[t/x] => B[t/x]) • Pi(t1, .. tn) [t/x] is Pi(t1[t/x], .. tn[t/x]) for predicate symbol Pi. SWE 623Classical Logic Duminda Wijesekera 25 Predicate Logic: Semantics • A model consists of – A set (of individuals), say A = { ai : i >= 0 }. – A set of total functions Fn = { fni : i >= 0 } on A. • I.e. fni(aj) is some ak for every aj. – A set of predicates Pr = { pri : i >= 0 } over A. • Do not have to be total. • Can have many arities. SWE 623Classical Logic Duminda Wijesekera 26 Interpreting Syntax • Mapping from Syntax to Semantics: – A mapping mCons : { ci : I >= 0 } to A={ai: i >= 0}. • Need not be ONTO A. I.e. there could be unnamed individuals in the semantic domain. – A mapping mFun : { fi : I >= 0 } to Fn={fni: i >= 0}. • Need not be onto. I.e. there could be unnamed functions in the semantic domain. – A mapping mPred: { pi : I >= 0 } to Pr={pri: i >= 0}. • Need not be onto. I.e. there could be unnamed predicates in the semantic domain. SWE 623Classical Logic Duminda Wijesekera 27 Interpreting Formulas: naming • We do not interpret formulas with free variables. • In order to interpret quantified formulas, need to expand the syntax by adding a constant in the syntax for each unnamed individual in the model. – I.e. for each ai for which there is no cj such that Fn(cj ) is ai, add a new constant Cai to the syntax. – Now expand the definition of terms to include these new constants. Let newT = { Nti : i >= 0} be the collection of new terms so defined. SWE 623Classical Logic Duminda Wijesekera 28 Interpreting Formulas • Let M be a model. We define M |= F for every quantified formula as follows. – For every n-ary predicate symbol pi , and every sequence of new variable free terms Nt1, … Ntn define M |= pi(Nt1, … Ntn ) if and only if mPred(pi)(Nt1, … Ntn ). – I.e. pi(Nt1, … Ntn ) is true in M if and only if its image under the map mPred holds with parameters Nt1, … Ntn . SWE 623Classical Logic Duminda Wijesekera 29 Interpreting Formulas: Continued – For every formula A , M |= y A(y) if and only if M |= A(Nti) for every Nti e newT. – For every formula A , M |= y A(y) if and only if there is some Nti e newT satisfying M |= A(Nti). – M |= A ^ B if M |= A and M |= B . – M |= A v B if M |= A or M |= B. – M |= A => B if when M |= A then M |= B. – M |= A if it is not the case that M |= A. SWE 623Classical Logic Duminda Wijesekera 30 Proof Rules for Predicate Logic • Proof rules of introduction and elimination of ^, v, =>, and . • New rules required for introduction and elimination of and quantifiers. SWE 623Classical Logic Duminda Wijesekera 31 Proof Rules for • Introduction A(x) provided x is not free in the x A(x) assumptions of A • Elimination x A(x) A[t/x] SWE 623Classical Logic Duminda Wijesekera 32 Proof Rules for • Introduction A[t/x] xA(x) • Elimination [A] provided x is not free xA(x) B in B nor in the B assumptions of B apart from A SWE 623Classical Logic Duminda Wijesekera 33 An Example Proof • Prove ((x A(x)) ^ B) => (x (A(x)^ B)) provided that x is not free in B. • Plan: – Since outer connective is =>, need to use => introduction at the last step. Hence can use (x A(x)) ^ B as an assumption for the steps above. – Now in order to get x (A(x)^ B) using introduction, we need to get A[t/x] )^ B. – Can use ^ elimination to (x A(x)) ^ B and obtain B – Can use x elimination to get A[t/x]. SWE 623Classical Logic Duminda Wijesekera 34 Example Proof x A(x) ^ B x A(x) ^ B x A(x) [A(t/x)] B A(t/x) ^ B x(A(x) ^ B • The other direction of the proof appears in the handout page 32. SWE 623Classical Logic Duminda Wijesekera 35 Induction Rule [A(x)] A[0/x] A[x+1/x] A(x) Proviso: x is not free in the assumptions of A[x+1/x] apart from A(x). SWE 623Classical Logic Duminda Wijesekera 36 Equality Reasoning • Rules for equality – Reflexivity axiom: t = t. – Symmetry rule: t= u. u=t – Transitivity rule: s = t t=u. s=u • Congruence laws for each function and predicate symbol, or substitution rules. SWE 623Classical Logic Duminda Wijesekera 37 Equality Reasoning: Continued • Congruence Law for functions: t1 = u1 …. tn = un f(t1, …., tn) = f(u1, ….,un) • Congruence Law for Predicates: t1 = u1 …. tn = un p(t1, …., tn) p(u1, ….,un) • Substitution Rule: t = u S[t/x] = S[u/x] SWE 623Classical Logic Duminda Wijesekera 38 Equality Reasoning: An Example This example is from Page 37, of the Logic handout. x f(x,x) = x f(g(z), g(z)) = g(z) p(f(g(z), g(z)) p(g(z)) p(f(g(z), g(z)) p(g(z)) SWE 623Classical Logic Duminda Wijesekera 39 Logic: Suggested Exercises • Go thorough all proofs and suggested exercises in the handout. • Take the midterm and final exams from last semester and attempt the proofs. • Go through the second homework from last semester. • Reference: Chapter 2 of Logic and Computation by L.C. Paulson. SWE 623Classical Logic Duminda Wijesekera 40

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