classical logic

SWE 623Classical Logic Duminda Wijesekera 1

Classical Logic

Lecture Notes for SWE 623

by

Duminda Wijesekera


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.


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.


Propositional Logic: Semantics• A model M M of a propositional language consists

of – a collection of atoms, say B = { bi : 0 < i },

whereis excluded from B, and

– a partial mapping M from A = { ai : 0 < i } to B = { bi : 0 < i }.

– If M(ai) B, we say that ai is true in M. M. We write “ai is true in MM” as M M |= ai. (Read MM satisfies ai).

– |= is referred to as the satisfaction relation.


Propositional Logic: Example

• B = { a1, a3} where M given as M(a1) = a1 and M(a2) = a2 has the following properties.– M M |= a1

– M M |= a1 ^ a3

– M M |= a2

– M M |= a2 => a4

• M M does not satisfy the following propositions.– M M |= a4

– M M |= a1 => a4


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.


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.


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.


Rules for Conjunction

• IntroductionA B

A ^ B

• Elimination

A ^ B A ^ B

A B


Rules for Disjunction• Introduction

A B

A v B A v B

• Elimination[A] [B]

A v B C C

C

• [X] denotes discharged assumption X.


Rules for Implication

• Introduction [A]

B

A => B

• Elimination (Modus Ponens) A => B A

B


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


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.


Prop. Proof: Example Continued

• Prove: ( A ^ B ) => (A v B)

[A ^ B ]

A ^ elimination

A v B 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.


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


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 M |= A1 , …… , M M |= An, then M M |= B.

• A collection of rules are sound if all rules in the collection is sound.

• A collection of rules is complete if M M |= A for all models M, 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.


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.


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).


Predicate LogicRecursive 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)


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.


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.


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).


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).


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.


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.


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.


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.


Interpreting Formulas

• Let M M be a model. We define M 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 M |= pi(Nt1, … Ntn ) if and only if mPred(pi)(Nt1, … Ntn ).

– I.e. pi(Nt1, … Ntn ) is true in M M if and only if its image under the map mPred holds with parameters Nt1, … Ntn .


Proof Rules for Predicate Logic

• Proof rules of introduction and elimination of ^, v, =>, and .

• New rules required for introduction and elimination of and quantifiers.


Proof Rules for

IntroductionA(x) provided x is not free in the

x A(x) assumptions of A

Elimination x A(x)

A[t/x]


Proof Rules for Introduction

A[t/x]xA(x)

Elimination[A] provided x is not

freexA(x) B in B nor in the

B assumptions of B apart from A


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].


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.


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).


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.


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]


Equality Reasoning: An ExampleThis 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))


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.

classical logic

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