1. Lecture Notes for SWE 623 by Duminda Wijesekera
Classical Logic
Lecture Notes for SWE 623 by
Duminda Wijesekera
SWE 623Classical Logic
Duminda Wijesekera

2. 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.
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3. 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.
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4. 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.
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5. 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
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6. 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
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7. 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.
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8. 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.
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9. 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
B B2  Applications of rules
D  Theorem being proved
There are introduction and elimination rules for each connective in Natural Deduction proof systems.
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10. Rules for Conjunction Introduction Elimination A B A ^ B A ^ B A ^ B
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11. Rules for Disjunction Introduction Elimination
A B
A v B A v B
Elimination
[A] [B]
A v B C C C
[X] denotes discharged assumption X.
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12. Rules for Implication Introduction Elimination (Modus Ponens) [A] B
A => B A
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13. 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
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14. 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.
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15. 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.
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16. 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
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17. 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.
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18. 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.
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19. 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).
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20. 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)
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21. 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.
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22. 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.
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23. 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).
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24. 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).
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25. 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.
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26. 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.
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27. 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.
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28. 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.
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29. 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 .
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30. 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.
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31. Proof Rules for Predicate Logic
Proof rules of introduction and elimination of ^, v, =>, and .
New rules required for introduction and elimination of  and  quantifiers.
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32. Proof Rules for   Introduction  Elimination
A(x) provided x is not free in the
x A(x) assumptions of A
 Elimination
x A(x)
A[t/x]
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33. 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
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34. 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].
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35. Example Proof x A(x) ^ B x A(x) ^ B x A(x) [A(t/x)] B A(t/x) ^ B
The other direction of the proof appears in the handout page 32.
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36. 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).
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37. 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.
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38. Equality Reasoning: Continued
Congruence Law for functions:
t1 = u1 …. tn = un
f(t1, …., tn) = f(u1, ….,un)
Congruence Law for Predicates:
p(t1, …., tn)  p(u1, ….,un)
Substitution Rule:
t = u
S[t/x] = S[u/x]
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39. 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))
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40. 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.
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