chapter 2: the logic of quantified statements. predicate calculus

1Introduction to Abstract Mathematics

Chapter 2: The Logic of Quantified Statements. Predicate Calculus

Instructor: Hayk Melikya [email protected]

2.3 Arguments with Quantified Statements


Argument #1 Universal instantiation

All men are mortal.Socrates is a man.Therefore, Socrates is mortal

(xU)P(x)├ P(a)

Universal instantiation is the fundamental tool for deductive reasoning


Arguments with Quantified Statements

Rule of universal instantiation: if some property is true ofeverything in the domain, then this property is true for anysubset in the domain

Universal Modus Ponens: Premises: (x, if P(x) then Q(x)); (major)

– P(a) for some a (minor) Conclusion: Q(a)

Universal Modus Tollens: Premises: (x, if P(x) then Q(x));

~Q(a) for some a Conclusion: ~P(a)

Converse and inverse errors

Definition: An argument form is valid if no matter what particular predicates are substituted for predicate symbols in it promises,if resulting promise statements are true, then the conclusion is also true

An argument is called valid iff its form is valid


Validity of Arguments using Diagrams

Premises: All human beings are mortal; Zeus is not mortal. Conclusion: Zeus is not a human being

Premises: All human beings are mortal; Felix is mortal. Conclusion: Felix is a human being

Premises: No polynomial functions have horizontal asymptotes; This

function has a horizontal asymptote. Conclusion: This function is not a polynomial


Diagrams for Validity (p. 104)

Diagrams can sometimes be used to:– support a validity of an argument– or, show that an argument is invalid

Diagrams are not a formal proof!

mortalmortal

human

SocrateSocratess

SocrateSocratess

pneumoniapneumonia

fever

patientpatientpatientpatient


True no matter what

• the Domain is,

• or the predicates are.

z [Q(z) P(z)] → [x.Q(x) y.P(y)]

Predicate Calculus Validity

True no matter what the truth values of A and B are

A B B A

Propositional validity

Predicate calculus validity

That is, logically correct, independent of the specific content.


Arguments with Quantified Statements

Universal instantiation:

Universal modus ponens:

Universal modus tollens:


├ (xU)P(x)

To prove a theorem of the form (xU)P(x)which states “for all elements x in a given universe U, theproposition P(x) is true” we select an arbitrary aU from theuniverse, and then prove the assertion P(a) .

Let a be an arbitrary constant from the universe U. If P(a)contains no particular constant from U then P(a)├ (xU)P(x)This is called Universal Generalization


Example 1 (Universal Direct Proof)

Show that all integers divisible by 6 are even.Proof: In the language of predicate logic, we write

(x Z) ( 6 divides x x is even)

where Z = {0,±1,± 2,...} is the universe of integers. Letting a be

an integer, we assume a is divisible by 6, which means there

exists an integer y which satisfies a = 6y . Rewriting this as a= 2(3y) we have a = 2k for some integer k = 3y , which proves that a is an even integer. ▌


Universal Instantiation (UI) (xU)P(x)├

P(a) If a is an arbitrary constant from the universe U then

(xU)P(x)├ P(a)This is refered as Universal Instantiation rule.

Existential Instantiation (EI) (xU)P(x)├ P(a)

where a is a paricular contant from univers

Existantial Generalization(EG) P(a) ├ (xU)P(x)

This rule says that if P(a) true for some constan from the universe U then (xU)P(x) is true


├ (xU)P(x)

To prove a theorem of the form (xU)P(x) which states “thereexists an element x in a given universe U that satisfies theproposition P(x) ” the strategy is to show one or more elementsxU satisfy the assertion P(x).

Example 2 (Proof by Demonstration or construction) Show there exists an even prime integer.

Proof: In language of predicate logic, we would write

(x N)(x is even x is prime)The proof is simple because 2 is both prime and even


Proof of (x)P(x) by Contradiction:

To prove the theorem (x)P(x) which says “for all x , P(x) istrue” by contradiction, assume the contrary; i.e.

~(x)P(x) which is equivalent to (x) ~P(x) , which says “there

exists an x such that P(x) is not true”. One then continues theProof until arriving at a contradiction.


Example 4: (Proof by Contradiction)

Show if m, n are integers, then 5m+ 20n 1.

Proof: In the language of predicate logic this theorem becomes

(mZ)( nZ) (5m+ 20n 1).Assuming the contrary, we have

(mZ)( nZ) (5m+ 20n = 1).But the equation 5m+ 20n = 1 cannot hold since 5 divides theleft side of the equation and not the right. Hence, the denial isfalse so the theorem is true. ▌


Proof of (x)P(x) by Contradiction:

To prove “there exists an x such that P(x) is true” assume thetheorem is not true: i.e.

~(x) P(x) (x) ~P(x) which states“for all x the assertion P(x) is not true. You thencontinue the proof until you arrive at a contradiction of somekind.


Proving Unique Existential Theorems:

To prove a theorem of the form(!x U )P(x)

which states “there exists a unique element x such that P(x) istrue” the strategy is to show first that some element x satisfiesP(x) , then show that if two elements y, z U satisfy theassertion, then in fact they are the same; i.e. y = z. In predicatelogic language, we must show

(xU )P(x) (yU )(zU )[P( y) P(z) y = x]


Important Relations in Predicate Logic

a) (x)(y)P(x, y) (y)(x)P(x, y)

b) (x)(y)P(x, y) (y)(x)P(x, y)

c) (x)[P(x) Q(x)] [ (x)P(x) (x)Q(x) ]

d) (x)[P(x) Q(x)] [(x)P(x) (x)Q(x)]

e) [(x)P(x) (x)Q(x)] (x)[P(x) Q(x)]

f) (x)(y)P(x, y) (y)(x)P(x, y)


Proof: Give countermodel, where

z [Q(z) P(z)] is true,

but x.Q(x) y.P(y) is false.

In this example, let domain be integers,

Q(z) be true if z is an even number, i.e. Q(z)=even(z)

P(z) be true if z is an odd number, i.e. P(z)=odd(z)

z [Q(z) P(z)] → [x.Q(x) y.P(y)]Not Valid

Find a domain,

and a predicate.

Validity z [Q(z) P(z)] → [x.Q(x) y.P(y)]

Proof strategy: We assume z [Q(z) P(z)]

and prove x.Q(x) y.P(y)


Proof and Logic

We prove mathematical statement by using logic.

, ,P Q Q R R P

P Q R

not valid

To prove something is true, we need to assume some axioms!

This is invented by Euclid in 300 BC,

who begins with 5 assumptions about geometry,

and derive many theorems as logical consequences.


Validity of Arguments

Hence validity of arguments is defined in the same wayThe difference is:

– in predicate logic it is not always possible to go through all interpretations to prove that P logically implies Q

Why?– The number of interpretations can be infinite

Thus, proving arguments with inference rules becomes the method of choice

We can also derive new inference rules for our toolbox


Thm: For any “reasonable” theory that proves basic arithemetic truth, an arithmetic statement that is true, but not provable in the theory, can be constructed.

Gödel's Incompleteness Theorem for Arithmetic

Power and Limits of Logic

No hope to find a complete and consistent set of axioms!

Thm : Given a set of axioms, there is no procedure that decide whether quantified assertions are valid. (unlike propositional formulas)

That is, starting from a few propositional & simple

predicate

validities, every valid assertion can be proved using just

universal generalization and modus ponens repeatedly!

Good news: Gödel's Completeness Theorem


Practice problems

1. Study the Sections 3.3 and 3.4 from your textbook.2. Be sure that you understand all the examples discussed in class and in

textbook.3. Do the following problems from the textbook:

Exercise 3.3 # 1, 11, 16, 19, 21, 24, 30, 41, 55, 57. Exercise 3.4 # 1, 4, 11, 14, 22, 26, 32, 34.

chapter 2: the logic of quantified statements. predicate calculus

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