Discrete Mathematics Transparency No. 1-0 Chapter 1. The Foundations: Logic and Proofs Cheng-Chia Chen

Discrete Mathematics

Transparency No. 1-0

Chapter 1. The Foundations: Logic and Proofs

Cheng-Chia Chen

The Foundations


1. The foundations : logic and proofs

Logic and proofs: The basis of all mathematical reasoning has practical applications to the design of

computing machine give the precise meaning of math. statements.

The Foundations


Foundations of Logic (§§1.1-1.4)

Mathematical Logic is a tool for working with complicated compound statements.

It includes:A language(Vocabulary) for expressing them.A concise notation(Syntax) for writing them.A methodology for objectively reasoning about

their truth or falsity (Semantics and Axiomatics).It is the foundation for expressing formal proofs

in all branches of mathematics.

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Foundations of Logic: Overview

Propositional logic (§1.1-1.2): Basic definitions. (§1.1) Equivalence rules & derivations. (§1.2)

Predicate logic (§1.3-1.4) Predicates. Quantified predicate expressions. Equivalences & derivations.

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§1.1 Propositional Logic

Propositional Logic is the logic of compound statements built from simpler statements using so-called Boolean connectives.

Some applications in computer science:Design of digital electronic circuits.Expressing conditions in programs.Queries to databases & search engines.

George Boole(1815-1864)

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PropositionProposition: A statement that is either true or false (,but

not both) even if we do not know it is true or false now.

Example:1. Taipei is the capital of ROC. 2. Tokyo is the capital of Korea.3. 1+1 = 2. 4. 2+2 = 3.5. There is no integer x,y,z and n > 2 s.t.

xn + yn = zn ---- Fermat’s last theorem6. “Every even number > 2 is the sum of two prime numbers.” ----

Goldbach’s conjecture (1742)==>1. Statement (1)~(6) are propositions,2. (1),(3) are true; (2),(4) are false.3. Nobody knew (5) is true or false until 1994, but it must be true or false,

hence it is a proposition. (6) open problem.

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More Examples

Examples:1. What time is it? (interrogative)

2. Read this carefully! (command)

3. x+1 = 2.

4. x+y = z.

5. 李登輝是中華民國總統==> 1. (1) and (2) are not statements

2. (3),(4) are not propositions:

neither true nor false!

3. How about (5) ?

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Propositional variables and propositional constants In algebra, we use variables to denote an item with

unspecified value. e.g. 10 chickens and rabbits in a cage with 16 feet. ==>

#rabbits = ? and #chickens=? ==> let x = #rabbits, y = #chickens ==> x + y = 10 /\ 2x + 4y = 16.

Likewise, In logic, we can use propositional variable to denote an arbitrary proposition with unspecified truth value. e.g., If it is raining, the ground is wet. since the ground is

wet, it is raining now. ==> Let p =def “It is raining”, q =def “the ground is wet” ==> ((p ->q) /\ q) -> p.

Propositional constants(literals): only two values: T(true), F(false).

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An operator or connective combines one or more operand expressions into a larger expression. (E.g., “+” in numeric exprs.)

Unary operators take 1 operand (e.g., −3);Binary operators take 2 operands (eg 3

4). ternary: p? a:b In C and Java

Propositional or Boolean operators operate on propositions or truth values instead of on numbers.

Operators / Connectives Topic #1.0 – Propositional Logic: Operators

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Some Popular Boolean Operators

Formal Name Nickname Arity Symbol

Negation operator NOT Unary ¬ (~)

Conjunction operator AND Binary Disjunction operator OR Binary Exclusive-OR operator

XOR Binary

Implication operator IMPLIES Binary Biconditional operator

IFF Binary ↔

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Forming Compound propositions from simpler ones using logical connectives

Primitive (or atomic) propositions: propositional variables propositional constants

P, Q: two known propositions Then the followings are also propositions:

1. “P and Q” : ( P /\ Q) conjunction of P and Q2. “It is not the case that p”: (~P) negation of P3. “P or Q” : (P \/ Q) disjunction of P and Q4. “P implies Q” : ( P Q) implication

P : antecedent, premise, hypotheses; Q: consequence, conclusion equivalent phrase:

– “P entails Q”, “Q if P”, “P only if Q”

5. “P iff Q” (P ↔ Q) equivalence of P and Q

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Truth conditions for compound propositions

1. P /\ Q is true iff both P and Q are true

2. P \/ Q is true iff either P or Q is true

3. ~P is true iff P is not true (i.e., false)

4. P ↔ Q is true iff both are true or both are false.

5. PQ is true iff whenever P is true, Q is true

( i.e., it is not the case that P is true but Q is false ) In the propositions, /\, \/, ↔ and are called binary

connectives(or operations) and ~ is called unary connective(or operations).

Other binary connectives are possible.

What about the truth condition for primitive propositions?

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From natural language statements to formal logic expressions

Natural language statementToday is FridayI have a test todayIt is not the case that today is

FridayToday is not FridayToday is Friday and

I have a test todayIf today is Friday, then I have

a test today.Today is Friday or I have a

test today.

symbolized logic expr.pq~p

~p(P q)∧

(pq)

(P q)∨

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Truth conditions for propositions

ProblemProblem: : when will you say that the sentence:

It_is_raining is true ?

=> The proposition:” It_is_raining” is true if the meaning (or fact) that the proposition is intended to represent occurs(happens, exists) in the situation which the proposition is intended to describe.

=>Example: Since it is not raining now(the current situation), the statement It_is_raining is false (in the current situation). But if it were raining now, then I would say that It_is_raining is true.

Factors affecting the truth value of a proposition: the situation in which the proposition is used. the meaning of the proposition.

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Truth table

A B ~A A/\B A\/B A->B A<->B

0 0 1 0 0 1 1

0 1 1 0 1 1 0

1 0 0 0 1 0 0

1 1 0 1 1 1 1

•The meaning of logical connectives can also be represented by a truth table.

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Other usual connectives

Besides /\,\/, and <->, there are also some other connectives which you will study in digital system.

xor , nand, nor.Rules: let I(x) denote the truth value of x.

I(A B)=def I(~(A<->B)) [ = 1 iff not I(A)=I(B) ]

I(A nand B) =def I(~(A/\B)).

I(A nor B) =def I(~( A \/B)).

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Nested Propositional Expressions

Use parentheses to group sub-expressions:“I just saw my old friend, and either he’s grown or I’ve shrunk.” = f (g s) (f g) s would mean something different f g s would be ambiguous

By convention, “¬” takes precedence over “” and “” takes precedence over “”. ¬s f means (¬s) f , not ¬ (s f)

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The Implication Operator

The implication p q states that p implies q.

I.e., If p is true, then q is true;

but if p is not true, then q could be either true or false.

E.g., let p = “You study hard.” q = “You will get a good grade.”

p q = “If you study hard, then you will get a good grade.” (else, it could go either way)

in p q , p is the antecedent and q is the consequent of the implication.

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Implication Truth Table

p q is false only whenp is true but q is not true.

p q does not saythat p causes q!

p q does not requirethat p or q are ever true!

E.g. “(1=0) pigs can fly” is TRUE!

p q p q F F T F T T T F F T T T

The onlyFalsecase!

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Examples of Implications

“If this lecture ends, then the sun will rise tomorrow.” True or False?

“If Tuesday is a day of the week, then I am a penguin.” True or False?

“If 1+1=6, then ChenShuaBen is president. ” True or False?

“If the moon is made of green cheese, then I am richer than Bill Gates.” True or False?

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English Phrases Meaning p q

“p implies q”“if p, then q”“if p, q”“when p, q”“whenever p, q”“q if p”“q when p”“q whenever p”

“p only if q”“p is sufficient for q”“q is necessary for p”“q follows from p”“q is implied by p”We will see some equivalent

logic expressions later.

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Converse, Inverse, Contrapositive

Some terminology, for an implication p q:Its converse is: q p.Its inverse is: ¬p ¬q.Its contrapositive: ¬q ¬ p.Which of the above three has the same

meaning (logical equivalent; same truth table) as p q ? Ans: ¬q ¬ p.

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How do we know for sure?

Proving the equivalence of p q and its contrapositive using truth tables:

p q q p p q q pF F T T T TF T F T T TT F T F F FT T F F T T

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Boolean Operations Summary

We have seen 1 unary operator (out of the 4 possible) and 5 binary operators (out of the 16 possible). Their truth tables are below.p q p p q p q pq p q pqF F T F F F T TF T T F T T T FT F F F T T F FT T F T T F T T

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Some Alternative Notations

Name: notandorxorimplies iffPropositional logic: Boolean algebra: ppq+C/C++/Java (wordwise):!&&||!= ==C/C++/Java (bitwise): ~&|^Logic gates:

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Bits and Bit Operations

A bit is a binary (base 2) digit: 0 or 1.Bits may be used to represent truth

values.By convention:

0 represents “false”; 1 represents “true”.

Boolean algebra is like ordinary algebra except that variables stand for bits, + means “or”, and multiplication means “and”. See chapter 10 for more details.

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Bit Strings

A Bit string of length n is an ordered series or sequence of n0 bits.

By convention, bit strings are written left to right: e.g. the first bit of “1001101010” is 1.

When a bit string represents a base-2 number, by convention the first bit is the most significant bit. Ex. 11012=8+4+1=13.

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Bitwise Operations

Boolean operations can be extended to operate on bit strings as well as single bits.

E.g.:01 1011 011011 0001 110111 1011 Bit-wise OR01 0001 Bit-wise AND10 1010 Bit-wise XOR

Topic #2 – Bits

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End of §1.1

You have learned about:Propositions: What

they are.Propositional logic

operators’ Symbolic notations. English equivalents. Logical meaning. Truth tables.

Atomic vs. compound propositions.

Alternative notations.Bits and bit-strings.Next section: §1.2

Propositional equivalences.

How to prove them.

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§1.2 Propositional Equivalence

Two syntactically (i.e., textually) different compound propositions may be semantically identical (i.e., have the same meaning). We call them equivalent.

Learn:Various equivalence rules or laws.How to prove equivalences using symbolic

derivations.Analogy:

x * (5 + y) and xy + 5x are always equal in algebra. But how about

p /\ (q \/ r) and (p/\q) \/ (p /\ r) ?

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Tautology, contradiction and contingency

Some definitions: Let A be any proposition, then

1. A is a tautology(contradiction) iff A is true (false) no matter what the truth values of the prop. variables inside A are given.

2. If A is neither a tautology nor a contradiction, then A is called a contingency.

The geography of propositions

Tautologies(valid)

Contradictions(unsatisfiable)

Contingencies

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Examples:

p \/ ~p, p p : a tautologyp /\ ~p : a contradictionp, p ~p, ~p p : a contingency

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Logical Equivalence

proposition A is logically equivalent to proposition B, written A B, IFF the compound proposition AB is a tautology.

Note:1. is a logical symbol (connective) whereas is a meta logical symbol.

2. The textbook uses instead of .Propositions A and B are logically

equivalent to each other iff A and B contain the same truth values as

each other in all rows of their truth tables.

Topic #1.1 – Propositional Logic: Equivalences

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Logical Equivalence

Theorem1: Logical equivalence relation is reflexive, symmetric and transitive.

I.e., for all propositions A,B and C, 1. A A2. A B implies B A and3. A B and B C imply A C.

Theorem2[substitution theorem]: If A B and C[X] is a proposition containing X as a subproposition, then C[A] and C[B] are logically equivalent, where C[A] is the result of C with X in C replaced by A. ex: (p q) ∨ (q p), C[X] =∨ def ~(p X) ∧ => ~(p (p q) )∧ ∨ ~(p (q p))∧ ∨

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Determine tautologies, contradictions and contingencies

Problems: Given a proposition A, how can you determine whether it is a tautology, a contradiction or a contingency?

=>Exhausted evaluation! (by using truth table)

e.g., A = (p->q) <-> (~p \/q) a tautology ?

1. there are two primitive props: p, q.

2. So there are 4 possible assignments of truth values to p and q.

p=0,q=0: p=0,q=1: p=1,q=0: p=1,q=1:

3. Under each assignment, we can evaluate the truth value of A (in bottom up way) by the truth condition rule of connecting connectives.

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Determine tautologies

p:0q:0

p->q:1

~p:1

~p \/q:1

(pq)<->~p\/q :1

The evaluation tree for (p->q)<-> (~p\/q)

p q ~p p->q ~p\/q p->q <-> ~p\/q

0 0 1 1 1 1

0 1

1 0

1 1

T

The truth table for pq <-> ~p\/q

11

1

1

00

0

1

1

1

01

A tautology since all are 1’s

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Determine other properties

1. logical equivalence: A B iff A<->B is a tautology.

So we can apply the procedure for determining tautology to determine if A B. Or A and B have the same truth values in all rows of their truth

table.

2. contradiction: A is a contradiction iff ~A is a tautology

3. Satisfiability: A is satisfiable iff ~A is not a tautology.

4. contingence: A is a contingence iff neither A nor ~A is a tautology.

5. logical consequence: B is a logical consequence of A [denoted A |= B] iff AB is a tautology.

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Ex. Prove that pq (p q).

p q pp qq pp qq pp qq ((pp qq))F FF TT FT T

Proving Equivalence via Truth Tables

FT

TT

T

T

T

TTT

FF

F

F

FFF

F

TT


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Equivalence Laws

In algebra, we make use of equality reasoning to replace equal by equal: Ex: Given the know equality: x + y = z, we have x + (x + y) = x + z.

Thanks to substitution theorem, we can do also equality reasoning for logical expressions. Ex: since p \/ p p, q /\ ( p \/ p) q \/ p


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Equivalence Laws - Examples

Identity: pT p pF pDomination: pT T pF FIdempotent: pp p pp pDouble negation: p pCommutative: pq qp pq qpAssociative: (pq)r p(qr)

(pq)r p(qr)

Note. The textbook uses instead of .


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More Equivalence Laws

Distributive: p(qr) (pq)(pr) p(qr) (pq)(pr)

De Morgan’s:(pq) p q

(pq) p q generalization: ~ (p /\ q /\ r …) ~p \/ ~q \/ ~r \/ …

Trivial tautology/contradiction: p p T p p F


AugustusDe Morgan(1806-1871)

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One Example:

Show that ~(p \/ (~p /\ q)) ~p /\ ~q

Proof:1. By truth table (omitted)2. ~(p \/ (~p /\ q) deM

~p /\ ~(~p /\ q) deM

~p /\ (~(~p) \/ ~q) DbNeg [+substitution]

~p /\ (p \/ ~q) Distr

(~p /\ p) \/ (~p /\ ~q) TrivalContradition

F \/ (~p /\ ~q) identity

(~p /\ ~q)

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Defining Operators via Equivalences

Using equivalences, we can define operators in terms of other operators.

Exclusive or: pq (pq)(pq) pq (pq)(qp)

Implies: pq p qBiconditional: pq (pq) (qp)

pq (pq)Boolean Algebra uses as operationsBoolean Ring uses T as operations


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An Example Problem

Check using a symbolic derivation whether (p q) (p r) p q r.

(p q) (p r) [definition of ]

(p q) (p r) [Defn. of ]

(p q) ((p r) (p r)) [DeMorgan’s Law + DoubleNeg]

(p q) ((p r) (p r)) [ commutes]

(q p) ((p r) (p r))


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Example Continued...

(q p) ((p r) (p r)) [ associative] q (p ((p r) (p r))) [distrib. over ] q (((p (p r)) (p (p r))) [assoc.]

q (((p p) r) (p (p r))) [trival taut.]

q ((T r) (p (p r))) [domination]

q (T (p (p r))) [identity]

q (p (p r)) cont.


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End of Long Example

q (p (p r)) [DeMorgan’s] q (p (p r)) [Assoc.] q ((p p) r) [Idempotent] q (p r) [Assoc.] (q p) r [Commut.] p q r

Q.E.D. (quod erat demonstrandum)


(Which was to be shown.)

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Review: Propositional Logic (§§1.1-1.2)

Atomic propositions: p, q, r, … Boolean operators: (~) Compound propositions: (p q) rEquivalences: pq (p q)Proving equivalences using:

Truth tables. Symbolic derivations. p q r …

Topic #1 – Propositional Logic

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Logical puzzles

Knight or knave:Two kinds of inhabitants lived in an island

Knights : always tell the truth Knaves : always lie.

Dialog of two people A and B lived in the island: A : “You are a knight”, B: “ We are opposite types”

Question: What are the kinds of A and B ?

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Solution:Formalization :

Let p = “ A is a knight” ; ~p “A is a knave” Let q = “B is a knight” ; ~q “B is a knave” Observed facts: 1. since A says q, we have p q (1) /\ ~p ~q (2)

2. since B says R where R is (p /\ ~q) \/ ( ~p /\ q) ) (5) , we have q R (3) /\ ~q ~R (4).

Analysis: p true =(1)=> q true =(3)=> R true =(5)=> R false (a contradiction!) P false =(2)=>q false =(4)=> ~R => ~p /\ ~q. (consistent). Conclusion: Both are knaves!!

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Muddy Children puzzle ( reasoning about knowledge)

Two children Son and Daughter getting dirty mud on their foreheads. Father says to them: “At lest one of you has a muddy

forehead.” (1) F2S: Do you know whether you have a muddy

forehead? S : No, Then, F2D: Do you know whether you have a muddy

forehead? D: Yes.

Problem: Why did S and D answer No and Yes, respectively?

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Analysis

Let d = “D has a muddy forehead”

s = “S has a muddy forehead”Observed facts:

1. d \/ s. Hence know(S, d \/ s) and know(D, d \/ s)

2. Since both children can see each other’s forehead but cannot see their own foreheads.

we have know(S, d), Hence S only knows

{d, d \/ s}, which cannot make him infer that s is true. Hence S answered no when he was asked.

3. D’s inference when asked:

~d => Know(S, ~d) => Know(S, {~d, d \/ s }) => know(S, s)

=> S answered Yes.

Now since S did not answer Yes, ~d must be false (i.e., d must be true), => D answered yes.

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1.3 Predicates (述詞 ) and quantifiers(量詞 )

Problems of Propositional logic: Cannot analyze the content of primitive

propositions. cannot group similar propositions cannot analyze quantifiers.

Ex: Consider the following situations:

1. Family = {f,m,c1,c2,c3 } 2. We are interested in who is whose

brother. 3. Suppose in Family, ci is cj’s brother if i

j for i,j =1..3.

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Inadequacy of propositional logic

Primitive propositions needed: x_is_y’s_brother, x,y Family So we need 25 propositions.

New approach: Parameterized proposition(Predicate). Use a (predicate) symbol B to symbolize the

statement: B(x,y) x is y ‘s brother. e.g., B(f,c1) : f is c1’s brother, B(c2,c2) : c2 is c2’s brother, ...

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Benefits of Predicates

Advantage of using predicates:==>

1. need only 6 (1 predicate and 5 names) symbols.

2. can group related propositions together. (B(x,y) and others, for instance,

q(x) x_isOld. )3.The most important : allow representation of

quantified statement like: c1_has_a_brother (or, there is an object x which is c1’s brother, x B(x, c1))

to be inferred from

c2_is_c1’s brother ( B(c2,c1) )

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PredicatesIn the expression: B(x,y)

B : predicate symbol (or proposition function) x (y) : first (2nd) argument B(x,y) : [atomic] proposition(or formula) #arguments in B(X,Y): the arity of the predicate

symbol B.Notes:1. Each predicate symbol has a fixed arity.

2. B(x,y,c1) and B(x) are meaningless.

because used #arguments different from B’s arity.

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Predicates (cont’d)

3. The truth value of a proposition p(x1,...,xn) depends on both p and real values of x1,...,xn.Ex1: Let p(x) “x >3 “. then

P(4) “4 > 3” has truth value T, P(2) “2 > 3” has truth value F.

Ex2: Let Q(x,y) “x = y +3” then Q(1,2) = “1 = 2 + 3” is false. Q(3,0) = “3 = 0 + 3” is true.

Mathematically, we can define (the meaning of) P as a function [P] [P]: Un (n is P’s arity) --> {T,F}with the intention that “P(x1,...,xn)” holds (is true) iff [P]([x1],...,[xn]) = T.

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Set representations of truth-functionsEx:

[B]: Family2 -->{T,F} s.t. [B](x,y) = T iff x =[ci], y=[cj], i j. P(x) “x >3 “ , So [P]: N --> {T,F } s.t. [P](x) = T iff x > 3. Q(x,y) “x = y +3” , so [Q] : N2 --> {T,F} with [Q](x,y) = T iff x = y +3.

Note: [P] partition N into two sets, one with members mapped to T and one mapped to F.

N

{4,5,...}{1,2,3}

[P](x) = T.

[P](x) = F

So, instead of representP as a truth function, we equivalently represent Pas the set {x | [P](x) = T } = {4,5,6,...}

Similarly, we can represent Q as {(x,y) | x = y + 3 }

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constant, function, variables and terms

In the statement: “y > min(x, 3)” --- (s1) x,y are called variables, 3 is a constant, min is a function symbol with arity 2 “min(3,2)” behaves more like x, 3 than “x >y”. So if let P(x,y) “x > y”, then s1 can be represented as

P(y, min(x,3))we call any expression that can be put on the

argument position of an atomic proposition a termObviously, constants and variables are terms;

moreover,if f is a function of arity n, and t1,...tn are n terms, then so is f(t1,...,tn).

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Quantifiers and universe of discourse

Consider the sentence: p(x) = “x > 0”. 1. as stated before, p(x) is true or false depending on

the value of x. 2. p(1), p(2),... : true; p(0) , p(-1),...: false

Universe of Discourse (U,論域 ): The set of objects (domain) from which all variables

are allowed to have values.Ex: consider the sentence:

s2: ”all numbers are greater than 0”.

(for-all x “x > 0” ) or ( x p(x) ) U = N+={1,2,3,...} => s2 is true. U = Z={...,-1,0,1,...}=>

s2 is false.

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Universal and existential quantificationsA: any statement (or called formula)

e.g., A = p(x,y) /\ q(x,f(3)),...

x: any variable, Then 1. (x A) is a new formula called the universal

quantification of A, 2. (x A) means:

“A(x) is true for all values of x in the universe of discourse.”

3. (x A) is a new formula called the existential quantification of A,

4. (x A) means:“A(x) is true for some value of x in the universe of

discourse.”

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Truth conditions for quantifications

quantification When ture ? When false ?

(x A) A is true for every x

There is an x for which A is false.

(x A) There is an x for which A is true.

A is false for every x.

Logical equivalences: 1. ~xxxx

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The Universal Quantifier

Example: Let the u.d. of x be {1,2,3,4} Let P(x) be the predicate “x2 > 10.”

Then the universal quantification of P(x), x P(x), is the proposition:

All numbers x (in u.d.) satisfy the predicate x2 > 10 which is ture if P(1), P(2),P(3), P(4) all hold, i.e., (P(1) /\ p(2) /\(3) /\

p(4) ). But since P(2) = “22 > 10 “ is false, the universal

quantification is thus false.

Topic #3 – Predicate Logic

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The Existential Quantifier

Example: Let the u.d. of x be all natural numbers N = {0,1,2,3,4,…} Let P(x) be the predicate “x2 > 10.”

Then the existential quantification of P(x), x P(x), is the proposition:

Some number x (in u.d.) satisfy the predicate x2 > 10

which is ture iff one of P(0),P(1), P(2),P(3), P(4), … holds, i.e., (P(0) \/ p(1) \/ (2) \/ … ). since P(4) = “42 > 10 “ is true, the universal quantification is thus

true.


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Translation of NL statements into logical ones

Translate the following sentences into logical expressions:

1. Everybody likes Wu2. Everybody loves somebody.3. There is somebody loved by every one.4. Nobody likes everybody.5.There is somebody Wu don’t like.6. there is somebody no one likes.7.there is exactly one person everybody likes.8 Wu likes more than one person9. Everyone likes himself.10. There is someone who likes no one who does not

like himself.

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Solutions 1. Everybody likes Wu

x like(x, Wu)2. Everybody loves somebody.

x y love(x, y) 3. There is somebody loved by every one.

yx love(x, y) 4. Nobody likes everybody.

~xy like(x, y) or x y ~love(x, y) 5.There is somebody Wu don’t like.6. There is somebody no one likes.

y~ x like(x, y) 7.there is exactly one person everybody likes.

y ( (x like(x, y) /\ ~z (~z = y /\ w like(w, z) ) ) or y ( (x like(x, y) /\ z (w like(w, z) z = y ) )

8 John likes more than one person. xy ( ~(x = y) /\ like(john,x) /\ like(john, y).

9. Everyone likes himself. y like(y, y)

10. There is someone who likes no one who does not like himself.

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Summary of the syntax of predicate logic

Suppose we are given the following sets of symbols (Vocabulary) : 1. Object variables: x,y,z,… 2. Predicate symbols: p,q,r,… 3. function symbols : f, g, h,… 4. constant symbols: a,b,c,…

Notes: 1. (2,3,4) together is called a signature, which can be

designated by users according to their application while (1) is usually predefined by the logic.

2. Constant symbols may be viewed as function symbols of zero-arity.

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The syntax of predicate logicA signature is a triple( F, P, ar) where F is a set of

function symbols and P is a set of predicate symbols, and ar: P U F N is the arity fuction.(1) If f F, then f is a function symbol of of arity ar(f).

(2) If n = 0, then f is a constant.

(3) If p P, => p is a predicate symbol of arity ar(p).

(4) If n = 0 => p is an atom or proposition.

(5) Fun() = F is the set of function symbols

(6) Pred() = P is the set of predicate symbols

Ex: Arith =({0, 1, suc + *} , { >, =}, ar )

String=({e, *},{ < }, ar) ...

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Terms

Terms (object expressions) : Every variable x or constant symbol b is a term if f is a function symbol of arity k>0, and t1,…,tk

are known terms, then f(t1,…,tk) is a term.

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Hierarchy of definitions: primitive symbols terms atomic formula

(general) formula.Atomic Formuls:

if p is a predicate symbol of arity k≥0 , and t1,…tk are terms, then p(t1,…,tk) is an atomic formula.

note: zero-arity predicate symbols are propositional is variables in propositional logic.

Formula: 1. Every atomic formula is a formula. 2. If A and B are formulas, then (A/\B), (A\/B), (A -> B), (A B), A<->B, and ~A are formulas. 3. if x is a variable and A is a formula, then X A and X A are formula.

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-structures and variable assignments : a signature.A (first order) -structure M is a triple:

(D, {[f]}f Fun(∈ ), {[p]}p Pred(∈ ) ) where D is a nonempty set (of objects), called the domain of M

(,written DM). [ f ]: Dar(f) -> D is a function on D of arity n. [f] is called the

denotation(or meaning) of f in M (, or written fM). [p] Dar(p) is a ar(p)-ary relations on D (a subset of Dar(p)). [p] is

called the denotation (or meaning) of p in M (,written pM).

Variable M-assignment: Any mapping : Var -> DM from variables to the domain of a -

structure M is called a variable M-assignment (or simply an assignment).

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Denotations of terms in structures

syntactic objects semantic objects (M, )

-----------------------------------------------------------------------

function symbol(f) function ([f] or fM)

constant(c) domain object ([c] or cM)

predicate symbol(p) relation ([p] or pM)

variable (X) domain object ((X))

Terms ?

Formula ?

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Denotation of terms in a structure

Given the pair (M,), we can define a meaning function (M,) : T() -> DM inductively as follows:1. (M, ) (X) = (X) for any variable X.

2. (M, )(f(t1,..,tn)) = fM( (M, )(t1),...,(M, )(tn)),

where f is a function symbol of arity n, and t1,..,tn are terms.

Note: (M,)(c) = cM for constant symbol c (function of arity zero)

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: a variable assignment; d: an object D; X: a variable,

=> [X/d] is a new variable assignment agreeing with on all variables but X, for which it is assigned d. Formally,

[X/d] is defined as follows: for all var Y :

[X/d] (Y) =def if Y = X then d else (Y)

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Truth condition of formulasThe relation M, |= A (A is true (satisfied)

under interpretation M and assignment ) is defined inductively as follows: 1. M, |= p(t1,..,tn), iff pM((M, )(t1),..,(M, )(tn)) holds

i.e, ((M, )(t1),..,(M,)(tn)) pM.

2. M, |= ~A iff not M, |= A 3. M, |= A \/ B iff M, |= A or M, |= B 4. M, |= A /\ B iff M, |= A and M, |= B 5. M, |= X A iff M, [x/d] |= A for all d D. 6. M, |= X A iff M, [x/d] |= A for some d DM.

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Logical notions A: a -formula; T: a set of formulas

1. A is satisfiable iff there is -structure M and an M-assignment s.t. M, |= A.

2. A is valid iff for all -structures M and for all M-assignment , M, |= A.

3. A is valid in M iff for all M-assignments , M, |=A.4. A is satisfiabe in M iff M, |= A for some M-assignment . 1' T is satisfiable iff there is a -structure M and an M-

assignment s.t. M, |= A for all A in T.2' T is valid in M iff for all M-assignments , M, |=A for all A in

T.3' T is satisfiabe in M iff there is an assignment s.t. M, |= A

for all A in T.

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Semantical conclusionA [or T ] is valid iff for all -structures M and for all M-

assignment , M, |= A [for all A T ].∈

If A is valid in M,(denoted M |= A) we say M is a model of A.

T: a (set of) formulas; A: a formula T |= A (,A is a logical consequence of T) iff Every model of T is a model of A.

A and B are logically equivalent (denoted A B) iff A<-> B is valid. (i.e., for any structure assignment pair (M, ), M, |= A iff M, |= B.)

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syntactic objects and their meaning

syntactic object examples semantical meaning

constant a,b,c domain object

[a]

function symbol f,g,h domain function

[f] : Uarity U

predicate symbol p,q, r,… domain relations

(or truth function)

[p] : Uarity {t,f }

variable x,y,z,… ? (determined by assignment)

term f(t1,…,tn) domain object

[f] ([t1],…[tn])

atomic formula f(t1,…,tn) proposition (trut value) [p] ([t1],…[tn])

(general) formula A,B,C truth value [A] dertermined by components.

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Some Terminology

Occurrences of variablesFree/Bound Variables

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Occurrences of variables in a formula

Let A = p(f(x),c) \/ q(x,y)

A contains 2 variables x, y, where x occurs twice and y occurs once.

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Free and Bound Variables

An expression like p(x) , where p is a predicate symbol, is said to have a free variable x (meaning, x is undefined).

A quantifier (either or ) operates on an expression having one or more free variables, and binds one or more of those variables, to produce an expression having one or more bound variables. ex: x p(x,y).


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Example of Binding

p(x,y) has 2 free variables, x and y.x p(x,y) has 1 free variable, and one bound

variable. [Which is which?]

An formula with zero free variables is a sentence (or proposition).

An formula with one or more free variables is still only a predicate: x P(x,y)

Note: we use expression and formula interchangeably as logical expressions are usually called formulas.


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Formalization of free variablesA: a formula (or expression)fv(A) = the set of all free variables in A is defined as

follows: If A is atomic => fv(A) is the set of all variables in

A fv(~A) = fv(A) fv(A/\B)=fv(A\/B)=fv(A->B)=fv(A<->B)= fv(A) U

fv(B). fv(x A) = fv(xA) = fv(A) \ {x}.

What is fv(A) where A is an atomic formula? can be defined by recursion also: 1. var(t) = {t} if t is var 2. var(t) = var(t1) U var(t2)…U var(tn) if t = f(t1,t2,…tn) or p(t1,…tn). Finally let fv(A) = var(A) if A is an atomic formula.

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Formalization of bound variablesbd(A): the set of bound variables in A is defined as

follows; 1. If A is atomic => bd(A) = {} 2. bd(~A) = bd(A) 3. bd(A/\B) = d(A\/B) = bd(AB) =bd(A<->B) = bd(A) U bd(B). 4. bd( x A) = ∀

bd(A) if x ∉ fv(A). bd(A) U {x} if x fv(A). ∈Example A = ( x p(x,y)) \/ q(x) => fv(A) = {x,y}; bd(A) = {x} x occurs twice in A, one occurs free and the other

occurs bound.Def: A is a sentence iff fv(A) = {}

Note: A is a proposition if A is a sentence

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Conventions

Sometimes the universe of discourse is restricted within the quantification, e.g.,

x>0 P(x) is shorthand for“For all x that are greater than zero, P(x).”=x (x>0 P(x))

x>0 P(x) is shorthand for“There is an x greater than zero such that P(x).”=x (x>0 P(x))


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More to Know About Binding

x x P(x) x is not a free variable in

x P(x), therefore the x binding isn’t used.

(x P(x)) Q(x) The variable x is outside of the scope of the x

quantifier, and is therefore free. Not a proposition!

(x P(x)) (x Q(x)) This is legal, because there are 2 different x’s!


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Quantifier Equivalence Laws

Definitions of quantifiers: If u.d.=a,b,c,… x P(x) P(a) P(b) P(c) … x P(x) P(a) P(b) P(c) …

From those, we can prove the laws:x P(x) x P(x)x P(x) x P(x)

Which propositional equivalence laws can be used to prove this? DeMorgan’s Law


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More Equivalence Laws

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

x (P(x) Q(x)) (x P(x)) (x Q(x))x (P(x) Q(x)) (x P(x)) (x Q(x))


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Some Number Theory Examples

Let u.d. = the natural numbers 0, 1, 2, … “A number x is even, E(x), if and only if it

is equal to 2 times some other number.”x (E(x) (y x = 2 * y ))

“A number is prime, P(x), iff it’s greater than 1 and it isn’t the product of two non-unity numbers.”

x (P(x) (x>1 yz x=yz y1 z1))


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Goldbach’s Conjecture (unproven)

“Every even number greater than 2 is the sum of two primes.”

x [x>2 E(x)] →

y z ( P(y) P(z) y+z = x ).

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Calculus Example

One way of precisely defining the calculus concept of a limit, using quantifiers:

|)(|||

::0:0

)(lim

Lxfax

x

Lxfax


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End of §1.3-1.4, Predicate Logic

From these sections you should have learned: Predicate logic notation & conventions Conversions: predicate logic clear English Meaning of quantifiers, equivalences Simple reasoning with quantifiers

Upcoming topics: Introduction to proof-writing. Set theory –

a language for talking about collections of objects.


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1.5 Rules of inference

The following 3 sections are about proofs.Argument (論述 ):

Valid arguments Argument form (pattern)

Rules of inference For propositional reasoning For quantifier reasoning

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Arguments

An argument is a sequence of statements made to convince people of the truth of the final conclusion.

Definition 1.1: An argument in logic is a sequence of propositions: A1, A2,…,An (n > 0) The final one An is called the conclusion of the

argument, Some of {A1,…,An} are called the premises of the

statements. An argument is basic if all propositions but the final

one are premises.Note: Our definition of argument is more general than that

given in the textbook, which corresponds to our basic argument.

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Example of an argument

1. Have a passwd can log onto the network (premise)

2. Can log onto the network can change grade (premise)

3. Have a passwd (premise)

Therefore :

4. can log onto the network (lemma)

5. can change the grade (conclusion)

Notes: 1. 1..4 is a basic argument;

2. 1..5 is a non-basic argument.

Definition 1.2: An argument is valid if the truth of all its premises implies the conclusion is true.

Note: Basic argument A1, A2,…,An is valid iff A1, A2,…,An-1

|= An or (A1/\ A2/\…An-1) An is a tautology.

Ex: the above argument (1..4 or 1..5) if valid since the conclusion is true whenever all premises are true.

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Valid arguments and truth of conclusions

Note: An argument is valid does not mean that its conclusion must be valid since it is possible that some of the premises are false.

Ex: The following argument is valid

1. have access to the grading system can Change Grade

2. have access to the grading system

Therefore

3. can change the grade.

But (in the actual condition) the conclusion (3) is false since premise (2) is false.

Note: A valid argument can guarantee the truth of its conclusion only when all its premises are true.

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Argument forms

In the basic argument :

1. have access to the grading system can Change Grade

2. have access to the grading system

Therefore

3. can change the grade.

Let p represent “have access to the grading system “ and

q represent “can Change Grade”

Then the basic argument has the form (or pattern):

1. p q

2. p

3. qTo save space, we may write the above form as :

pq, p // q

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Argument form

Definition 1.3 : An argument form in logic is a basic argument of which

some propositions contain propositional variables. If A is an argument form and A’ is the result of

replacing all occurrences of each variable by a same proposition, then A’ is said to be an instance of the argument form A .

Ex: The argument form : p q, p // q has instances: raining wet, raining // wet, studyHard GoodGrade, studyHard // goodGrade pq , p // q p \/ q ~r, p \/ q // ~r, but not raining wet, raining // cloudy (why?)

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Argument forms and arguments

Def: An argument form A1,A2,…,An-1 // An in logic is valid if (A1/\A2…/\An A) is a tautology.

Theorem : If an argument form A = (A1, .., //An) is valid then all of its instances are valid.

Pf: This is a simple result of the substitution theorem of propositional logic.

Let A’ = (A’1, .., A’n) be any instance of A. Then

A is valid iff B= (A1/\ A2/\…An-1) An is a tautology

=> B’ = (A’1/\ A’2/\…A’n-1) A’n is a tautology (substitution theorem)

=> argument A’ is valid.

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Analogy: When we say the equality x + x = 2 * x is valid, it implies : #Students + #students = 2 * #Students ; #sons + #sons = 2

#sons ( #students / #classe) + ( #students / #classe) = 2 * + ( #students /

#classe) ( x – z = 5) + (x – z + 5) = 2 * (x – z + 5), …

Similarly, if pq, p // q is valid then so are raining wet, raining // wet, studyHard GoodGrade, stydyHard // goodGrade p \/ q ~r, p \/ q // ~r

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Rules of inferences

How to show the validity of an argument form:

Solution 1: Truth table If (A1, .., An) is an argument form with variables

p1,…,pm, then establish a truth table with 2m rows for the target formula (A1/\A2…/\An-1) An and then check if it is a tautology.

Problem: 2m is too large a number of rows to check validity even for a small m (e.g., 20).

Solution 2: deduction: Establish the validity of a (complicated)

argument forms from simpler known valid ones.

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Application of rules of inference

A rule of inference is an argument form that is chosen to build the validity of other argument forms. A rule of inference is valid if it is a valid

argument form. An invalid rule of inference is called a fallacy.

Ex: Modus ponens (MP): pq, p // q Conjunction : p, q // p /\ q Abduction: pq, q // p (fallacy)

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Application of rules of inferences

Known assumptions (premises): snow go skiing snow

Then1.Snow

2.Snow go skiing

3.Skiing

Is a valid argument. 1, 2 are premises 3 is the result of applying rule MP to 1 and 2.

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Modus Ponens & Tollens

p Rule of modus ponenspq (a.k.a. law of detachment)q

qpq Rule of modus tollens p

“the mode of affirming”

“the mode of denying”

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Some Inference Rules

p Rule of Addition pq

pq Rule of Simplification p

p Rule of Conjunction q pq

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Syllogism Inference Rules

pq hypothetical qr syllogism pr

p q disjunctive p syllogism q

p \/ q, ~p \/ r // q \/ r Resolution

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Build arguments using rules of inference

Suppose we have the following assumptions (premises):“It is not sunny and it is cold.”“We will swim only if it is sunny.”“If we do not swim, then we will dance.”“If we dance, then we will be home early.”

Given these premises, try to give a valid argument with the conclusin “We will be home early” using inference rules.

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Example cont.

Abbreviations: sunny = “It is sunny”; cold = “It is cold”;

swim = “We will swim”; dance = “We will dance”; early = “We will be home early”.

Then, the premises can be written as:(1) sunny cold

(2) swim sunny (3) swim dance

(4) dance early

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Valid argument Example cont.

Step Proved by1. sunny cold Premise #1.2. sunny Simplification of 1.3. swimsunny Premise #2.4. swim Modus tollens on 2,3.5. swimdance Premise #3.6. dance Modus ponens on 4,5.7. danceearly Premise #4.8. early Modus ponens on 6,7.

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Fallacies

A fallacy is an inference rule which is not valid.

(I.e., It is possible that all premises are true but the conclusion is false).

Ex: fallacy of affirming the premise: pq, q // p // q true, pq true but p false is possible Fallacy of denying the conclusion: pq, ~p // ~q ;

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Inference Rules for Quantifiersx P(x) universal instantiation (UI) P(t) (substitute any term t for x in P(-))

Ex: from x x2 > 0 infer (y+2)2 > 0.

P(t) existential generalization (EG)x P(x) (substitute a new var x for term t in P(-) )

ex: from (y + 2 )2 > (y+2) infer x x2 > (y+2) or x x2 > x.

Note: Term t is either a constant c or a variable x or function expression f(t1,…,tn)

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More Inference Rules for Quantifiers

universal generalization(UG)P(g)x P(x)

Existential instantiation (EI)x P(x)

P(c) (substitute a new (Skolem)

constant c )Note: c is an object restricted to property P

(where g is a constant referring to any general (or universal) element of u.d. I.e., g cannot be restricted to any property except being an element of u.d.)

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Example

D(x) = “x has taken Discrete math course”C(X) = “ x has taken a cs course”Premises:

D(Chen), x (D(x) C(x))

Argument D(Chen) premise x (D(x) C(x)) premise D(Chen) C(Chen) UI C(Chen) MP

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Example

Premises: 1. A student in this class has not read the

book, 2. Everyone in this class passed the exam

Expected conclusion: Someone passing the exam has not read the

book.

Abbreviations: C(x) : x is in this class R(x) : x has read the book P(x) : x has passed the exam.

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The arguments1 x (C(x) /\ ~R(x) ) premise 1.

2. C(a) /\ ~R(a) EI from 1

3. C(a) Simplification from 2

4 x (C(x) P(x) ) premise 2

5. C(a) P(a) UI 4

6. P(a) MP from 3,5

7. ~R(a) Simplification from 2

8. P(a) /\ ~R(a) Conjunction 6,7

9 x (P(X) /\ ~R(x)). EG from 8

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Example (skipped)

Definitions:c :≡ Confucius ;H(x) :≡ “x is human”;M(x) :≡ “x is mortal”.

Premises:H(c) Confucius is human.

x H(x)M(x) All humans are mortal.Conclusion:

M(c) Confucius is mortal.


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The argument (skipped)

Some valid conclusions you can draw:H(c) premise

H(c)M(c) [Instantiate universal.] If Confucius is human then he is mortal.

H(c) M(c) Socrates is inhuman or mortal.

H(c) (H(c) M(c)) Confucius is human, and also either inhuman or mortal.

(H(c) H(c)) (H(c) M(c)) [Apply distributive law.]

F (H(c) M(c)) [Trivial contradiction.]

H(c) M(c) [Use identity law.]

M(c) Confucius is mortal.


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Example (universal genealization (UG )The square of all odd numbers are odd.

x Odd(x) Odd(x2)

pf: Let g be any number. // note: cannot assume additional property for g except for its being an integer; g is a universal constants.

[ Assume g is odd, then,

g = 2k + 1 for some k,

g2 = 4kk + 4k + 1 = 2(2kk + 2k) + 1,

g2 is odd. ] // [ … ] is a subproof for

=> Odd(g) Odd(g2).

=> x Odd(x) Odd(x2).

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Example2

xy x < y. // no largest numberpf: Let a be any number. =>a < a + 1 ( number property; true of all numbers) (but not a2 < a + 10 ; not a universal number prop ) => y a < y (EG) => xy x < y (UG)

xy x y //least numberpf: Let a be any natural number. => 0 a ( property of natural numbers) => y 0 y (UG) => xy x y (EG)

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a fallacy

Find an argument for x P(x) |- x P(x) What’s wrong with the proof: ?

1. x P(x) --- premise 2. P(c) --- EI, where c is a restricted object. 3. x P(x) --- UG (x) why?

Ans: 23 is wrong! c is not a general element! It requires to satisfy property P and thus is not a general element.

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Proofs and proof methods

Nature and importance of proofsApplications of proofsProof terminologyInference rulesSome inference rulesSoundness of inference rulesFormal proofsFormal proof examplesInference rules for quantifierscommon fallaciesproof methods

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Proof Terminology

Theorem [定理 ] A statement that can be proven to be true.

Axioms[公設 ], postulates, hypotheses, premises Assumptions (often unproven) defining the

structures about which we are reasoning.

Rules of inference [推論法則 ] Patterns of logically valid deductions from

hypotheses to conclusions.

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More Proof Terminology

Lemma[輔助定理 ] - A minor theorem used as a stepping-stone to proving a major theorem.

Corollary - A minor theorem proved as an easy consequence of a major theorem.

Conjecture - A statement whose truth value has not been proven. (A conjecture may be widely believed to be true, regardless.)

Theory – The set of all theorems that can be proven from a given set of axioms.

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Graphical Visualization

A Particular TheoryA Particular Theory

…

Various TheoremsVarious TheoremsThe AxiomsThe Axiomsof the Theoryof the Theory

A proofA proof

inference ruleinference rule

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Formal Proofs

A formal proof of a conclusion C, given premises p1, p2,…,pn is a sequence of statements : A1,…,An=C,

each of which either is a premise or is the result of applying some inference rule to premises or to previously-proven statements.

I.e., for each k = 1 .. n, either

1. Ak is some premise Pj, or

2. there is an instance (D1,…,Dm, E) of a rule with {D1,…,Dm} a subset of {A1,…,Ak-1} and Ak = E.

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More about proofs

Notes: 1. We use B |-Th A (or B1,…,Bm |-Th A)

to mean the existence of a proof of A with B1,…,Bm as premises. (Th can be omitted if there is no worry of confusion)

2. If {} |-Th A (or simply |-Th A) , we say A is a theorem and each Ai (i <n) is called a lemma.

3. If B can be inferred from A directly, it is called a corollary of theorem A; both lemmas and corollaries are theorems.

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Soundness of a proof system

Soundness theorem: If

1. there is a proof of C from a set of premises P= {P1,… Pn } (I.e., P1,…Pn |- C ), and

2. all inference rules used are sound,

then the proof is valid, i.e.,

conclusion C must be a consequence of {P1,… Pn}.

(i.e., P1,P2,…Pn |= C )

This theorem holds not only for propositional logic and predicate logic but also for all logical systems.

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Corollary : If |-Th C, then |= C. I.e., Every theorem is valid).

pf: |-Th C means {} |-Th C.

Hence {} |= C.

But all interpretations are model of {},

C thus is valid.

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Example

Definitions:s :≡ Socrates (ancient Greek philosopher);H(x) :≡ “x is human”;M(x) :≡ “x is mortal”.

Premises:H(s) Socrates is human.

x H(x)M(x) All humans are mortal.Conclusion:

M(s) Socrates is mortal.


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The proof

Some valid conclusions you can draw:H(s)M(s) [Instantiate universal.] If Socrates is human

then he is mortal.

H(s) M(s) Socrates is inhuman or mortal.

H(s) (H(s) M(s)) Socrates is human, and also either inhuman or mortal.

(H(s) H(s)) (H(s) M(s)) [Apply distributive law.]

F (H(s) M(s)) [Trivial contradiction.]

H(s) M(s) [Use identity law.]

M(s) Socrates is mortal.


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Techniques for proving implication theorems : pq

Different ways of proving a theorem: p implies q. Vacuous proof: Prove that ~p. used rule : [ ~p //p->q ]

Trivial proof: Prove that q. [q // p->q ]

Direct proof: Prove that if p then q. [p |- q implies |- p->q]suppose p, ..., conclude q --------------------- (I)hence pq

Indirect proof: (proof by contraposition) Prove that "~q implies ~P" [~q->~p // p->q]

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Proof by contradiction and proof by cases

Proof by contradiction: To prove P, it suffices to show that ~P -> F (false)

rules used: [~p ->F // p]

Proof by cases: To prove that "p \/ q implies r " it suffices to show that p->r and q -> r.

rules used: [p->r, q->r // (p\/ q) ->r.]

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Direct Proof Example

Definition: An integer n is called odd iff n=2k+1 for some integer k; n is even iff n=2k for some k. Axiom: Every integer is either odd or even.

Theorem: (For all numbers n) If n is an odd integer, then n2 is an odd integer.

Proof: Let n be any integer. [Assume n is odd, then n = 2k+1 for some integer k. Thus, n2 = (2k+1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Therefore n2 is of the form 2j + 1 (with j the integer

2k2 + 2k), thus n2 is odd. ]

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Another example of direct proof

Show that |- (p(q r)) ((pq) (pr)).

Pf:

1. [Assume p (q r)

2. [assume pq

3. [assume p

4. q MP+2,3

5. q r MP+1,3

6. r ] MP+4,5

7. pr ] 3~6 is a proof of pr

8. (pq ) (pr) ] 2~7 is its proof

9. (p(q r)) ((pq) (pr)). 1~8 is its proof.

The Foundations


Indirect Proof Example

Theorem: (For all integers n) If 3n+2 is odd, then n is odd.

Proof: Suppose that the conclusion is false, i.e., that n is even.

Then n=2k for some integer k. Then 3n+2 = 3(2k)+2 = 6k+2 = 2(3k+1). Thus 3n+2 is even, because it equals 2j for

integer j = 3k+1. So 3n+2 is not odd. We have shown that ¬(n is odd)→¬(3n+2 is odd),

thus its contra-positive (3n+2 is odd) → (n is odd) is also true. □

The Foundations


Vacuous Proof Example

Theorem: (For all n) If n is both odd and even, then n2 = n + n.

Proof: The statement “n is both odd and even” is necessarily false, since no number can be both odd and even.

So, the theorem is vacuously true. □

lemma : n could not be both even and odd.

pf: Suppose n is both odd and even, then n = 2k = 2m + 1 for some integer m and k. Then 2(k-m) = 1.

Then k != m and hence 2 | 1. But this is impossible.

The Foundations


Trivial Proof Example

Theorem: (For integers n) If n is the sum of two prime numbers, then either n is odd or n is even.

Proof: Any integer n is either odd or even.

So the conclusion of the implication is true regardless of the truth of the antecedent.

Thus the implication is true trivially. □

The Foundations


Proving existence theorem

Methods for proving x p(x):Constructive proof: find an object (or term)

a, s.t. P(a). rules used : p(a) // x p(x) ;EG

Nonconstructive proof: a proof of x P(x) w/o knowing what object satisfies p. Ex: proof by contradiction: Show that ~x p(x) F. or (x ~p(X)) F.

The Foundations


Example of existence proofs

Ex 20: [constructive proof] Show that there are n consecutive composite integers for every integer n >0. (I.e. for all n x (x+1,x+2,...x+n) are all composites.

Sol: Let x = (n+1)! +1.

=> x+i = (n+1)! + (i+1) = (i+1)( (n+1)!/(i+1) +1) is composite for i = 1,..,n. QED.

Ex 21: [nonconstructive proof] For all n >0 prime number > n.

Sol: by contradiction. Assume n s.t. all prime number < n.

Let m = n! +1. ==> gcd(k, m) = 1 for all k ≤ n.

=> all primes cannot divide m

=> m is a prime > n

=> a contradiction. QED.

Note: We cannot know a prime > n from the proof.

The Foundations


Example of proof by cases...

[No largest prime: ] Given n>0, prove there is a prime p>n.

pf: Consider x = n!+1. Since x>1, we know (x is prime)(x is composite).

Case 1: x is prime. Obviously x>n, so let p=x and we’re done.

Case 2: x has a prime factor p. But if pn, then

x = (n!/p) x p + 1 => x 1 (mod p).

Then p does not divide x, a contradiction.

So p>n, and we’re done.

The Foundations


Limits on Proofs

Some very simple statements of number theory haven’t been proved or disproved! E.g. Goldbach’s conjecture: Every integer n≥2 is

exactly the average of some two primes. n≥2 primes p,q: n=(p+q)/2.

For every formal proof system, there are true (or false )statements of number theory which can never be proved (or disproved) (Gödel incompleteness theorem).

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Discrete Mathematics Transparency No. 1-0 Chapter 1. The Foundations: Logic and Proofs Cheng-Chia Chen