 # Chapter 1: The Foundations: Logic and Proofs

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Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic 1.2 Propositional Equivalences 1.3 Predicates and Quantifiers 1.4 Nested Quantifiers 1.5 Rules of Inference 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy

1.1: Propositional Logic Propositions: A proposition is a declarative sentence (that is, a sentence that declares a fact) that is either true or false, but not both.

Example 1: All the following declarative sentences are propositions: Washington D.C., is the capital of the USA. 2. Toronto is the capital of Canada 3. 1+1=2. =3.

Example 2: Consider the following sentences. Are they propositions? 1. What time is it? 2. Read this carefully. 3. x+1=2. 4. x+y=z

We use letters to denote propositional variables (or statement variables).
T: the value of a proposition is true. F: the value of a proposition is false. The area of logic that deals with propositions is called the propositional calculus or propositional logic.

Let p and q are propositions: Definition 1: Negation (Not) Symbol: ¬
Statement: “it is not the case that p”. Example: P: I am going to town ¬P: It is not the case that I am going to town; I am not going to town; I ain’t goin’.

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Definition 2: Conjunction (And)
Symbol: The conjunction pq is true when both p and q are true and is false otherwise. Example: P - ‘I am going to town’ Q - ‘It is going to rain’ PQ: ‘I am going to town and it is going to rain.’

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Definition 3: Disjunction (Or)
Symbol: The disjunction pq is false when both p and q are false and is true otherwise. Example: P - ‘I am going to town’ Q - ‘It is going to rain’ P  Q: ‘I am going to town or it is going to rain.’

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Definition 4: Exclusive OR
Symbol: The exclusive or of p and q, denote pq, is true when exactly one of p and q is true and is false otherwise. Example: P - ‘I am going to town’ Q - ‘It is going to rain’ P  Q: ‘Either I am going to town or it is going to rain.’

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Definition 5: Implication
If…. Then…. Symbol: The conditional statement pq is false when p is true and q is false, and true P is called the hypothesis and q is called the conclusion. Example: P - ‘I am going to town’ Q - ‘It is going to rain’ P Q: ‘If I am going to town then it is going to rain.’

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Equivalent Forms If P, then Q P implies Q If P, Q P only if Q
P is a sufficient condition for Q Q if P Q whenever P Q is a necessary condition for P

Note: The implication is false only when P is true and Q is false!
‘If the moon is made of green cheese then I have more money than Bill Gates’ (?) ‘If the moon is made of green cheese then I’m on welfare’ (?) ‘If 1+1=3 then your grandma wears combat boots’ (?) ‘If I’m wealthy then the moon is not made of green cheese.’ (?) ‘If I’m not wealthy then the moon is not made of green cheese.’ (?)

More terminology QP is the CONVERSE of P  Q
¬ Q  ¬ P is the CONTRAPOSITIVE of P  Q ¬ P ¬ Q is the inverse of P  Q Example: Find the converse of the following statement: R: ‘Raining tomorrow is a sufficient condition for my not going to town.’

Procedure Step 1: Assign propositional variables to component propositions P: It will rain tomorrow Q: I will not go to town Step 2: Symbolize the assertion R: P  Q Step 3: Symbolize the converse Q  P Step 4: Convert the symbols back into words ‘If I don’t go to town then it will rain tomorrow’ Homework: Find inverse and contrapositive of statements above.

Definition 6: Biconditional
‘if and only if’, ‘iff’ Symbol: The biconditional statement pq is true when p and q have the same truth value, and is false otherwise. Biconditional statements are also called bi-implications. Example: P - ‘I am going to town’ Q - ‘It is going to rain’ P Q: ‘I am going to town if and only if it is going to rain.’

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Translating English Breaking assertions into component propositions - look for the logical operators! Example: ‘If I go to Harry’s or go to the country I will not go shopping.’ P: I go to Harry’s Q: I go to the country R: I will go shopping If......P......or.....Q.....then....not.....R (P Q)  ¬ R

Constructing a truth table
one column for each propositional variable one for the compound proposition count in binary n propositional variables = 2n rows Construct the truth table for (P  ¬ Q)  (PQ) HW: Construct the truth table for (P Q)  ¬ R

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What is the real meaning of ¬ PQ ?
a) (¬ P) Q b) ¬ (PQ) What is the real meaning of PQR ? a) (PQ)R b) P(QR) What is the real meaning of P  QR ? a) (P  Q)R b) P  (QR)

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Logic and Bit Operations
Example 20 Find the bitwise OR, bitwise AND, and bitwise XOR of the bit strings and

Logic Puzzles Example 18:
There are two kind of inhabitants, knights, who always tell the truth, and their opposites, knaves, who always lie. You encounter two people A and B. What are A and B if A says “B is a knight” and B says “The two of us are opposite type”?

Proposition Negation Conjection Disjunction Exclusive OR Implication
Terms Proposition Negation Conjection Disjunction Exclusive OR Implication Inverse Converse Contrapositive

Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic 1.2 Propositional Equivalences 1.3 Predicates and Quantifiers 1.4 Nested Quantifiers 1.5 Rules of Inference 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy

1.2: Propositional Equivalences
Definition: Tautology: A compound proposition that is always true. Contradiction: A compound proposition that is always false. Contingency: A compound proposition that is neither a tautology nor a contradiction.

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Logical Equivalences Compound propositions that have the same truth values in all possible cases are called logically equivalent. Definition: The compound propositions p and q are called logically equivalent if pq is a tautology. Denote pq.

Logical Equivalences One way to determine whether two compound propositions are equivalent is to use a truth table. Symbol: PQ

Logical Equivalences Prove the De Morgan’s Laws.
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Logical Equivalences HW: Prove the other one (De Morgan’s Laws).
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Logical Equivalences Example:
Show that pq and ¬pq are logically equivalent. HW: example 4 of page 23

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Logical Equivalences Example 5: Use De Morgan’s laws to express the negations of “Miguel has a cellphone and he has a laptop computer”. Example 5: Use De Morgan’s laws to express the negations of “Heather will go to the concert or Steve will go to the concert”.

Logical Equivalences Example 6: Show that ¬(pq) and p ¬q are logically equivalent. Example 7: Show that ¬(p(¬p  q)) and ¬p  ¬q are logically equivalent by developing a series of logical equivalences. Example 8: Show that (p  q) ( pq) is a tautology.

Terms Tautology Contradiction Contingency Logical Equivalence
De Morgan’s Laws Commutative Law Associative Law Distributive Law

Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic 1.2 Propositional Equivalences 1.3 Predicates and Quantifiers 1.4 Nested Quantifiers 1.5 Rules of Inference 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy

Predicates Predicate: A generalization of propositions ; A propositions which contain variables Predicates become propositions once every variable is bound- by assigning it a value from the Universe of Discourse U or quantifying it P. 1

Predicates Examples: Let U = Z, the integers = { , -1, 0 , 1, 2, . . .} P(x): x > 0 is the predicate. It has no truth value until the variable x is bound. Examples of propositions where x is assigned a value: P(-3) (?, true or false); P(0)(?); (c) P(3)(?). The collection of integers for which P(x) is true are the positive integers. P(y) ν ¬ P(0) is not a proposition. The variable y has not been bound. However, P(3) ν ¬ P(0) is a proposition which is true. P. 1

Predicates Example: Let R be the three-variable predicate R(x, y, z): x + y = z Find the truth value of R(2, -1, 5), R(3, 4, 7), R(x, 3, z) P. 1

Quantifiers: Universal
P(x) is true for every x in the universe of discourse. Notation: universal quantifier ∀xP(x) ‘For all x, P(x)’, ‘For every x, P(x)’ The variable x is bound by the universal quantifier producing a proposition. An element for which P(x) is false is called a counterexample of ∀xP(x). Example: U={1,2,3} ∀ xP(x)  P(1) Λ P(2) Λ P(3) P. 1

Quantifiers: Universal
Example 8: Let P(x) be the statement “x+1>x.” What is the truth value of the quantification ∀ xP(x) where the domain consists of all real number. HW: P36, example 13 P. 1

Quantifiers: Existential
P(x) is true for some x in the universe of discourse. Notation: existential quantifier ∃xP(x) ‘There is an x such that P(x),’ ‘For some x, P(x)’, ‘For at least one x, P(x)’, ‘I can find an x such that P(x).’ Example: U={1,2,3} ∃xP(x)  P(1) ν P(2) ν P(3) P. 1

Quantifiers: Existential
Example 14: Let P(x) denote the statement “x>3.” What is the truth value of the quantification ∃ xP(x), where the domain consists of all real numbers. HW: Page 37, Example 16. P. 1

Quantifiers P. 1

Quantifiers: Unique Existential
P(x) is true for one and only one x in the universe of discourse. Notation: unique existential ∃!xP(x) ‘There is a unique x such that P(x),’ ‘There is one and only one x such that P(x),’ ‘One can find only one x such that P(x).’ P. 1

Quantifiers Example: U={1,2,3} Truth Table: P(1) P(2) P(3) ∃!xP(x)
P. 1

Quantifiers Note: A predicate is not a proposition until all variables have been bound either by quantification or assignment of a value! P. 1

Precedence of Quantifiers
The quantifiers  and  have higher precedence than all logical operators from propositional calculus. Example:  xP(x)Q(x) means (a) ( xP(x))Q(x) (b)  x(P(x)Q(x)) P. 1

Binding Variables Example: x(x+y=1) Bound variable: Free variable:
x(P(x)Q(x)) xR(x) The scope of : The scope of : The same meaning of x(P(x)Q(x)) yR(y) P. 1

Logical Equivalences Involving Quantifiers
Example 19: Show that x(P(x)Q(x)) and xP(x) xQ(x) are logically equivalent. P. 1

Logical Equivalences Involving Quantifiers
Statement involving predicates and quantifiers are logically equivalent if and only if they have the same truth value no matter which predicates are substituted into these statements and which domain of discourse is used for the variables in these propositional functions. Symbol: S T P. 1

Equivalences Involving the Negation Operator
¬∀xP(x)  ∃x ¬ P(x) ¬∃xP(x)  ∀ x ¬ P(x) Distributing a negation operator across a quantifier changes a universal to an existential and vice versa. P. 1

Equivalences Involving the Negation Operator
Example 20: What are the negations of the statements: “There is an honest politician” “All Americans eat cheeseburgers”. HW: Example 21 in P41. P. 1

Translating from English into Logical Expressions (single quantifier)
Example 23: Express the statement “Every student in this class has studied calculus” using predicates and quantifiers. P. 1

Using Quantifiers in System Specifications
Example 26: Consider these statements. “All lions are fierce.” “Some lions do not drink coffee.” “Some fierce creatures do not drink coffee.” Let P(x): x is a lion. Q(x): x is fierce. R(x): x drinks coffee. Assuming that the domain consists of all creatures, express the statements in the argument using quantifiers and P(x), Q(x), and R(x). P. 1

Logic Programming Prolog(Programming in Logic, developed in the 1970s.
Working in AI. Prolog programs including a set of declarations consisting of two types of statements: Facts : define predicates by specifying the elements that satisfy these predicates. Rules: define new predicates using those already defined by facts. P. 1

Logic Programming Example 28: Facts: Rules: instructor(chan, math273)
instructor(patel, ee222) instructor(grossman, cs301) enrolled(kevin, math273) enrolled(juana, ee222) enrolled(juana, cs301) enrolled(kiko, math273) enrolled(kiko, cs301) Rules: teacher(P,S) :- instructor(P,C), enrolled(S,C) Queries: ?enrolled(kevin, math273) ?enrolled(X,math273) ?teacher(X,juana) Uppercase letters are variables. The “” represents by “,” and the “” represents by “;” in Prolog. P. 1

Quantifiers Multiple Quantifiers: read left to right . . .
Example: Let U = R, the real numbers, P(x,y): xy= 0 ∀x∀yP(x, y) ∀x∃yP(x, y) ∃x∀yP(x, y) ∃x∃yP(x, y) The only one that is false is the first one. Suppose P(x,y) is the predicate x/y=1? Assume U=R-{0}. P. 1

∃yP(1, y) Λ ∃ yP(2, y) Λ ∃ yP(3, y)
Quantifiers Example: Let U = {1,2,3}. Find an expression equivalent to ∀x∃yP(x, y) where the variables are bound by substitution instead: Expand from inside out or outside in. Outside in: ∃yP(1, y) Λ ∃ yP(2, y) Λ ∃ yP(3, y) [P(1,1) ν P(1,2) ν P(1,3)] Λ [P(2,1) ν P(2,2) ν P(2,3)] Λ [P(3,1) ν P(3,2) ν P(3,3)] HW: Inside Out: P. 1

Quantifiers De Morgan’s Laws for Quantifiers P. 1

Quantifiers: Converting from English
Examples: F(x): x is a fleegle S(x): x is a snurd T(x): x is a thingamabob U={fleegles, snurds, thingamabobs} Note: the equivalent form using the existential quantifier is also given P. 1

Quantifiers: Converting from English
Everything is a fleegle ∀xF( x)  ¬∃x¬F(x) Nothing is a snurd. ∀x¬S(x)  ¬∃xS( x) All fleegles are snurds. ∀x[F(x) → S(x)] ∀ x[¬ F(x) ν S(x)]  ∀ x ¬[F(x) Λ ¬ S(x)]  ¬∃x[F(x) Λ ¬ S( x)] P. 1

Quantifiers: Converting from English
Some fleegles are thingamabobs. ∃x[F(x) Λ T(x)]  ¬ ∀ x[¬ F(x) ν ¬ T(x)] No snurd is a thingamabob. ∀ x[S(x) → ¬ T(x)]  ¬ ∃ x[S(x ) Λ T(x)] If any fleegle is a snurd then it's also a thingamabob ∀ x[(F(x) Λ S(x)) → T(x)]  ¬ ∃ x[F(x) Λ S(x) Λ ¬ T( x)] P. 1

Quantifiers:Dangerous situations
Commutativity of quantifiers ∀x∀yP(x, y)  ∀y∀xP(x, y)? YES! ∀x∃yP(x, y)  ∃y∀xP(x, y)? NO! DIFFERENT MEANING! Example: P(x,y): x+y=0, U: Integers ∀x∃yP(x, y) is Ture ∃y∀xP(x, y) is False P. 1

Quantifiers:Dangerous situations
Distributivity of quantifiers over operators ∀x[P(x)ΛQ(x)]  ∀xP( x)Λ∀xQ(x)? YES! ∀x[P(x)→Q(x)] [∀xP(x)→∀xQ(x)]? NO! Let P(x) sometimes true, sometimes false, and Q(x) is always false, then ∀x[P(x)→Q(x)] is False [∀xP(x)→∀xQ(x)] is True P. 1

Terms Proposition Predicate Universal Quantifier
Existential Quantifier Unique Existential Quantifier De Morgan’s Laws for Quantifiers Binding Variables Logic Programming P. 1

Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic 1.2 Propositional Equivalences 1.3 Predicates and Quantifiers 1.4 Nested Quantifiers 1.5 Rules of Inference 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy

Introduction: Nested Quantifiers
Nested quantifiers: Two quantifiers are nested if one is within the scope of the other. Example: xy(x+y=0)

Introduction: Nested Quantifiers
Example: Domain: real number. Addition inverse: xy(x+y=0) Commutative law for addition: xy(x+y=y+x) Associative law for addition: xyz (x+(y+z)=(x+y)+z)

The Order of Quantifiers
Example: Let Q(x,y) denote “x+y=0.” What are the truth values of the quantifications yxQ(x,y) and xyQ(x,y), where the domain for all variables consists of all real numbers?

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Translating Mathematical Statements into Statements Involving Nested Quantifiers
Example 7: Translate the statement “Every real umber except zero has a multiplicative inverse.” (A multiplicative inverse of a real number x is a real number y such that xy=1) HW: Example 8, p54

Translating from Nested Quantifiers into English
Example 9: Translate the statement x(C(x)y(C(y)F(x,y))) into English, where C(x) is “x has a computer,” F(x,y) is “x and y are friends,” and the domain for both x and y consists of all students in your school. HW: Example 10, p55.

Negating Nested Quantifiers
Example 14: Express the statement xy(xy=1) negation of the statement so that no negation precedes a quantifier. HW: Example 15, p57

Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic 1.2 Propositional Equivalences 1.3 Predicates and Quantifiers 1.4 Nested Quantifiers 1.5 Rules of Inference 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy

Rules of Inference Definition:
An argument in propositional logic is a sequence of propositions. All but the final proposition in the argument are called premises and the final proposition is called the conclusion. An argument is valid if the truth of all its premises implies that the conclusion is true.

Rules of Inference Definition: A theorem is a valid logical assertion which can be proved using other theorems axioms (statements which are given to be true) and rules of inference (logical rules which allow the deduction of conclusions from premises). A lemma (not a “lemon”) is a 'pre-theorem' or a result which is needed to prove a theorem. A corollary is a 'post-theorem' or a result which follows directly from a theorem.

Rules of Inference Many of the tautologies in Chapter 1 are rules of inference. They have the form H1 Λ H2 Λ..... Λ Hn →C Where Hi are called the hypotheses and C is the conclusion. As a rule of inference they take the symbolic form: H1 H2 . Hn C where  means 'therefore' or 'it follows that.'

Rules of Inference Examples:The tautology P Λ(P → Q) → Q becomes P
This means that whenever P is true and P → Q is true we can conclude logically that Q is true. This rule of inference is the most famous and has the name modus ponens or the law of detachment.

Rules of Inference

Rules of Inference for Quantifiers
Note: In Universal Generalization, x must be arbitrary. In Universal Instantiation, c need not be arbitrary but often is assumed to be. In Existential Instantiation, c must be an element of the universe which makes P(x) true.

J: John Smith, a member of the universe
Rules of Inference Example: Every man has two legs. John Smith is a man. Therefore, John Smith has two legs. Define the predicates: M(x): x is a man L(x): x has two legs J: John Smith, a member of the universe

Rules of Inference Example: The argument becomes 1.x[M(x) → L(x)]
2.M( J ) L( J) The proof is 1. x[M(x) → L(x)] Hypothesis 1 2.M( J ) → L(J ) step 1 and UI 3.M( J ) Hypothesis 2 4.L( J) steps 2 and 3 and modus ponens Q. E. D. Note: Using the rules of inference requires lots of practice.

Rule of Inference for Quantified Statement
Example 12: Show that the premises “Everyone in this discrete mathematics class has taken a course in computer science” and “Marla is a student in this class” imply the conclusion “Marla has taken a course in computer science.” HW: Example 13, p71.

Fallacies Fallacies are incorrect inferences. Some common fallacies:
The Fallacy of Affirming the Consequent The Fallacy of Denying the Antecedent (or the hypothesis) Begging the question or circular reasoning

Fallacies: The Fallacy of Affirming the Consequent
Example: If the butler did it he has blood on his hands. The butler had blood on his hands. Therefore, the butler did it. This argument has the form P → Q Q P or [(P → Q)ΛQ] → P which is not a tautology and therefore not a rule of inference!

Fallacies: The Fallacy of Denying the Antecedent (or the hypothesis)
Example: if the butler is nervous, he did it. The butler is really mellow. Therefore, the butler didn't do it. This argument has the form P → Q ¬ P  ¬ Q or [(P → Q)Λ ¬ P] → ¬ Q which is also not a tautology and hence not a rule of inference.

Fallacies: Begging the question or circular reasoning
This occurs when we use the truth of statement being proved (or something equivalent) in the proof itself. Example: Conjecture: if x2 is even then x is even. Proof: If x2 is even then x2 = 2k for some k. Then x = 2l for some l. Hence, x must be even.

Fallacies Example: P (x): x注射疫苗 Q(x): x死亡

Terms Theorem Argument Premises Axioms Lemma Conslusion Valid
Corollary Rule of inference Modus ponens Modus tollensva fallacies

Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic 1.2 Propositional Equivalences 1.3 Predicates and Quantifiers 1.4 Nested Quantifiers 1.5 Rules of Inference 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy

Formal Proofs Formal Proofs
To prove an argument is valid or the conclusion follows logically from the hypotheses: Assume the hypotheses are true Use the rules of inference and logical equivalences to determine that the conclusion is true. P. 1

Formal Proofs Formal Proofs
Example: Consider the following logical argument: If horses fly or cows eat artichokes, then the mosquito is the national bird. If the mosquito is the national bird then peanut butter takes good on hot dogs. But peanut butter tastes terrible on hot dogs. Therefore, cows don't eat artichokes. 1. Assign propositional variables to the component propositions in the argument: F Horses fly A Cows eat artichokes M The mosquito is the national bird P Peanut butter tastes good on hot dogs P. 1

Formal Proofs Formal Proofs
2. Represent the formal argument using the variables 1.(F ν A) → M 2.M →P 3. ¬ P  ¬ A P. 1

Formal Proofs Formal Proofs
3. Use the hypotheses 1., 2., and 3. and the above rules of inference and any logical equivalences to construct the proof. Assertion Reasons 1.(F ν A) → M Hypothesis 1. 2.M → P Hypothesis 2. 3.(F ν A) → P` steps 1 and 2 and hypothetical syll. 4. ¬ P Hypothesis 3. 5. ¬(F ν A) steps 3 and 4 and modus tollens 6. ¬F Λ¬A step 5 and DeMorgan 7. ¬A Λ¬F step 6 and commutativity of 'and' 8. ¬A step 7 and simplification Q. E. D. P. 1

Methods of Proof We wish to establish the truth of the 'theorem‘ P→Q.
P may be a conjunction of other hypotheses. P → Q is a conjecture until a proof is produced. P. 1

Methods of Proof Trivial Proof Vacuous Proof Direct Proof
Indirect Proof Proof by Contradiction Proof by Cases P. 1

Methods of Proof: Trivial Proof
If we know Q is true then P → Q is true. Example: If it's raining today then the void set is a subset of every set. The assertion is trivially true independent of the truth of P. P. 1

Methods of Proof: Vacuous Proof
If we know one of the hypotheses in P is false then P → Q is vacuously true. Example: If I am both rich and poor then hurricane Fran was a mild breeze. This is of the form (P Λ ¬ P) → Q and the hypotheses form a contradiction. Hence Q follows from the hypotheses vacuously. 歐亞書局 P. 1

Methods of Proof: Direct Proof
assumes the hypotheses are true uses the rules of inference, axioms and any logical equivalences to establish the truth of the conclusion. Example: Theorem: If 6x + 9y = 101, then x or y is not an integer. Proof: Assume 6x + 9y = 101 is true. Then from the rules of algebra 3(2x + 3y) = 101. But 101/3 is not an integer so it must be the case that one of 2x or 3y is not an integer (maybe both). Therefore, one of x or y must not be an integer. Q.E.D. 歐亞書局 P. 1

Methods of Proof: Indirect Proof
A direct proof of the contrapositive: assumes the conclusion of P → Q is false (¬ Q is true) uses the rules of inference, axioms and any logical equivalences to establish the premise P is false. Note, in order to show that a conjunction of hypotheses is false is suffices to show just one of the hypotheses is false. P. 1

Methods of Proof: Indirect Proof
Example: A perfect number is one which is the sum of all its divisors except itself. For example, 6 is perfect since = 6. So is 28. Theorem: A perfect number is not a prime. Proof: (Indirect). We assume the number p is a prime and show it is not perfect. But the only divisors of a prime are 1 and itself. Hence the sum of the divisors less than p is 1 which is not equal to p. Hence p cannot be perfect. Q. E. D. P. 1

assumes the conclusion Q is false derives a contradiction.. P. 1

Example: Theorem: There is no largest prime number. (Note that there are no formal hypotheses here.) We assume the conclusion 'there is no largest prime number' is false. There is a largest prime number. Call it p. Hence, the set of all primes lie between 1 and p. Form the product of these primes: r = 2•3•5•7•11•....•p. But r + 1 is a prime larger than p. (Why?). This contradicts the assumption that there is a largest prime. Q.E.D. P. 1

The formal structure of the above proof is as follows: Let P be the assertion that there is no largest prime. Let Q be the assertion that p is the largest prime. Assume ¬ P is true. Then (for some p) Q is true so ¬ P→Q is true. We then construct a prime greater than p so Q → ¬ Q. Applying hypothetical syllogism we get ¬ P → ¬ Q. From two applications of modus ponens we conclude that Q is true and ¬ Q is true so by conjunction ¬ QΛQ or a contradiction is true. Hence the assumption must be false and the theorem is true. 歐亞書局 P. 1

Methods of Proof: Proof by Cases
1. Break the premise of P→Q into an equivalent disjunction of the form P1 ν P2 ν... ν Pn . 2. Then use the tautology [(P1 → Q) Λ (P2 → Q) Λ... Λ(Pn → Q)]↔[(P1 ν P2 ν... ν Pn ) → Q] Each of the implications Pi → Q is a case. You must Convince the reader that the cases are inclusive, i.e., they exhaust all possibilities Establish all implications P. 1

Methods of Proof: Proof by Cases
Example: Let  be the operation 'max' on the set of integers: if a  b then a  b = max {a, b} = a = b  a. Theorem: The operation  is associative. For all a, b, c (a  b)  c = a (b  c). 歐亞書局 P. 1

Methods of Proof: Proof by Cases
Example:Proof: Let a, b, c be arbitrary integers. Then one of the following 6 cases must hold (are exhaustive): 1. a  b  c 2. a  c  b 3. b  a  c 4. b  c  a 5. c  a  b 6. c  b  a Case 1: a  b = a, a  c = a, and b  c = b. Hence (a  b)  c = a = a (b  c). Therefore the equality holds for the first case. The proofs of the remaining cases are similar (and are left for the student). Q. E. D. 歐亞書局 P. 1

Mistakes in Proofs Example 15:
What is wrong with the “proof” that 1=2? “Proof:” We use these steps where a and b are two equal positive integers. a=b (Given) a2=ab a2-b2=ab-b2 (a-b)(a+b)=b(a-b) a+b=b 2b=b 2=1 HW: Example 15, p83 P. 1

Terms Conjunction Disjunction Conjecture P. 1

Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic 1.2 Propositional Equivalences 1.3 Predicates and Quantifiers 1.4 Nested Quantifiers 1.5 Rules of Inference 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy

Existence Proofs We wish to establish the truth of xP( x).
Constructive existence proof: Establish P(c) is true for some c in the universe. Then  xP( x) is true by Existential Generalization (EG). Example: Theorem: There exists an integer solution to the equation x2+y2=z2. Proof: Choose x = 3, y = 4, z = 5. P. 1

Existence Proofs Example:
Theorem: There exists a bijection from A= [0,1] to B= [0, 2]. Proof: we could have chosen g(x) = x/2 and obtained a bijection directly. HW: Prove that g(x) above is a bijection function. 歐亞書局 P. 1

Nonconstructive Existence Proof
Assume no c exists which makes P(c) true and derive a contradiction. Example: Theorem: There exists an irrational number. Proof: Assume there doesn’t exist an irrational number. Then all numbers must be rational. Then the set of all numbers must be countable. Then the real numbers in the interval [0, 1] is a countable set. But we have already known this set is not countable. Hence, we have a contradiction (The set [0,1] is countable and not countable). Therefore, there must exist an irrational number. Q. E. D. Note: we have not produced such a number! HW: Prove that [0,1] is not countable. P. 1

Disproof by Counterexample
Recall that x ¬ P(x)↔¬ xP(x ). To establish that ¬xP(x ) is true (or xP(x) is false) construct a c such that ¬ P(c) is true or P(c) is false. In this case c is called a counterexample to the assertion xP(x) P. 1

Disproof by Counterexample
Prove or disprove that “every positive integer is the sum of the squares of two integers. Example: 5=12+22; 34=32+52. Counterexample: P. 1

Nonexistence Proofs Nonexistence Proofs
We wish to establish the truth of ¬  xP( x) (which is equivalent to x ¬ P(x)). Use a proof by contradiction by assuming there is a c which makes P(c) true. P. 1

Universally Quantified Assertions
We wish to establish the truth of xP(x) We assume that x is an arbitrary member of the universe and show P(x) must be true. Using UG it follows that xP(x) . P. 1

Universally Quantified Assertions
Example: Theorem: For the universe of integers, x is even iff x2 is even. Proof: The quantified assertion is  x[x is even ↔ x2 is even] We assume x is arbitrary. Recall that P ↔ Q is equivalent to (P→Q) Λ (Q → P). P. 1

Universally Quantified Assertions
Example: Case 1 . We show if x is even then x2 is even using a direct proof If x is even then x = 2k for some integer k. Hence, x2 = 4k2 which is even since it is an integer which is divisible by 2. This completes the proof of case 1. P. 1

Universally Quantified Assertions
Example: Case 2 . We show that if x2 is even then x must be even . We use an indirect proof: Assume x is not even and show x2 is not even. If x is not even then it must be odd. So, x = 2k + 1 for some k. Then x2 = (2k+1) 2 = 2(2k 2+2k)+1 which is odd and hence not even. This completes the proof of the second case. Therefore we have shown x is even iff x2 is even. Since x was arbitrary, the result follows by UG. Q.E.D. P. 1

Rosen to his student Dear students: Learning how to construct proofs is probably one of the most difficult things you will face in life. Few of us are gifted enough to do it with ease. One only learns how to do it by practicing . P. 1