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# FoCS Week 2 CSCI 2200

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This 51 page Class Notes was uploaded by thersh on Thursday February 11, 2016. The Class Notes belongs to CSCI 2200 at Rensselaer Polytechnic Institute taught by Petros Drineas in Spring 2016. Since its upload, it has received 57 views. For similar materials see Foundations of Computer Science in ComputerScienence at Rensselaer Polytechnic Institute.

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Date Created: 02/11/16

The Foundations: Proofs Revisiting the Socrates Example • We have the two premises: • “All men are mortal.” • “Socrates is a man.” • And the conclusion: • “Socrates is mortal.” • How do we get the conclusion from the premises? The Argument • We can express the premises (above the line) and the conclusion (below the line) in predicate logic as an argument: • We will see shortly that this is a valid argument. Arguments in Propositional Logic • A argument in propositional logic is a sequence of propositions. All but the final proposition are called premises. The last statement is the conclusion. • The argument is valid if the premises imply the conclusion. An argument form is an argument that is valid no matter what propositions are substituted into its propositional variables. • If the premises are p ,p , …,p and the conclusion is q then 1 2 n (p1∧ p 2 … ∧ p ) n q is a tautology. • Inference rules are all simple argument forms that will be used to construct more complex argument forms. Rules of Inference for Propositional Logic: Modus Ponens Corresponding Tautology: (p ∧ (p →q)) → q Example: Let p be “It is snowing.” Let q be “I will study discrete math.” “If it is snowing, then I will study discrete math.” “It is snowing.” “Therefore , I will study discrete math.” Modus Tollens Corresponding Tautology: (¬q∧(p →q))→¬p Example: Let p be “it is snowing.” Let q be “I will study discrete math.” “If it is snowing, then I will study discrete math.” “I will not study discrete math.” “Therefore , it is not snowing.” Hypothetical Syllogism Corresponding Tautology: ((p →q) ∧ (q→r))→(p→ r) Example: Let p be “it snows.” Let q be “I will study discrete math.” Let r be “I will get an A.” “If it snows, then I will study discrete math.” “If I study discrete math, I will get an A.” “Therefore , If it snows, I will get an A.” Disjunctive Syllogism Corresponding Tautology: (¬p∧(p ∨q))→q Example: Let p be “I will study discrete math.” Let q be “I will study English literature.” “I will study discrete math or I will study English literature.” “I will not study discrete math.” “Therefore , I will study English literature.” Addition Corresponding Tautology: p →(p ∨q) Example: Let p be “I will study discrete math.” Let q be “I will visit Las Vegas.” “I will study discrete math.” “Therefore, I will study discrete math or I will visit Las Vegas.” Simplification Corresponding Tautology: (p∧q) →p Example: Let p be “I will study discrete math.” Let q be “I will study English literature.” “I will study discrete math and English literature” “Therefore, I will study discrete math.” Conjunction Corresponding Tautology: ((p) ∧ (q)) →(p ∧ q) Example: Let p be “I will study discrete math.” Let q be “I will study English literature.” “I will study discrete math.” “I will study English literature.” “Therefore, I will study discrete math and I will study English literature.” Resolution Resolution plays an important role in AI and is used in Prolog. Corresponding Tautology: ((¬p ∨ r ) ∧ (p ∨ q)) →(q ∨ r) Example: Let p be “I will study discrete math.” Let r be “I will study English literature.” Let q be “I will study databases.” “I will not study discrete math or I will study English literature.” “I will study discrete math or I will study databases.” “Therefore, I will study databases or I will English literature.” Using the Rules of Inference to Build Valid Arguments • premise or follows from previous statements by rules of inference. The last statement is called conclusion. • A valid argument takes the following form: S1 S2 . . . Sn C Valid Arguments Example 1: From the single proposition Show that q is a conclusion. Solution : Handling Quantified Statements • Valid arguments for quantified statements are a sequence of statements. Each statement is either a premise or follows from previous statements by rules of inference which include: • Rules of Inference for Propositional Logic • Rules of Inference for Quantified Statements • The rules of inference for quantified statements are introduced in the next several slides. Universal Instantiation (UI) Example: Our domain consists of all dogs and Fido is a dog. “All dogs are cuddly.” “Therefore, Fido is cuddly.” Universal Generalization (UG) Used often implicitly in Mathematical Proofs. Existential Instantiation (EI) Example: “There is someone who got an A in the course.” “Let’s call her a and say that a got an A” Existential Generalization (EG) Example: “Michelle got an A in the class.” “Therefore, someone got an A in the class.” Using Rules of Inference Exshow thatUsing the rules of inference, construct a valid argument to “John Smith has two legs” is a consequence of the premises: “Every man has two legs.” “John Smith is a man.” Solution: Let M(x) denote “x is a man” and L(x) “ x has two legs” and let John Smith be a member of the domain. Valid Argument: Returning to the Socrates Example Solution for Socrates Example Valid Argument Universal Modus Ponens Universal Modus Ponens combines universal instantiation and modus ponens into one rule. This rule could be used in the Socrates example. Proofs of Mathematical Statements • A proof is a valid argument that establishes the truth of a statement. • In math, CS, and other disciplines, informal proofs which are generally shorter, are generally used. • More than one rule of inference are often used in a step. • Steps may be skipped. • The rules of inference used are not explicitly stated. • Easier for to understand and to explain to people. • But it is also easier to introduce errors. Definitions • A theorem is a statement that can be shown to be true using: • definitions • other theorems • axioms (statements which are given as true) • rules of inference • A lemma is a ‘helping theorem’ or a result which is needed to prove a theorem. • A corollary is a result which follows directly from a theorem. • Less important theorems are sometimes called propositions. • A conjecture is a statement that is being proposed to be true. Once a proof of a conjecture is found, it becomes a theorem. It may turn out to be false. Forms of Theorems • Many theorems assert that a property holds for all elements in a domain, such as the integers, the real numbers, or some of the discrete structures that we will study in this class. • Often the universal quantifier (needed for a precise statement of a theorem) is omitted by standard mathematical convention. For example, the statement: “If x > y, where x and y are positive integers, then x > y ” really means “For all positive integers x and y, if x > y, then x > y .” Proving Theorems • Many theorems have the form: • To prove them, we show that where c is an arbitrary element of the domain, • By universal generalization the truth of the original formula follows. • So, we must prove something of the form: Even and Odd Integers Definition: The integer n is even if there exists an integer k such that n = 2k, and n is odd if there exists an integer k, such that n = 2k + 1. Note that every integer is either even or odd and no integer is both even and odd. We will need this basic fact about the integers in some of the example proofs to follow. Proving Conditional Statements: p → q • Direct Proof: Assume that p is true. Use rules of inference, axioms, and logical equivalences to show that q must also be true. Example: Give a direct proof of the theorem “If n is an odd integer, then n is odd.” Solution: Assume that n is odd. Then n = 2k + 1 for an integer k. Squaring both sides of the equation, we get: n = (2k + 1) = 4k + 4k +1 = 2(2k + 2k) + 1= 2r + 1, 2 where r = 2k + 2k , an integer. We have proved that if n is an odd integer, then n is an odd integer. ( marks the end of the proof. Sometimes QED is used instead. ) Proving Conditional Statements: p → q Definition: The real number r is rational if there exist integers p and q where q≠0 such that r = p/q Example: Prove that the sum of two rational numbers is rational. Solution: Assume r and s are two rational numbers. Then there must be integers p, q and also t, u such that Thus the sum is rational. where v = pu + qt w = qu ≠ 0 Proving Conditional Statements: p → q • Proof by Contraposition: Assume ¬q and show ¬p is true also. This is sometimes called an indirect proof method. If we give a direct proof of ¬q → ¬p then we have a proof of p → q. Why does this work? Example: Prove that if n is an integer and 3n + 2 is odd, then n is odd. Solution: Assume n is even. So, n = 2k for some integer k. Thus 3n + 2 = 3(2k) + 2 =6k +2 = 2(3k + 1) = 2j for j = 3k +1 Therefore 3n + 2 is even. Since we have shown ¬q → ¬p , p → q must hold as well. If n is an integer and 3n + 2 is odd (not even) , then n is odd (not even). Proving Conditional Statements: p → q Example: Prove that for an integer n, if n is odd, then n is odd. Solution: Use proof by contraposition. Assume n is even (i.e., not odd). Therefore, there exists an integer k such that n = 2k. Hence, n = 4k = 2 (2k ) and n is even(i.e., not odd). We have shown that if n is an even integer, then n is even. Therefore by contraposition, for an integer n, if n is odd, then n is odd. Proving Conditional Statements: p → q • Proof by Contradiction: (AKA reductio ad absurdum). To prove p, assume ¬p and derive a contradiction such as p ∧ ¬p. (an indirect form of proof). Since we have shown that ¬p →F is true , it follows that the contrapositive T→p also holds. Proof by Contradiction Example: Use a proof by contradiction to give a proof that √2 is irrational. Solution:Suppose √2 is rational. Then there exists integers a and b with √2 = a/b, where b≠ 0 and a and b have no common factors(we will not prove that).Then Thereforea must be even. If a is even then a must be even (an exercise). Since a is even, a = 2c for some integer c. Thus, 2 Thereforeb is even. Again then b must be even as well. But then 2 must divide botha and b. This contradictsour assumption that our initial assumptionmust be false and therefore √2 is irrational . Theorems that are Biconditional Statements • To prove a theorem that is a biconditional statement, that is, a statement of the form p ↔ q, we show that p → q and q →p are both true. Example:2rove the theorem: “If n is an integer, then n is odd if and only if n is odd.” Solution: We have already shown (previous slides) that both p →q and q →p. Therefore we can conclude p ↔ q. Sometimes iff is used as an abbreviation for “if an only if,” as in 2 “If n is an integer, then n is odd iif n is odd.” What is wrong with this? “Proof” that 1 = 2 Solution: Step 5. a - b = 0 by the premise and division by 0 is undefined. Looking Ahead • If direct methods of proof do not work: • We may need a clever use of a proof by contraposition. • Or a proof by contradiction. • We will see mathematical induction and related techniques. • We will see combinatorial proofs Proof by Cases • To prove a conditional statement of the form: • Use the tautology • Each of the implications is a case. Proof by Cases Example: Let a @ b = max{a, b} = a if a ≥ b, otherwise a @ b = max{a, b} = b. Show that for all real numbers a, b, c (a @b) @ c = a @ (b @ c) (This means the operation @ is associative.) Proof: Let a, b, and c be arbitrary real numbers. Then one of the following 6 cases must hold. 1. a ≥ b ≥ c 2. a ≥ c ≥ b 3. b ≥ a ≥c 4. b ≥ c ≥a 5. c ≥ a ≥ b 6. c ≥ b ≥ a Continued on next slide Proof by Cases Case 1: a ≥ b ≥ c (a @ b) = a, a @ c = a, b @ c = b Hence (a @ b) @ c = a = a @ (b @ c) Therefore the equality holds for the first case. A complete proof requires that the equality be shown to hold for all 6 cases. But the proofs of the remaining cases are similar. Try them. Without Loss of Generality Example: Show that if x and y are integers and both x∙y and x+y are even, then both x and y are even. Proof: Use a proof by contraposition. Suppose x and y are not both even. Then, one or both are odd. Without loss of generality, assume that x is odd. Then x = 2m + 1 for some integer k. x + y = (2m + 1) + 2n = 2(m + n) + 1 is odd.ger n, so Case 2: y is odd. Then y = 2n + 1 for some integer n, so x ∙ y = (2m + 1) (2n + 1) = 2(2m ∙ n +m + n) + 1 is odd. odd is similar. The use phrase without loss of generality (WLOG)s indicates this. Existence Proofs Srinivasa Ramanujan (1887-1920) • Proof of theorems of the form . • Constructive existence proof: • Find an explicit value of c, for which P(c) is true. • Then is true by Existential Generalization (EG). Example: Show that there is a positive integer that can be written as the sum of cubes of positive integers in two different ways: Proof: 1729 is such a number since 3 3 3 3 1729 = 10 + 9 = 12 + 1 Godfrey Harold Hardy (1877-1947) Nonconstructive Existence Proofs • In a nonconstructive existence proof, we assume no c exists which makes P(c) true and derive a contradiction. Example: Show that there exist irrational numbers x and y such that x is rational. Proof: We know that √2 is irrational. Consider the number √2 √2 . If it is rational, we have two irrational numbers x and y with x y rational, namely x = √2 and y = √2. But if √2 √2 is irrational, √2 then we yan let x√2 √√2 an(√2 √2)2 so 2hat aaaaa x = (√2 ) = √2 = √2 = 2. Counterexamples • Recall . • To establish that is true (or is false) find a c such that P(c) is true or P(c) is false. • In this case c is called a counterexample to the assertion . Example: “Every positive integer is the sum of the squares of 3 integers.” The integer 7 is a counterexample. So the claim is false. Uniqueness Proofs • Some theorems asset the existence of a unique element with a particular property, !x P(x). The two parts of a uniqueness proof are • Existence: We show that an element x with the property exists. • Uniqueness: We show that if y≠x, then y does not have the property. Example:Show that if a and b are real numbers and a ≠0, then there is a unique real number r such that ar + b = 0. Solution: • a(−b/a) + b = −b + b =0.er r = −b/a is a solution of ar + b = 0 because • Uniqueness: Suppose that s is a real number such that as + b = 0. Then dividing by a shows that r = s.. Subtracting b from both sides and Proof Strategies for proving p → q • Choose a method. 1. First try a direct method of proof. 2. If this does not work, try an indirect method (e.g., try to prove the contrapositive). • For whichever method you are trying, choose a strategy. 1. First try forward reasoning. Start with the axioms and known theorems and construct a sequence of steps that end in the conclusion. Start with p and prove q, or start with ¬q and prove ¬p. 2. If this doesn’t work, try backward reasoning. When trying to prove q, find a statement p that we can prove with the property p → q. Universally Quantified Assertions • To prove theorems of the form ,assume x is an arbitrary member of the domain and show that P(x) must be true. Using UG it follows that . 2 Example: An integer x is even if and only if x is even. Solution: The quantified assertion is 2 x [x is even x is even] We assume x is arbitrary. Recall that is equivalent to So, we have two cases to consider. These are considered in turn. Continued on next slide Universally Quantified Assertions Case 1. We show that if x is even then x is even using a direct proof (the only if part or necessity). If x is even then x = 2k for some integer k. Hence x = 4k = 2(2k ) which is even since it is an integer divisible by 2. This completes the proof of case 1. Case 2 on next slide Universally Quantified Assertions 2 Case 2. We show that if x is even then x must be even (the if part or sufficiency). We use a proof by contraposition. Assume x is not even and then show that x is not even. If x is not even then it must be odd. So, x = 2k + 1 for some k. Then x = (2k + 1) = 4k + 4k + 1 = 2(2k + 2k) + 1 which is odd and hence not even. This completes the proof of case 2. Since x was arbitrary, the result follows by UG. Therefore we have shown that x is even if and only if x is even. The Role of Open Problems • Unsolved problems have motivated much work in mathematics. Fermat’s Last Theorem was conjectured more than 300 years ago. It has only recently been finally solved. n n n Fermat’s Last Theorem: The equation x + y = z has no solutions in integers x, y, and z, with xyz≠0 whenever n is an integer with n > 2. A proof was found by Andrew Wiles in the 1990s. Additional Proof Methods • Later we will see many other proof methods: • Mathematical induction, which is a useful method for proving statements of the form n P(n), where the domain consists of all positive integers. • Structural induction, which can be used to prove such results about recursively defined sets. • Cantor diagonalization is used to prove results about the size of infinite sets. • Combinatorial proofs use counting arguments.

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