Professor: Munehiro Fukuda CSS342: Proofs Professor: Munehiro Fukuda CSS342: Proofs

Terminologies Axioms: are assumed true. Ex: Given two distinct points, there is exactly one line that contains them. Undefined terms: implicitly defined (and used) by the axioms. Ex: Points, lines Definitions: used to create new concepts. Ex: Two lines are parallel if they never cross each other. Theorem: a proposition that has been proved to be true. If two sides of a triangle are equal, the angles opposite them are equal. Corollary: a theorem that follows quickly from another theorem. Ex: If a triangle is equilateral, then it is equiangular. Lemma: a theorem not interesting but useful in proving another theorem. – Ex: A positive integer – 1 ≥ 0 CSS342: Proofs

Axioms, Definitions, and Undefined Terms Examples Euclidean geometry: Axiom 1: Given two distinct points, there is exactly one line that contains them. Axiom 2: If three points are not collinear, then there is exactly one plane that contains them. Definition 1: Two angles are supplementary if the sum of their measures is 180. Definition 2: Two lines are parallel if they never cross each other. Undefined terms: points, lines, planes, and angles Real numbers: Axiom 1: The commutative law stands up right for +-*/ operations. Axiom 2: If x and y are in a subset P, -x and –y are not in P, and x+y and xy are in P. Definition 1: P is called positive real numbers. Definition 2: Given a nonnegative real number, x and a positive integer, n, x1/n is y satisfying yn = x Undefined terms: numbers and 0 CSS342: Proofs

Theorem and Corollary Examples If two sides of a triangle are equal, then the angles opposite them are equal. Proof. Corollary: If a triangle is equilateral, then it is equiangular. Draw a line from the top vertex down to the bottom side so that the side is divided into two halves equally. Then, we can have two sub triangles, both having all equal sides. Thus, they are congruent. Therefore, the angles opposite them are equal. CSS342: Proofs

Lemma Example If n is a positive integer number, the either n – 1 is a positive integer or n – 1 = 0. Proof: The minimal positive integer is 1. Thus, n – 1 cannot be smaller than 0. Therefore, the lemma is true. This lemma is not interesting in its own right, but can be used to prove other results. CSS342: Proofs

Direct Proof When proving a universally quantified statement: x1, …, xn, P(x1, …, xn)→q(x1, …, xn) If P(x1, …, xn) is false, this statement is always true. Thus, focus on only the case when P(x1, …, xn) is true. Using P(x1, …, xn) for a proof is called A direct proof A CSS342: Proofs

Direct Proof Example Universally quantified statement: Proof: If d = min{d1, d2} and x ≤ d, x ≤ d1 and x ≤ d2 Proof: Assume that d = min {d1, d2} and x ≤ d. From the definition of min, d ≤ d1 and d ≤ d2 From x ≤ d and d ≤ d1 , x ≤ d1 From x ≤ d and d ≤ d2 , x ≤ d2 Thus, the statement is true. CSS342: Proofs

Proof by contradiction Assume that the hypothesis p is true but the conclusion q is false. Use p, !q and r (= other axioms, definitions, theorems). Derive r && !r = false. In other words, p && !q→ r && !r p q (!q) r p → q p&&!q r&&!r p && !q → r && !r T T (F) F F (T) CSS342: Proofs

Proof by Contradiction Example 1 If xy = 0, then either x = 0 or y = 0 Proof. p: xy = 0 q: x = 0 || y = 0 !q: !(x=0 || y=0) ≡ x ≠ 0 && y ≠ 0 r: if ab = ac and a ≠ 0, b = c (Let’s assume it has been proved) p && !q: xy = x * 0 = 0 and x ≠ 0 From r: y must be 0 This contradicts !q, which thus means r is wrong. This derives r && !r Thus, this statement must be true. CSS342: Proofs

Proof by Contradiction Example 2 For all real numbers x and y, if x + y ≥ 2, then either x ≥ 1 or y ≥ 1. Proof. p: x + y ≥ 2 q: x ≥ 1 || y ≥ 1 !q: !(x ≥ 1 || y ≥ 1) ≡ !(x ≥ 1) && !(y ≥ 1) ≡ x < 1 && y < 1 ≡ x + y < 2 ≡ !p This derived p && !p Thus, the statement must be true. Therefore, the statement must be true. Instead of r && !r, we derived p && !p, (i.e., r = p) Special case: proof by contrapositive CSS342: Proofs

Deductive Reasoning Drawing a conclusion from a sequence of propositions. Hypothesis: P1: The bug is either in module 17 or in 81. P2: The bug is a numerical error. P3: Module 81 has no numerical error. Conclusion: ∴ Q: The bug is in module 17 CSS342: Proofs

p→q p ∴ q p q p → q p → q && p T F P: 1 * 2 = 2 Q: I ate candy. P is true Thus Q is true (= I ate candy) p q p → q p → q && p T F Note: p →q, p /∴ q does not mean p →q && p ≡ ∴ q In fact, truth values do not match perfectly. It means that if p →q && p, then q = true. CSS342: Proofs

p→q q ∴ p p q p → q p → q && q T F P: 1 + 2 = 2 Q: I ate my hat. Q is true (= I ate my hat.) Then, is 1 + 2 = 2 true? p q p → q p → q && q T F When p → q && q = true, p can be true or false. Thus, this deductive argument is wrong. CSS342: Proofs

Rules of inference for propositions Rule of Inference Name p → q p ∴q Modus ponens q ∴p && q Conjunction !q ∴!P Modus tollens q → r ∴p → r Hypothetical syllogism ∴p || q Addition p || q !p Disjunctive syllogism p && q ∴p Simplification CSS342: Proofs

Inference Example 1 If you pass CSS342, then you can take CSS343. (p → q) If you can take CSS343, then you’ll learn binary trees. (q → r) If you can take CSS343, then you’ll learn inheritance. (q → s) You passed CSS342. (p) Applying the hypothetical syllogism: Thus, r && s You’ll learn both binary trees and inheritance p → q q → r ∴p → r p → q q → s ∴p → s CSS342: Proofs

Rules of Inference for Quantified Statements Name ∀x ∈ D P(x) ∴P(d) if d ∈ D Universal instantiation P(d) for any d ∈ D ∴∀x P(x) Universal generalization ∃x ∈ D P(x) ∴P(d) for some d ∈ D Existential instantiation P(d) for some d ∈ D ∴ ∃ x P(x) Existential generalization CSS342: Proofs

Inference Example 2 Given two statements: P(x): x loves Microsoft. Everyone loves either Microsoft or Apple. Lynn does not love Microsoft. P(x): x loves Microsoft. Q(x): x loves Apple. From universal generalization, ∀x P(x) || Q(x): Everyone loves either MicroSoft or Apple. !P(Lynn): Lynn does not love Microsoft. From disjunctive syllogism: p || q, !p / ∴q Q(Lynn) is true. Q(Lynn): Lynn loves Apple. CSS342: Proofs

Final Review Why Logic is So Important in CS? Inference is quite often used in knowledge database. Knowledge database is a core of expert system. Thus, inference is a core of expert system.  Example in CS: Prolog likes(mary, food). likes(mary, wine). likes(john, wine). likes(john, mary). ?- likes(mary, X), likes(john, X). X=wine CSS342: Proofs