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Shawn Zhong

 钟万祥
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Home / 2018 / February / 6

1.7 Introduction to Proofs

  • Feb 06, 2018
  • Shawn
  • Math 240
  • No comments yet
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 xy, where x and y are positive real numbers, then x^2y^2” ○ really means ○ “For all positive real numbers x and y, if xy, then x^2y^2.” Proving Theorems • Many theorems have the form: ∀x(P(x)→Q(x)) • To prove them, we show that P(c)→Q(c) 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: p→q Proving Conditional Statements: p→q • Trivial Proof ○ If we know q is true, then p→q is true as well. ○ “If it is raining then 1=1. • Vacuous Proof ○ If we know p is false then p→q is true as well. ○ “If I am both rich and poor then 2 + 2 = 5.” • Even though these examples seem silly, both trivial and vacuous proofs are often used in mathematical induction, as we will see in Chapter 5 • Direct Proof Assume that p is true. Use rules of inference, axioms, and logical equivalences to show that q must also be true. • Example 1 of Direct Proof ○ Give a direct proof of the theorem “If n is an odd integer, then n^2 is odd.” ○ Assume that n is odd. Then n=2k+1 for an integer k. ○ Squaring both sides of the equation, we get: § n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1=2r+1, § where r=2k^2+2k , an integer. ○ We have proved that if n is an odd integer, then n^2 is an odd integer. • Example 2 of Direct Proof ○ Prove that the sum of two rational numbers is rational. ○ Assume x and y are two rational numbers. ○ Then there must be integers p,q,r,s such that x=p/q, y=r/s and s≠0, p≠0 ○ x+y=p/q+r/s=(ps+qr)/qs, where q,s≠0, and (ps+qr),qs are integers ○ Hence, x+y is rational • 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. • Example of Proof by Contraposition ○ Prove that for an integer n, if n^2 is odd, then n is odd. ○ Use proof by contraposition. ○ Assume n is even (i.e., not odd). ○ Therefore, there exists an integer k such that n=2k. ○ Hence, n^2=4k^2=2(2k^2), and n^2 is even(i.e., not odd). ○ We have shown that if n is an even integer, then n^2 is even. ○ Therefore by contraposition, for an integer n, if n^2 is odd, then n is odd. • 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. • Example of Proof by Contradiction ○ Use a proof by contradiction to give a proof that √2 is irrational. ○ Towards a contradiction assume that √2 is rational ○ Let a,b be such that √2=a/b, b≠0, and a,b have no common factors ○ 2=a^2/b^2 ⇒2b^2=a^2, so a^2 is even and a is even ○ Let a≔2k for some k∈Z, then a^2=4k^2 ○ Then 2b^2=4k^2⇒b^2=2k, so b^2 is even, and b is also even ○ So 2 divides a and b, which makes a contradiction ∎ 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 ○ Prove the theorem: “If n is an integer, then n is odd if and only if n^2 is odd.” ○ We have already shown (previous slides) that both p→q and q→p. ○ Therefore we can conclude p ↔ q. • Note ○ Sometimes iff is used as an abbreviation for “if an only if,” as in “If n is an integer, then n is odd iif n^2 is odd.” 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. • In the next section, we will see strategies that can be used when straightforward approaches do not work. • In Chapter 5, we will see mathematical induction and related techniques. • In Chapter 6, we will see combinatorial proofs
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1.6 Rules of Inference

  • Feb 06, 2018
  • Shawn
  • Math 240
  • No comments yet
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 • An 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_1, p_2, …,p_n and the conclusion is q then ○ (p_1∧p_2∧…∧p_n)→q is a tautology. • Inference rules are all argument simple argument forms that will be used to construct more complex argument forms. Rules of Inference for Propositional Logic: • Modus Ponens ○ Equation ○ 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 ○ Equation ○ 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 ○ Equation ○ 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 ○ Equation ○ 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 ○ Equation ○ 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 ○ Equation ○ 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 ○ Equation ○ 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 ○ Equation ○ 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 study English literature.” Using the Rules of Inference to Build Valid Arguments • A valid argument is a sequence of statements. • Each statement is either a premise or follows from previous statements by rules of inference. • The last statement is called conclusion. Valid Arguments Example 1 ○ From the single proposition p∧(p→q) ○ Show that q is a conclusion. Example 2 ○ With these hypotheses: § “It is not sunny this afternoon and it is colder than yesterday.” § “We will go swimming only if it is sunny.” § “If we do not go swimming, then we will take a canoe trip.” § “If we take a canoe trip, then we will be home by sunset.” ○ Using the inference rules, construct a valid argument for the conclusion: § “We will be home by sunset.” ○ Choose propositional variables: § p: “It is sunny this afternoon.” § r: “We will go swimming.” § t: “We will be home by sunset.” § q: “It is colder than yesterday.” § s: “We will take a canoe trip.” ○ Translation into propositional logic: § Hypotheses: ¬p∧q, r→p,¬r→s,s→t § Conclusion: t ○ Argument Handling Quantified Statements • 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 • Example 1 ○ Using the rules of inference, construct a valid argument to show that § “John Smith has two legs” ○ is a consequence of the premises § “Every man has two legs.” § “John Smith is a man.” ○ Notation and domain § Let M(x) denote “x is a man” § L(x) “x has two legs” § Let John Smith be a member of the domain. ○ Argument • Example 2 ○ Use the rules of inference to construct a valid argument showing that the conclusion § “Someone who passed the first exam has not read the book.” follows from the premises “A student in this class has not read the book.” “Everyone in this class passed the first exam.” ○ Notation § Let C(x) denote “x is in this class.” § B(x) denote “x has read the book.” § P(x) denote “x passed the first exam.” ○ First we translate the premises and conclusion into symbolic form. ○ Argument Returning to the Socrates Example • Premises and conclusion • Argument The Barber Example • Show that from the statements • Every barber in Jonesville shaves those and only those who don t shave themselves. and There is a barber in Jonesville • We can derive a contradiction ○ ∀x(B(x)→∀y(S(x,y)⟷¬S(y,y))) ○ ∃x B(x) ○ ∀c B(c) ○ B(c)→∀y(S(c,y)⟷¬S(y,y)) ○ ∀y (S(c,y)⟷¬S(y,y)) ○ S(c,c)⟷¬S(c,c) ○ Thus, we have a contradiction Lewis Carroll • The first three are called premises and the third is called the conclusion ○ “All hummingbirds are richly colored.” ○ “No large birds live on honey.” ○ “Birds that do not live on honey are dull in color.” ○ “Hummingbirds are small.” • Notation ○ H(x)≔x is a hummingbird ○ C(x)≔x is richly colored ○ L(x)≔x is large ○ Ho(x)≔x lives on honey • Here is one way to translate these statements to predicate logic ○ ∀x (H(x)→C(x)) ○ ∀x (L(x)→¬Ho(x)) ○ ∀x (¬Ho(x)→¬C(x)) ○ ∀x (H(x)→¬L(x)) • Let c be an arbilirary element of the universe (1) ∀x (H(x)→C(x)) (2) ∀x (L(x)→¬Ho(x)) (3) ∀x (¬Ho(x)→¬C(x)) (4) H(c)→C(c)≡¬H(c)∨C(c) (5) L(c)→¬Ho(c)≡¬L(c)∨¬Ho(c) (6) ¬Ho(c)→¬C(c)≡Ho(c)∨¬C(c) (7) By resolution of (4) and (6), ¬H(c)∨Ho(c) (8) By resolution of (5) and (7), ¬H(c)∨¬L(c)≡H(c)→¬L(c) (9) By (8), ∀x(H(x)→¬L(x))
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