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CS100 : Discrete Structures

CS100 : Discrete Structures. Proof Techniques(1) Dr.Saad Alabbad Department of Computer Science E-mail: salabbad@gmail.com Office # SR 068 Tel # 2581888. Rules of Inference – Valid Arguments in Propositional Logic.

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CS100 : Discrete Structures

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  1. CS100 : Discrete Structures Proof Techniques(1) Dr.Saad Alabbad Department of Computer Science E-mail: salabbad@gmail.com Office # SR 068 Tel # 2581888 Prepared by Dr.Saad Alabbad

  2. Rules of Inference – Valid Arguments in Propositional Logic • Argument: An argument is a sequence of statements that end with a conclusion. • Valid: An argument is valid if and only if it is impossible for all premises (preceding statements) to be true and the conclusion to be false. By valid, we mean that the conclusion of the argument must follow from the truth of the premises of the argument. • Consider the arguments: “If you have a current password, then you can log onto the network.” “You have a current password.” Therefore, “You can log onto the network.”

  3. Rules of Inference – Valid Arguments in Propositional Logic • The conclusion “You can log onto the network” must be true when the premises “If you have a current password, then you can log onto the network” and “You have a current password” are true. • Let p= “You have a current password.” and q=“You can log onto the network.” Then the argument has the form where  is the symbol that denotes “therefore.” • The statement ((p → q) ⌃p) → q is a tautology. p → q p  q

  4. Rules of Inference – Rule for Propositional Logic • The argument form with premises p1, p2, …., pn and conclusion q is valid, when (p1⌃p2⌃….⌃pn) → q is a tautology. • To show that an argument is valid, instead of showing by truth table, we can establish the validity of some relatively simple argument forms, called rules of inference, which can be used as building blocks to construct more complicated valid argument forms. • The tautology ((p → q) ⌃p) → q is the basis of the rule of inference called modusponens or law of detachment.

  5. Rules of Inference – Rules for Propositional Logic • Example: Suppose that the conditional statement “If it snows today, then we will go skiing” and its hypothesis “It is snowing today”, are true. Then by modus ponens, it follows that the conclusion of the conditional statement, “We will go skiing” is true. • Q1: Determine whether the argument given here is valid and determine whether its conclusion must be true because of the validity of the argument.

  6. Rules of Inference – Rules for Propositional Logic

  7. Rules of Inference – Rule for Propositional Logic • Example: State which rule of inference is the basis of the following argument: “It is below freezing now. Therefore, it is either below freezing or raining now”. • Sol: Let p= “It is below freezing now.” and q = “It is raining now.” Then this argument is of the form: p p v q This is an argument that uses the addition rule. • Q2: State which rule of inference is the basis of the following argument: “It is below freezing and raining now. Therefore, it is below freezing”.

  8. Introduction to Proofs • A theorem is a statement that can be shown to be true (usually important statement) • Less important theorem sometimes are called propositions • A proof is a sequence of statements (valid argument) to show that a theorem is true • The statements to be used in proofs include: • Axioms (statement assumed to be true without proof) • Ex: If x is positive integer then x+1 is positive integer. • Hypothesis (premises) of the theorem • Previously proven theorems • Rules of inference used to draw conclusions and to move from one step to another

  9. Introduction to Proofs • A theorem is a statement that can be shown to be true (usually important statement) • Less important theorem sometimes are called propositions • A proof is a sequence of statements (valid argument) to show that a theorem is true • The statements to be used in proofs include: • Axioms (statement assumed to be true without proof) • Ex:If x is positive integer then x+1 is positive integer. • Hypothesis (premises) of the theorem • Previously proven theorems • Rules of inference used to draw conclusions and to move from one step to another Axioms Rules of inference New theorem Hypothesis proven theorems Prepared by Dr.Saad Alabbad

  10. Introduction to Proofs • Example: • If I have a car (C) I will drive to Makkah(M). My boss gave me 40,000 (G) or Fired me(F). If I have SR40,000 (H) then I have a car(C). My boss did not fire me. Therefore I will drive to Makkah(M). • GF Hypothesis • F Hypothesis • G Disjunctive syllogism rule using 1 and 2 • GH Axiom  • H C Hypothesis • G C Hypo. syllogism using 4,5 • C Modus ponens using 3 and 6 • C M Hypothesis • M Modus ponens using 7 and 8 Prepared by Dr.Saad Alabbad

  11. Methods of proving theorems:Direct proofs • A direct proofof a conditional statement H1 H2 … Hn C is established by assuming the truth of all hypothesis and using rules of inference, axioms and proven theorems to assert the truth of the conclusion • Example: prove that “if n is even then n2 is even” • Assume that n is even (hypothesis) • Therefore n=2k where k is integer (definition of even number) • By squaring n=2k we get n2=(2k)2=4k2=2(2k2) • Since 2K2 is integer (Axiom. See p.A-5) • Therefore n2=2r is even Prepared by Dr.Saad Alabbad

  12. Methods of proving theorems:Proof by contraposition • An indirect proofof a conditional statement p q is a direct proof of its contraposition qp. • Example 1: prove that “if n2 is even then n is even” The contraposition is “if n is not even then n2 is not even” • Assume that n is not even i.e nis odd (hypothesis) • Therefore n=2k+1 where k is integer (definition of odd number) • By squaring n=2k+1 we get • n2=(2k+1)2 • =4k2+4k+1 • =2(2k2+2k)+1= 2r+1 • Since 2k2+2k is integer (Axiom. See p.A-5) • Therefore n2=2r+1 is not even (odd) Prepared by Dr.Saad Alabbad

  13. Methods of proving theorems:Proof by contraposition • Example 2: prove that if n=ab then an or bn where a and b are positive integers • Let p be an , q be bn and r be n=ab • We want to prove that r pq • The contraposition is(pq )   r (By definition) • Which is equivalent top  q  r (De Morgan’s law) • Now assume thatan and bn (p  q) • Then a.bn.n=n (by multiplying the two inequalities) • Therefore abn • Which is the same as  r Prepared by Dr.Saad Alabbad

  14. Methods of proving theorems:Vacuous and Trivial Proofs • Vacuous Proof If we know that p is false then pq is vacuously true. • Example 1: prove that if x20 then 1=2 where x is a real number • Since x20 for every real number then the implication is vacuously true • Example 2: prove that if he is alive and he is dead then the sun is ice cold. • Since the hypothesis is always false the implication is vacuously true. • Trivial Proof If we know q is true then pq is trivially true. • Example 1: prove that if x=2 then x2  0 for all real numbers • Since x2  0 is true then the implication is trivially true.(we didn’t use the fact x=2) Prepared by Dr.Saad Alabbad

  15. Methods of proving theorems:Proofs by Contradiction(1) • In proof by contradiction it is shown that if some statement were false, a logical contradiction occurs, hence the statement must be true • Show that 2 is irrational • Let p be 2 is irrational • Assume p is true “2 is rational” • Then 2=a/b where gcd(a,b)=1 , b0 and a0 (Definition of rational numbers) • Squaring both sides we get 2=a2/b2 • It follows that 2a2=b2 • Hence b2 is even (Definition of even numbers) • Which means that b is even (Proved theorem. see slide 5) • But a2=b2/2=(2k)2/2=2k2 so a is even also (Definition of even numbers and proved thoeorm) • But this is a contradiction since 7 and 8 contradict the fact that gcd(a,b)=1 since if a and b are even then 2 is a common divisor. Prepared by Dr.Saad Alabbad

  16. Methods of proving theorems:Proofs by Contradiction(2) • Why is this method valid ? • The contradiction forces us to reject our assumption because our other steps based on that assumption are logical and justified. The only “mistake” that we could have made was the assumption itself. • Be careful! • Sometimes the contradiction comes from a mistake in the steps of the proof and not from the assumption. This makes the proof invalid. • Example: prove that 1=2 • Suppose that 21 and a=b for some a. • Then 2b b [multiply by b] • a+b b [2b=b+b=a+b by hypothesis] • (a-b)(a+b)  b(a-b) [multiply by a-b] • a2-b2  ab-b2 • a2 ab [subtract b2 from both sides] • a b which contradicts our assumption that a=b • Hence it follows that 1=2 • Can you find the error  Prepared by Dr.Saad Alabbad

  17. Methods of proving theorems:Proofs by Contradiction(3) • To prove a conditional statement p q by contradiction we prove thatp  q F is true which is equivalent top q . • Example: • If 3n+2 is odd then n is odd • Suppose that 3n+2 is odd and n is even [p  q] Then • n=2r [hypothesis q and definition of even numbers] • 3n=6r [multiply 1 by 3] • 3n+2=2+6r [add 2 to both sides] • 3n+2=2(1+3r) • 3n+2=2k [let k=1+3r] • Thus 3n+2 is even which is false (a contradiction !) • Therefore the implication is true. Prepared by Dr.Saad Alabbad

  18. Methods of proving theorems:Proofs of Equivalence and Disproof by counter example • Proofs of Equivalence • To prove pq we have to prove p  q and q  p • Example: prove “n is even if and only if n2 is even” • “if n is even then n2 is even” proved in slide 4 • “n2 is even then n is even” proved in slide 5 • Therefore, n is even if and only if n2 is even • Proving equivalence of several propositions • If we want to prove that p1 p2  p3 …  pn • Then it is sufficient to prove p1 p2,, p2 p3 … pn  p1 • Disproof by Counterexample • Prove that “For all real numbers x2>x” is false • X=0.5 is a counterexample since 0.52>0.5 is not true Prepared by Dr.Saad Alabbad

  19. Methods of proving theorems:Exhaustive Proof and Proof by Cases(1) • Sometimes there is no single argument that works for all possible cases appearing in the statement. So we have to consider all different cases and/or instances . • (1) Exhaustive Proof In this type of proofs. All possible instances are relatively small. So we prove each instance separately. Example:Prove that (n+1)3 >= 3n if n is a positive integer with n 4. We only have 4 instances to consider, n = 1, 2, 3, and 4. Proof of case n = 1: (1+1)3 >= 31 i.e 4 >= 3 which is true. Check n = 2, 3, and 4 similarly. • Notes • Human can perform exhaustive proofs with limited number of possibilities. Computers can consider much larger number of instances/cases. However • Instances/cases must be finite and enumerable Prepared by Dr.Saad Alabbad

  20. Methods of proving theorems:Exhaustive Proof and Proof by Cases(2) • (2) Proof by Cases In this type of proofs. The proof covers each possible case. Each case may cover an infinite number of instances Example:Prove "if n is an integer then n2 >= n". Case 1: n = 0. Since 02 = 0, 02 >= 0. Case 2: n >= 1. n >= 1 implies n2 >= n. Case 3: n <= -1. Since n is negative, n2 is positive, so n2 > n. . • Question: prove that |xy|=|x||y| • Hint: Consider all 4 cases depending on whether x is positive or x is negative and whether or not y is negative. • Common errors in with exhaustive proof and proof by cases: • Draw conclusion from non-exhaustive examples • Not covering all possible cases Prepared by Dr.Saad Alabbad

  21. Methods of proving theorems:Existence Proofs • Many theorems state that an object with certain properties exists. I.e., xP(x) where P is a predicate. A proof of such a theorem is called an EXISTENCE PROOF. There are two kinds: • (1) Constructive Existence Proof The proof is established be giving example a such that P(a) is true Example:Prove "there is a positive integer that can be written as the sum of cubes in two different ways." Proof: consider 1729=103+93=123+13. Finding such examples may require computer assistance. Check n = 2, 3, and 4 similarly. Prepared by Dr.Saad Alabbad

  22. Methods of proving theorems:Existence Proofs (2) Non-constructive Existence Proof • The proof is established by showing that an object a with P(a) is true must exist without explicitly demonstrating one. Proofs by contradiction are usually used in such cases. • Example: Let x1,x2,..,xnbe positive integers such that their average is m. prove that there exists xi such that xi≥m • Proof: suppose that there is no such number.i,e x1m , x2m … xnm By adding these inequalities we get: x1+ x2+…+ xn  nm Dividing by n : (x1+ x2+…+ xn)/n  m But since the average is defined as(x1+ x2+…+ xn)/n Then we havemmwhich is a contradiction. Therefore there must be a numberxi such that xi≥m. But we can not specify which number is that. Prepared by Dr.Saad Alabbad

  23. Methods of proving theorems:UniquenessProofs • Some theorems state that there is exactly one element with a certain property. • A proof of such a theorem is called a UNIQUENESS PROOF. • Strategy here is (1) show that an element x with the desired property exists (2) show that any other y (y != x) does not have the property. I.e., if x and y both have the property, then x must equal y. • Example: prove that the equation 3x+5=9 has a unique solution. • (1) There exists a solution namely x=4/3 • (2) suppose that y and z are solutions then 3y+5=9=3z+5 So 3y=3z Dividing by 3 we get y=z This proves that the solution is unique Prepared by Dr.Saad Alabbad

  24. END Prepared by Dr.Saad Alabbad

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