![Example 1
Premise: p ^ (p → q)
Conclusion: q
1 p ^ (p → q) Premise
2 p Simplification on 1
∵(p ^ q)
∴ p
3 p → q Simplification on 1
4 q Modus Ponens on 2 & 3
∵[(p → q) ^ p ]
∴ q
Sahar Selim MATH211 Lecture 4 | Rules of Inference
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p → q
p____
∴q
modus
ponens
p→q
¬q___
∴¬p
modus tollens
p______
∴ p ∨ q
Addition
p ∧ q
∴ p
Simplification
p
q_____
∴p ^ q
Conjunction
p → q
q → r
∴ p → r
hypothetical
syllogism
p ∨ q
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disjunctive
syllogism
p ∨ q
¬p ∨ r
∴ q ∨ r
Resolution](https://image.slidesharecdn.com/l4inferencerules-250228190008-ffb662e4/85/Discrete-Mathematics-Lecture-4-Inference-Rules-pdf-25-320.jpg)
Discrete Mathematics: Lecture 4 - Inference Rules.pdf
- 1. Math211 Discrete Mathematics Rules of Inference SAHAR SELIM
- 2. Agenda 1.6 Rules of Inference 1.6.1 Rules of inference Modus ponens, addition, simplification, conjunction, modus tollens, contrapositive, hypothetical syllogism, disjunctive syllogism, resolution, 1.6.2 Proofs with quantifiers Discrete Mathematics and Its Applications Kenneth H. Rosen Sahar Selim MATH211 Lecture 4 | Rules of Inference 2
- 3. Motivation “Mathematical proofs, like diamonds, are hard and clear, and will be touched with nothing but strict reasoning.” -John Locke Mathematical proofs are, in a sense, the only true knowledge we have They provide us with a guarantee as well as an explanation (and hopefully some insight) Sahar Selim MATH211 Lecture 4 | Rules of Inference 3
- 4. Motivation Mathematical proofs are necessary in CS You must always (try to) prove that your algorithm terminates is sound, complete, optimal finds optimal solution You may also want to show that it is more efficient than another method Proving certain properties of data structures may lead to new, more efficient or simpler algorithms Arguments may entail assumptions. You may want to prove that the assumptions are valid Sahar Selim MATH211 Lecture 4 | Rules of Inference 4
- 5. 1.6.1 Rules of Inference Sahar Selim MATH211 Lecture 4 | Rules of Inference
- 6. Sahar Selim MATH211 Lecture 4 | Rules of Inference 6
- 7. Rules of Inference Inference is deriving conclusions from evidences. The rules of inference are the means used to draw conclusions from other assertions, and to derive an argument or a proof An argument is valid If, whenever all the hypotheses are true, Then, the conclusion also holds Sahar Selim MATH211 Lecture 4 | Rules of Inference 7
- 8. Example Consider the following argument involving propositions (which, by definition, is a sequence of propositions): “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.” Sahar Selim MATH211 Lecture 4 | Rules of Inference 8
- 9. Example Use p to represent “You have a current password” q to represent “You can log onto the network” Then, the argument has the form p → q p ∴ q Sahar Selim MATH211 Lecture 4 | Rules of Inference 9
- 10. Inference Rules - General Form Inference Rule Pattern establishing that if we know that a set of hypotheses are all true, then a certain related conclusion statement is true. “∴” means “therefore” Sahar Selim MATH211 Lecture 4 | Rules of Inference 10 Hypothesis 1 Hypothesis 2 … __________________________ ∴ conclusion Premises Conclusion
- 11. Inference Rules & Implications Each logical inference rule corresponds to an implication that is a tautology. Corresponding tautology: ((Hypoth. 1) ∧ (Hypoth. 2) ∧ …) → conclusion From a sequence of assumptions, p1, p2, …, pn, you draw the conclusion q. (p1 ∧ p2 ∧ … ∧ pn) → q Sahar Selim MATH211 Lecture 4 | Rules of Inference 11 p1 p2 … pn q Premises Conclusion
- 12. Rules Sahar Selim MATH211 Lecture 4 | Rules of Inference
- 13. Contrapositive The contrapositive is the following tautology (p → q) → (¬q→ ¬p) Usefulness If you are having trouble proving the p implies q in a direct manner You can try to prove the contrapositive instead! Sahar Selim MATH211 Lecture 4 | Rules of Inference 13
- 14. Modus Ponens p → q Rule of modus ponens p (a.k.a. law of detachment) ∴q In logic terminology, modus ponens is the tautology: (p ∧ (p → q)) → q Example “If it snows today, then we will go skiing” and “It is snowing today” imply “We will go skiing” “the mode of affirming” Sahar Selim MATH211 Lecture 4 | Rules of Inference 14
- 15. Modus Tollens p → q ¬q Rule of modus tollens ∴¬p Similar to the modus ponens, modus tollens is based on the following tautology ((p → q) ∧ ¬q) → ¬p Example “if you are in this class, you will get a grade” “you will not get a grade” By Modus Tollens, you can conclude that you are not in this class “the mode of denying” Let p = “you are in this class” Let q = “you will get a grade” Sahar Selim MATH211 Lecture 4 | Rules of Inference 15
- 16. Addition p Rule of Addition ∴ p ∨ q Addition involves the tautology p → (p ∨ q) Example: “It is below freezing now. Therefore, it is either below freezing or raining now.” Sahar Selim MATH211 Lecture 4 | Rules of Inference 16
- 17. Simplification p ∧ q Rule of Simplification ∴ p Simplification is based on the tautology (p ∧ q) → p (p ∧ q) → q Example: “It is below freezing and raining now. Therefore, it is below freezing now. Sahar Selim MATH211 Lecture 4 | Rules of Inference 17
- 18. Conjunction p Rule of Conjunction q ∴ p ∧ q The conjunction is almost trivially intuitive. It is based on the following tautology: ((p) ∧ (q)) → (p ∧ q) Sahar Selim MATH211 Lecture 4 | Rules of Inference 18
- 19. Hypothetical Syllogism p → q Rule of hypothetical syllogism q → r ∴ p → r Hypothetical syllogism is based on the following tautology ((p → q) ∧ (q → r)) → (p → r) Example: If you don’t get a job, you won’t have money If you don’t have money, you will starve. Therefore, if you don’t get a job, you’ll starve Sahar Selim MATH211 Lecture 4 | Rules of Inference 19
- 20. Disjunctive Syllogism p ∨ q Rule of disjunctive syllogism ¬p ∴ q A disjunctive syllogism is formed on the basis of the tautology ((p ∨ q) ∧ ¬p)→ q Example The sky is either blue or grey Well it isn’t blue Therefore, the sky is grey Sahar Selim MATH211 Lecture 4 | Rules of Inference 20
- 21. Resolution p ∨ q Resolution ¬p ∨ r ∴ q ∨ r For resolution, we have the following tautology ((p ∨ q) ∧ (¬p ∨ r)) → (q ∨ r) Essentially, If we have two true disjunctions that have mutually exclusive propositions Then we can conclude that the disjunction of the two non-mutually exclusive propositions is true Sahar Selim MATH211 Lecture 4 | Rules of Inference 21
- 22. Rules of Inference Sahar Selim MATH211 Lecture 4 | Rules of Inference 22
- 23. Rules of Inference Sahar Selim MATH211 Lecture 4 | Rules of Inference 23
- 24. Example 1 Premise: p ^ (p → q) Conclusion: q Sahar Selim MATH211 Lecture 4 | Rules of Inference 24 p → q p____ ∴q modus ponens p→q ¬q___ ∴¬p modus tollens p______ ∴ p ∨ q Addition p ∧ q ∴ p Simplification p q_____ ∴p ^ q Conjunction p → q q → r ∴ p → r hypothetical syllogism p ∨ q ¬p___ ∴ q disjunctive syllogism p ∨ q ¬p ∨ r ∴ q ∨ r Resolution
- 25. Example 1 Premise: p ^ (p → q) Conclusion: q 1 p ^ (p → q) Premise 2 p Simplification on 1 ∵(p ^ q) ∴ p 3 p → q Simplification on 1 4 q Modus Ponens on 2 & 3 ∵[(p → q) ^ p ] ∴ q Sahar Selim MATH211 Lecture 4 | Rules of Inference 25 p → q p____ ∴q modus ponens p→q ¬q___ ∴¬p modus tollens p______ ∴ p ∨ q Addition p ∧ q ∴ p Simplification p q_____ ∴p ^ q Conjunction p → q q → r ∴ p → r hypothetical syllogism p ∨ q ¬p___ ∴ q disjunctive syllogism p ∨ q ¬p ∨ r ∴ q ∨ r Resolution
- 26. Example 2 Assume that the statements below hold: • (p → q) • (r → s) • (r ∨ p) Assume that q is false Show that s must be true Premises (p → q) (r → s) (r ∨ p) ~q Conclusion s Sahar Selim MATH211 Lecture 4 | Rules of Inference 26
- 27. Sahar Selim MATH211 Lecture 9 | Mathematical Induction 27 p → q p____ ∴q modus ponens p → q ¬q___ ∴¬p modus tollens p______ ∴ p ∨ q Addition p ∧ q ∴ p Simplification p q_____ ∴p ^ q Conjunction p → q q → r ∴ p → r hypothetical syllogism p ∨ q ¬p___ ∴ q disjunctive syllogism p ∨ q ¬p ∨ r ∴ q ∨ r Resolution p → q r → s r ∨ p ¬q _______ s
- 28. Example 2 Premises (p → q) (r → s) (r ∨ p) ~q Conclusion s 1. (p → q) 2. ¬q 3. (¬q ∧ (p → q)) → ¬p 4. (r ∨ p) 5. (r ∨ p) ∧ ¬p) → r 6. (r → s) 7. (r ∧ (r → s)) → s Premise Premise by modus tollens on 1 + 2 Premise by disjunctive syllogism 3 + 4 Premise by modus ponens 5 + 6 Sahar Selim MATH211 Lecture 4 | Rules of Inference 28
- 29. Example 3 Suppose we have the following premises: “It is not sunny and it is cold.” “if it is not sunny, we will not swim” “If we do not swim, then we will canoe.” “If we canoe, then we will be home early.” Given these premises, prove the theorem “We will be home early” using inference rules. Sahar Selim MATH211 Lecture 4 | Rules of Inference 29
- 30. Example 3 “It is not sunny and it is cold.” “if it is not sunny, we will not swim” “If we do not swim, then we will canoe.” “If we canoe, then we will be home early.” Given these premises, prove the theorem “We will be home early” using inference rules. Sahar Selim MATH211 Lecture 4 | Rules of Inference 30 Let us adopt the following abbreviations: sunny = “It is sunny”; cold = “It is cold”; swim = “We will swim”; canoe = “We will canoe”; early = “We will be home early”.
- 31. Example 3 “It is not sunny and it is cold.” “if it is not sunny, we will not swim” “If we do not swim, then we will canoe.” “If we canoe, then we will be home early.” Given these premises, prove the theorem “We will be home early” using inference rules. Sahar Selim MATH211 Lecture 4 | Rules of Inference 31 Let us adopt the following abbreviations: sunny = “It is sunny”; cold = “It is cold”; swim = “We will swim”; canoe = “We will canoe”; early = “We will be home early”. Then, the premises can be written as: (1) ¬sunny ∧ cold (2) ¬sunny → ¬swim (3) ¬swim → canoe (4) canoe → early
- 32. Sahar Selim MATH211 Lecture 9 | Mathematical Induction 32 p → q p____ ∴q modus ponens p → q ¬q___ ∴¬p modus tollens p______ ∴ p ∨ q Addition p ∧ q ∴ p Simplification p q_____ ∴p ^ q Conjunction p → q q → r ∴ p → r hypothetical syllogism p ∨ q ¬p___ ∴ q disjunctive syllogism p ∨ q ¬p ∨ r ∴ q ∨ r Resolution ¬sunny ∧ cold ¬sunny → ¬swim ¬swim → canoe canoe → early ________________ early
- 33. Example 3 Step Proved by 1. ¬sunny ∧ cold Premise #1. 2. ¬sunny Simplification of 1. 3. ¬sunny → ¬swim Premise #2. 4. ¬swim Modus tollens on 2,3. 5. ¬swim→canoe Premise #3. 6. canoe Modus ponens on 4,5. 7. canoe→early Premise #4. 8. early Modus ponens on 6,7. Sahar Selim MATH211 Lecture 4 | Rules of Inference 33
- 34. 1.6.2 Rules of Inference for Quantified Statements Sahar Selim MATH211 Lecture 4 | Rules of Inference
- 35. Inference Rules for Quantifiers Rules of inference can be extended in a straightforward manner to quantified statements 1. Universal Instantiation (UI) 2. Universal Generalization (UG) 3. Existential Instantiation (EI) 4. Existential Generalization (EG) Sahar Selim MATH211 Lecture 4 | Rules of Inference 35
- 36. Universal Instantiation (UI) ∀x P(x) ∴P(o) (substitute any object o) Example All dogs are cute Oliver is a dog Therefore, Oliver is cute Sahar Selim MATH211 Lecture 4 | Rules of Inference 36 UI = ∀𝑥𝑥 𝑃𝑃(𝑥𝑥) ∴ 𝑃𝑃(𝑐𝑐)
- 37. Universal Generalization (UG) (for g a general element of u.d.) Example 1: Oliver is a dog that has 4 legs. Then, All dogs have 4 legs. Example 2: Ahmed is an Egyptian that loves falafel. Then, ALL Egyptians love falafel UG = 𝑃𝑃(𝑐𝑐) ∴∀𝑥𝑥 𝑃𝑃(𝑥𝑥) Sahar Selim MATH211 Lecture 4 | Rules of Inference 37 We will mainly focus on valid cases
- 38. Inference Rules for Quantifiers Existential Instantiation (EI) ∃x P(x) ∴P(c) (substitute a new constant c) Existential Generalization (EG) P(o) (substitute any extant object o) ∴∃x P(x) Sahar Selim MATH211 Lecture 4 | Rules of Inference 38
- 39. Rules of Inference for Quantified Statements Sahar Selim MATH211 Lecture 4 | Rules of Inference 39
- 40. Example 1 Show that “A car in the garage has an engine problem” and “Every car in the garage has been sold” imply the conclusion “A car has been sold has an engine problem” Let G(x): “x is in the garage” E(x): “x has an engine problem” S(x): “x has been sold” Let UoD be the set of all cars The premises are as follows: ∃x (G(x) ∧ E(x)) ∀x (G(x) → S(x)) The conclusion we want to show is: ∃x (S(x) ∧ E(x)) Sahar Selim MATH211 Lecture 4 | Rules of Inference 40
- 41. Sahar Selim MATH211 Lecture 9 | Mathematical Induction 41 p → q p____ ∴q modus ponens p → q ¬q___ ∴¬p modus tollens p______ ∴ p ∨ q Addition p ∧ q ∴ p Simplification p q_____ ∴p ^ q Conjunction p → q q → r ∴ p → r hypothetical syllogism p ∨ q ¬p___ ∴ q disjunctive syllogism p ∨ q ¬p ∨ r ∴ q ∨ r Resolution ∃x (G(x) ∧ E(x)) ∀x (G(x) → S(x)) ________________ ∃x (S(x) ∧ E(x)) ∀𝑥𝑥 𝑃𝑃(𝑥𝑥) ∴ 𝑃𝑃(𝑐𝑐) Universal Instantiation 𝑃𝑃(𝑐𝑐) ∴ ∀𝑥𝑥 𝑃𝑃(𝑥𝑥) Universal Generalization ∃𝑥𝑥 𝑃𝑃(𝑥𝑥) ∴ 𝑃𝑃(𝑐𝑐) Existential Instantiation 𝑃𝑃(𝑐𝑐) ∴ ∃𝑥𝑥 𝑃𝑃(𝑥𝑥) Existential Generalization
- 42. Example 1 – Solution 1. ∃x (G(x) ∧ E(x)) 1st premise 2. (G(c) ∧ E(c)) Existential instantiation of (1) 3. G(c) Simplification of (2) 4. ∀x (G(x) → S(x)) 2nd premise 5. G(c) → S(c) Universal instantiation of (4) 6. S(c) Modus ponens on (3) and (5) 7. E(c) Simplification from (2) 8. S(c) ∧ E(c) Conjunction of (6) and (7) 9. ∃x (S(x) ∧ E(x)) Existential generalization of (8) QED Sahar Selim MATH211 Lecture 4 | Rules of Inference 42
- 43. Example 2 “Everyone in this discrete math class has taken a course in computer science” and “Ali is a student in this class” imply “Ali has taken a course in computer science” D(x): “x is in discrete math class” C(x): “x has taken a course in computer science” ∀x (D(x) → C(x)) D(Ali) ∴ C(Ali) Sahar Selim MATH211 Lecture 4 | Rules of Inference 43
- 44. Sahar Selim MATH211 Lecture 9 | Mathematical Induction 44 p → q p____ ∴q modus ponens p → q ¬q___ ∴¬p modus tollens p______ ∴ p ∨ q Addition p ∧ q ∴ p Simplification p q_____ ∴p ^ q Conjunction p → q q → r ∴ p → r hypothetical syllogism p ∨ q ¬p___ ∴ q disjunctive syllogism p ∨ q ¬p ∨ r ∴ q ∨ r Resolution ∀x (D(x) → C(x)) D(Ali) ________________ ∴ C(Ali) ∀𝑥𝑥 𝑃𝑃(𝑥𝑥) ∴ 𝑃𝑃(𝑐𝑐) Universal Instantiation 𝑃𝑃(𝑐𝑐) ∴ ∀𝑥𝑥 𝑃𝑃(𝑥𝑥) Universal Generalization ∃𝑥𝑥 𝑃𝑃(𝑥𝑥) ∴ 𝑃𝑃(𝑐𝑐) Existential Instantiation 𝑃𝑃(𝑐𝑐) ∴ ∃𝑥𝑥 𝑃𝑃(𝑥𝑥) Existential Generalization
- 45. Example 2 – Solution Step Proved by 1. ∀x (D(x) → C(x)) Premise #1. 2. D(Ali) → C(Ali) Univ. instantiation. 3. D(Ali) Premise #2. 4. C(Ali) Modus ponens on 2,3. Sahar Selim MATH211 Lecture 4 | Rules of Inference 45
- 46. Example 3 “A student in this class has not read the book” and “Everyone in this class passed the first exam” imply “Someone who passed the first exam has not read the book” C(x): “x is in this class” B(x): “x has read the book” P(x): “x passed the first exam” Sahar Selim MATH211 Lecture 4 | Rules of Inference 46 ∃x (C(x) ∧ ¬B(x)) ∀x (C(x) → P(x)) ∴ ∃x(P(x) ∧ ¬B(x))
- 47. Sahar Selim MATH211 Lecture 9 | Mathematical Induction 47 p → q p____ ∴q modus ponens p → q ¬q___ ∴¬p modus tollens p______ ∴ p ∨ q Addition p ∧ q ∴ p Simplification p q_____ ∴p ^ q Conjunction p → q q → r ∴ p → r hypothetical syllogism p ∨ q ¬p___ ∴ q disjunctive syllogism p ∨ q ¬p ∨ r ∴ q ∨ r Resolution ∃x (C(x) ∧ ¬B(x)) ∀x (C(x) → P(x)) ________________ ∴ ∃x(P(x) ∧ ¬B(x)) ∀𝑥𝑥 𝑃𝑃(𝑥𝑥) ∴ 𝑃𝑃(𝑐𝑐) Universal Instantiation 𝑃𝑃(𝑐𝑐) ∴ ∀𝑥𝑥 𝑃𝑃(𝑥𝑥) Universal Generalization ∃𝑥𝑥 𝑃𝑃(𝑥𝑥) ∴ 𝑃𝑃(𝑐𝑐) Existential Instantiation 𝑃𝑃(𝑐𝑐) ∴ ∃𝑥𝑥 𝑃𝑃(𝑥𝑥) Existential Generalization
- 48. Example 3 – Solution Step Proved by 1. ∃x(C(x) ∧ ¬B(x)) Premise #1. 2. C(a) ∧ ¬B(a) Exist. Instantiation of 1. 3. C(a) Simplification on 2. 4. ∀x (C(x) → P(x)) Premise #2. 5. C(a) → P(a) Univ. instantiation of 4. 6. P(a) Modus ponens on 3,5 7. ¬B(a) Simplification on 2 8. P(a) ∧ ¬B(a) Conjunction on 6,7 9. ∃x(P(x) ∧ ¬B(x)) Exist. Generalization of 8 Sahar Selim MATH211 Lecture 4 | Rules of Inference 48
- 49. Supplementary Material Discrete Math - 1.6.1 Rules of Inference for Propositional Logic Discrete Math - 1.6.2 Rules of Inference for Quantified Statements Sahar Selim MATH211 Lecture 4 | Rules of Inference 49
- 50. Next Lecture Introduction to proofs direct proof, proof by contraposition, proof by contradiction, proof by cases. Sahar Selim MATH211 Lecture 4 | Rules of Inference 50
- 51. Sahar Selim MATH211 Lecture 4 | Rules of Inference
- 52. Problem 1 The premises “If you send me an e-mail message, then I will finish writing the program,” “If you do not send me an e-mail message, then I will go to sleep early,” “If I go to sleep early, then I will wake up feeling refreshed” The conclusion “If I do not finish writing the program, then I will wake up feeling refreshed.” Sahar Selim MATH211 Lecture 4 | Rules of Inference 52
- 53. Solution Step Reason 1. p → q 2. ￢q →￢p 3. ￢p → r 4. ￢q → r 5. r → s 6. ￢q → s Premise Contrapositive of (1) Premise Hypothetical syllogism using (2) and (3) Premise Hypothetical syllogism using (4) and (5) Sahar Selim MATH211 Lecture 4 | Rules of Inference 53
- 54. Problem 2 Is this argument correct or incorrect? “All TAs compose easy quizzes. Rana is a TA. Therefore, Rana composes easy quizzes.” Sahar Selim MATH211 Lecture 4 | Rules of Inference 54
- 55. Solution First, separate the premises from conclusions: Premise #1: All TAs compose easy quizzes. Premise #2: Rana is a TA. Conclusion: Rana composes easy quizzes. Next, re-render the example in logic notation. Premise #1: All TAs compose easy quizzes. Let U.D. = all people Let T(x) :≡ “x is a TA” Let E(x) :≡ “x composes easy quizzes” Then Premise #1 says: ∀x, T(x)→E(x) Sahar Selim MATH211 Lecture 4 | Rules of Inference 55
- 56. Solution cont… Premise #2: Rana is a TA. Let R :≡ Rana Then Premise #2 says: T(R) Conclusion says: E(R) The argument is correct, because it can be reduced to a sequence of applications of valid inference rules, as follows: Sahar Selim MATH211 Lecture 4 | Rules of Inference 56
- 57. The Proof in Detail Statement How obtained 1. ∀x, T(x) → E(x) (Premise #1) 2. T(Rana) → E(Rana) (Universal instantiation) 3. T(Rana) (Premise #2) 4. E(Rana) (Modus Ponens from statements #2 and #3) Sahar Selim MATH211 Lecture 4 | Rules of Inference 57