§1.5 Rules of Inference

1. §1.5 Rules of Inference • In mathematics, a proof is: • a correct (well-reasoned, logically valid) and complete (clear, detailed) argument that rigorously & undeniably establishes the truth of a mathematical statement. • Why must the proof be correct & complete? • Correctness: If the system (claims to) prove something is true, it really is true. • Completeness: If something really is true, the system is capable of proving it. • Proving a theorem allows us to rely upon on its correctness even in the most critical scenarios.

2. Proof Terminology • Axioms: assertions which are given to be true. • Postulates, Hypotheses, Premises • Assumptions (often unproven) defining the structures about which we are reasoning. • Rules of inference • Patterns of logically valid deductions from hypotheses to conclusions. • Theorem • A theorem is a valid logical assertion which can be proved using other theorems, axioms, postulates and rules of inference.

3. More Proof Terminology • Lemma - A minor theorem used as a stepping-stone to proving a major theorem. • Corollary - A minor theorem proved as an easy consequence of a major theorem. • Conjecture - A statement whose truth value has not been proven.(A conjecture may be widely believed to be true, regardless.) • Theory – The set of all theorems that can be proven from a given set of axioms.

4. Inference Rules - General Form • An Inference Rule is • A pattern establishing that if a set of antecedent statements (hypotheses) are all true, then we can validly deduce that a certain related consequent statement (conclusion) is true. • antecedent 1 antecedent 2 …  consequent “” means “therefore” • Corresponding tautology: • ((ante. 1)  (ante. 2)  …)  consequent • i.e., ((ante. 1)  (ante. 2)  …) consequent

5. Some Inference Rules • p Rule of Addition pq • pq Rule of Simplification p • p Rule of Conjunctionq  pq

6. Modus Ponens • pq Rule of modus ponens p (a.k.a. law of detachment)q Example: If I dance all night, then I get tired. I danced all night. Therefore I got tired.

7. Modus Tollens • pq qRule of modus tollensp • pq Rule of hypotheticalqr syllogismpr • p q Rule of disjunctive p syllogism q

8. Constructive dilemma • (P  Q)  (R  S) P R Constructive dilemma  Q S • (P  Q)  ( P  S) P   P Resolution  Q S

9. Formal Proofs • A formal proof of a conclusion C, given premises p1, p2,…,pnconsists of a sequence of steps, each of which applies some inference rule to premises or previously-proven statements (antecedents) to yield a new true statement (the consequent). • A proof demonstrates that if the premises are true, then the conclusion is true.

10. Formal Proof Example • Suppose we have the following premises:“It is not sunny and it is cold.”“We will swim if only if it is sunny.”“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.

11. Proof Example cont. • 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) swim ↔sunny(3) swim  canoe (4) canoe  early

12. Proof Example cont. StepProved by1. sunny  coldPremise #1.2. sunnySimplification of 1.3. swim ↔sunnyPremise #2.4. swimIFF Truth table on 2,3.5. swimcanoePremise #3.6. canoeModus ponens on 4,5.7. canoeearlyPremise #4.8. earlyModus ponens on 6,7.

13. Another Example Show that (RS) can be derived from (P (Q S)), (R  P), and Q StepInference Rule • R  P P • R P (assumed premise) • P disjunctive syllogism on (1), (2) • P(Q S) P • Q S Modus ponens on (3), (4) • Q P • S Modus ponens on (5), (6) • R  S conditional premise of (2), (7)

14. More Examples Show that S  R can be derived from (P  Q), (P  R) and (Q  S) StepInference Rule • P  Q P • P  Q T, (1) • Q  S P • P  S hypo. Syll. on (2), (3) • S P T, (4) • P  R P • S  R T, (5), (6) • S  R T, (7)

15. Inference Rules for Quantifiers • xP(x) Universal instantiation (US)P(o) (substitute any object o) • P(g) (for g a general element of u.d.)xP(x) Universal generalization (UG) • xP(x) Existential instantiation (ES)P(c) (substitute a newconstantc) • P(o) (substitute any extant object o) xP(x) Existential generalization (EG)

16. Counter Examples • (x)(y)G(x, y), where G(x, y)：y>x • (y)G(y, y) US,(1) Wrong! • (y)G(z, y) US,(1) • (y)(y)G(y, y) UG,(3) Wrong! • (z)(y)G(z, y) UG,(3)

17. An Example • Show that (x) (P(x)  Q(x))  (x) (Q(x)  R(x))⇒ (x) (P(x)  R(x)) StepInference Rule • (x) (P(x)  Q(x)) P • P(y)  Q(y) US, (1) • (x) (Q(x)  R(x)) P • Q(y)  R(y) US, (3) • P(y)  R(y) hypo. Syll. on (2), (4) • (x) (P(x)  R(x)) UG, (5)

18. More Examples Show ((x) (F(x)  S(x))  (y) (M(y)  W(y))) ((y) (M(y)  W(y)))⇒ (x) (F(x)   S(x)) StepInference Rule • (y) (M(y)  W(y)) P • M(z)  W(z) ES, (1) •  (M(z)  W(z)) T, (2) • (y)  (M(y)  W(y)) EG, (3) • (y)(M(y)  W(y)) T, (4) • (x) (F(x)  S(x))  (y) (M(y)  W(y)) P • (x) (F(x)  S(x)) T, (5), (6) • (x) (F(x)  S(x)) T, (7) • (F(x)  S(x)) US, (8) • F(x)   S(x) T, (9) • (x) (F(x)   S(x)) UG, (10)

19. Restriction • UG applicable variable should not be free in any of the given premises • UG should not be applied to the free variable resulting from ES making other variable free in a prior step. (x)(z) A(z,x) ⇒ (z)A(z,x) by US ⇒ A(z,x) by ES ⇒ (x)A(z,x) by UG(not allowed!) ⇒ (z) (x)A(z,x) by EG

20. Common Fallacies • A fallacy is an inference rule or other proof method that is not logically valid. • Fallacy of affirming the conclusion: • “pq is true, and q is true, so p must be true.” • Fallacy of denying the hypothesis: • “pq is true, and p is false, so q must be false.” • Circular reasoning: • This occurs when we use the truth of statement being proved (or something equivalent) in the proof itself.

21. Proof Methods for Implications For proving implications pq, we have: • Direct proof:Assume p is true, and prove q. • Indirect proof:Assume q, and prove p. • Vacuous proof:Prove p by itself. • Trivial proof:Prove q by itself.

22. Direct Proof Example • Definition: An integer n is called odd iff n=2k+1 for some integer k; n is even iff n=2k for some k. • Theorem: (For all numbers n) If n is an odd integer, then n2 is an odd integer. • Proof: If n is odd, then n = 2k+1 for some integer k. Thus, n2 = (2k+1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Therefore n2 is of the form 2j + 1 (with j the integer 2k2 + 2k), thus n2 is odd. □

23. Indirect Proof Example • Theorem: (For all integers n) If 3n+2 is odd, then n is odd. • Proof: Suppose that the conclusion is false, i.e., that n is even. Then n=2k for some integer k. Then 3n+2 = 3(2k)+2 = 6k+2 = 2(3k+1). Thus 3n+2 is even, because it equals 2j for integer j = 3k+1. So 3n+2 is not odd. We have shown that ¬(n is odd)→¬(3n+2 is odd), thus its contra-positive (3n+2 is odd) → (n is odd) is also true. □

24. Vacuous Proof Example • Theorem: (For all n) If n is both odd and even, then n2 = n + n. • Proof:The statement “n is both odd and even” is necessarily false, since no number can be both odd and even. So, the theorem is vacuously true. □

25. Trivial Proof Example • Theorem: (For integers n) If n is the sum of two prime numbers, then either n is odd or n is even. • Proof:Any integer n is either odd or even. So the conclusion of the implication is true regardless of the truth of the antecedent. Thus the implication is true trivially. □

26. Proof by Contradiction of any Statements • A method for proving p. • Assume p, and prove both q and q for some proposition q. (Can be anything!) • Thus p (q  q) • (q  q) is a trivial contradiction, equal to F • Thus pF, which is only true if p=F • Thus p is true.

27. Proof by Contradiction Example • Theorem: is irrational. • Proof: Assume 21/2 were rational. This means there are integers i,j with no common divisors such that 21/2 = i/j. Squaring both sides, 2 = i2/j2, so 2j2 = i2. So i2 is even; thus i is even. Let i=2k. So 2j2 = (2k)2 = 4k2. Dividing both sides by 2, j2 = 2k2. Thus j2 is even, so j is even. But then i and j have a common divisor, namely 2, so we have a contradiction. □

28. Blackboard Exercises for 1.5 Prove: If x2 is even, x is even • Proof: if x is not even, x is odd. Therefore x2 is odd. This is the contrapositive of the original assertion.

29. Review: Proof Methods So Far • Direct, indirect, vacuous, and trivial proofs of statements of the form pq. • Proof by contradiction of any statements.