1. March 23 rd

2. Four Additional Rules of Inference

3.  Constructive Dilemma (CD): (p  q) (r  s) p v r q v s

4.  Constructive Dilemma (CD): (p  q) (r  s) p v r q v s  Involves two modus ponens steps

5.  Constructive Dilemma (CD): (p  q) (r  s) p v r q v s  Involves two modus ponens steps ◦ If we have p then we have q; if we have r we have s. ◦ Since it is the case that p or r, it follows by modus ponens that we have either q or s.

6.  Constructive Dilemma (CD): (p  q) (r  s) p v r q v s

7.  Constructive Dilemma (CD): (p  q) (r  s) p v r q v s

8. Simplification (Simp) Also known as “Conjunction Elimination” p qp q pq When a conjunction is true, both conjuncts will be true. From a conjunction pq on one line, we may assert p, or alternatively assert q.

9. Simplification (Simp) Also known as “Conjunction Elimination” p qp q pq When a conjunction is true, both conjuncts will be true. From a conjunction pq on one line, we may assert p, or alternatively assert q. ◦ Yes, this is a divergence from the book – though the book does admit you can easily derive q the same way by flipping the order of the conjuncts.

10. Simplification (Simp) Example: 1, Simp

11. Simplification (Simp) Example: 1, Simp

12. Conjunction (Conj) AKA “Conjunction Introduction” / “AND Intro” p q. p q From two statements on separate lines, we may assert a conjunction on a new line that conjoins the two statements.

13. Conjunction (Conj) Example: p q p q 1, 2, Conj

14. Conjunction (Conj) Example: p q p q 1, 2, Conj

15. Addition (Add) AKA Disjunction Introduction / OR Intro p p v q From any statement, we may assert on a new line the disjunction of that statement and any additional statement.

16. Addition (Add) Example: 1, Add

17. Addition (Add) Example: 1, Add

18. Addition (Add) Example: 1, Add

19. Misapplication of Simp (AND Elimination) 1, Simp This is invalid. The problem was that to derive S via Simp, we would first need to put (S ∙ T) on a line by itself.

20. Misapplication of Add (Addition, AKA Disjunction Introduction / OR Intro) 1, Add This is an improper use of Add. J must be added to the whole line – we can’t just add it to the consequent. ◦ Note: The above inference is still valid, but we can’t get it using Add in this manner.

21. Common strategies involving the additional rules of inference

22. ◦ If the conclusion contains a letter that appears in a conjunctive statement in the premises, consider obtaining that letter via simplification. ◦ If the conclusion is a conjunctive statement, consider obtaining it via conjunction by first obtaining the individual conjuncts.

23. ◦ If the conclusion is a disjunctive statement, consider obtaining it via constructive dilemma or addition. ◦ If the conclusion contains a letter not found in the premises, addition must be used to introduce that letter. ◦ Conjunction can be used to set up constructive dilemma.

24. NOTE: There are also RULES OF REPLACEMENT, expressed as pairs of logically equivalent statement forms, either of which can replace each other in a proof sequence. ◦ Underlying the use of rules of replacement are Axioms of Replacement, which assert that within the context of a proof, logically equivalent expressions may replace each other. ◦ By Axioms of Replacement, the rules of replacement may be applied to an entire line or to any part of a line.

25.  A double colon (::) is used to designate logical equivalence. ◦ e.g., p v q :: q v p

26. DeMorgan’s Rule (DM) ~(p q) :: (~p v ~q) ~(p v q) :: (~p ~q)

27. Commutativity (Com) (p v q) :: (q v p) (p q) :: (q p)

28. Associativity (Assoc): [p v (q v r) ] :: [ (p v q) v r) ] [p (q r) ] :: [ (p q) r) ]

29. Distribution (Dist): [p (q v r) ] :: [ (p q ) v (p r) ] [p v (q r) ] :: [ (p v q ) (p v r) ]

30. Double Negation (DN): p :: ~ ~p

31. Transposition (Trans): (p  q) :: (~q  ~p) - You can use this to set up hypothetical syllogisms and constructive dilemmas. Material Implication (Impl): (p  q) :: (~q v p) - Material implication can be used to set up hypothetical syllogisms.

32. Material Equivalence (Equiv): (p ≡ q) :: [(p  q) (q  p)] (p ≡ q) :: [(p q) v (~q ~p) Exportation (Exp): [(p q)  r] :: [(p  (q  r)] - You can use this to set up modus ponens or modus tollens.

33. Tautology (Taut): p :: ( p v p ) p :: ( p p )

34.  Conditional Proof is a method for deriving a conditional statement (either the conclusion or some intermediate line) that offers the usual advantage of being both shorter and simpler than the direct method. For example: 1.A  (B C) 2.( B V D)  E / A  E We want to show that A  E…

35. To Construct a Conditional Proof: ◦ Begin by assuming the antecedent of the desired conditional statement on one line.

36. To Construct a Conditional Proof: ◦ Begin by assuming the antecedent of the desired conditional statement on one line. 1. A  (B C) 2. ( B V D)  E / A  E

37. To Construct a Conditional Proof: ◦ Begin by assuming the antecedent of the desired conditional statement on one line. 1. A  (B C) 2. ( B V D)  E / A  E 3. AACP (“assumption for conditional proof”)

38. To Construct a Conditional Proof: ◦ Derive the consequent on a subsequent line, showing the justification for each step.

39. To Construct a Conditional Proof: ◦ Derive the consequent on a subsequent line, showing the justification for each step. 1. A  (B C) 2. ( B V D)  E / A  E 3. AACP (“assumption for conditional proof”) 4. B C1, 3, MP

40. To Construct a Conditional Proof: ◦ Derive the consequent on a subsequent line, showing the justification for each step. 1. A  (B C) 2. ( B V D)  E / A  E 3. AACP (“assumption for conditional proof”) 4. B C1, 3, MP 5. B4, Simp

41. To Construct a Conditional Proof: ◦ Derive the consequent on a subsequent line, showing the justification for each step. 1. A  (B C) 2. ( B V D)  E / A  E 3. AACP (“assumption for conditional proof”) 4. B C1, 3, MP 5. B4, Simp 6. B V D5, Add

42. To Construct a Conditional Proof: ◦ Derive the consequent on a subsequent line, showing the justification for each step. 1. A  (B C) 2. ( B V D)  E / A  E 3. AACP (“assumption for conditional proof”) 4. B C1, 3, MP 5. B4, Simp 6. B V D5, Add 7. E2, 6, MP

43. To Complete a Conditional Proof: ◦ “Discharge” these lines in the desired conditional statement.  Every conditional proof must be discharged, otherwise any conclusion can be derived from any premises.

44. To Complete a Conditional Proof: ◦ “Discharge” these lines in the desired conditional statement.  Every conditional proof must be discharged, otherwise any conclusion can be derived from any premises. 1. A  (B C) 2. ( B V D)  E / A  E 3. AACP (“assumption for conditional proof”) 4. B C1, 3, MP 5. B4, Simp 6. B V D5, Add 7. E2, 6, MP 8. A  E 3-7, CP (“conditional proof”)

45.  Indirect proof is a technique similar to conditional proof that can be used on any argument to derive either the conclusion or some intermediate line leading to the conclusion.

46. To construct an indirect proof: 1.Begin by assuming the negation of the statement to be obtained. 2.Use this assumption to derive a contradiction. 3.From this contradiction, conclude that the original statement is false. 4.As in conditional proofs, every indirect proof must be discharged, otherwise any conclusion can be derived from any premises.

47.  Indirect and conditional proofs can be combined to derive either a line in a proof sequence or the conclusion of a proof.