March 23 rd

Four Additional Rules of Inference

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

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

 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.

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

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

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.

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.

Simplification (Simp) Example: 1, Simp

Simplification (Simp) Example: 1, Simp

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.

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

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

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.

Addition (Add) Example: 1, Add

Addition (Add) Example: 1, Add

Addition (Add) Example: 1, Add

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.

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.

Common strategies involving the additional rules of inference

◦ 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.

◦ 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.

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.

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

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

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

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

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

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

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.

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.

Tautology (Taut): p :: ( p v p ) p :: ( p p )

 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…

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

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

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”)

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

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

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

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

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

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.

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”)

 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.

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.

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