CS250: Discrete Math for Computer Science. L6: CNF and Natural Deduction for PropCalc
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1 CS250: Discrete Math for Computer Science L6: CNF and Natural Deduction for PropCalc
2 How to Simplify a PropCalc Formula: (p q) ((q r) p)
3 How to Simplify a PropCalc Formula: 1. Get rid of s using def. of implication. (p q) ((q r) p) ( p q) (( q r) p) ( p q) (( q r) p)
4 How to Simplify a PropCalc Formula: 1. Get rid of s using def. of implication. 2. Push s all the way inside using de Morgan. Now its in Negation Normal Form (NNF). (p q) ((q r) p) ( p q) (( q r) p) ( p q) (( q r) p) NNF (p q) (( q r) p)
5 How to Simplify a PropCalc Formula: 1. Get rid of s using def. of implication. 2. Push s all the way inside using de Morgan. Now its in Negation Normal Form (NNF). 3. Use distributive, associative, commutative to put in either Disjunctive Normal Form (DNF: s of s of literals) A literal is a prop variable or its negation, e.g., p, q. (p q) ((q r) p) ( p q) (( q r) p) ( p q) (( q r) p) NNF (p q) (( q r) p) NNF (p q) (( q p) (r p)) DNF (p q) (p r)
6 How to Simplify a PropCalc Formula: 1. Get rid of s using def. of implication. 2. Push s all the way inside using de Morgan. Now its in Negation Normal Form (NNF). 3. Use distributive, associative, commutative to put in either Disjunctive Normal Form (DNF: s of s of literals) Conjunctive Normal Form (CNF: s of s of literals) A literal is a prop variable or its negation, e.g., p, q. (p q) ((q r) p) ( p q) (( q r) p) ( p q) (( q r) p) NNF (p q) (( q r) p) NNF (p q) (( q p) (r p)) DNF (p q) (p r) CNF p ( q r)
7 iclicker 6.1 Which of the following formulas is in CNF (conjunction of disjunctions of literals)? A: (p q r) ( p q s) (q r s) B: ( p q r) (p q s) ( q r s) C: (p q) (( q p) (r p)) D: ( p q) (( q r) p)
8 Natural Deduction R6: Our PropCalc proof rules are slightly different from Epp s proof rules. F introduction p q p q p q p q p q p q p q p p F p p elimination p q p p q q p q p r q r r p q p p q q q p p F p p F p p p
9 Natural Deduction rule: -introduction 1 p 2 3 q p q p q 4 p q -i, 1 3
10 Natural Deduction rule: -introduction 1 p 2 3 q p q p q 4 p q -i, 1 3 Notation: p q (p proves q): From assumption p, can prove q.
11 Natural Deduction rule: -introduction 1 p 2 3 q p q p q 4 p q -i, 1 3 Notation: p q (p proves q): From assumption p, can prove q. Proposition: -i is sound, i.e., if from assumption p we can prove q, then every world satisfies p q.
12 Natural Deduction rule: -introduction 1 p 2 3 q p q p q 4 p q -i, 1 3 Notation: p q (p proves q): From assumption p, can prove q. Proposition: -i is sound, i.e., if from assumption p we can prove q, then every world satisfies p q. Proof. Since p q, and by the soundness of the proof rules so far, we know that every world that satisfies p must also satisfy q. Thus every world satisfies p q.
13 Natural Deduction rule: -introduction 1 p 2 3 q p q p q 4 p q -i, 1 3 Notation: p q (p proves q): From assumption p, can prove q. Proposition: -i is sound, i.e., if from assumption p we can prove q, then every world satisfies p q. Proof. Since p q, and by the soundness of the proof rules so far, we know that every world that satisfies p must also satisfy q. Thus every world satisfies p q. More about this once we have studied inductive proofs.
14 Example use of -introduction 1 p 2 r p -i, 1 3 p (r p) -i, 1 2
15 Example use of -introduction 1 p 2 r p -i, 1 3 p (r p) -i, 1 2 Thus, p (r p)
16 Natural Deduction rule: F-e Proof by Contradiction 1 p 2 3 F p F p p F p 4 p F-e, 1 3
17 Natural Deduction rule: F-e Proof by Contradiction 1 p 2 3 F p F p p F p 4 p F-e, 1 3 Proposition: F-e is sound, i.e., if p F then every world satisfies p.
18 Natural Deduction rule: F-e Proof by Contradiction 1 p 2 3 F p F p p F p 4 p F-e, 1 3 Proposition: F-e is sound, i.e., if p F then every world satisfies p. Proof. Since p F, by the soundness of the proof rules so far, we know that every world that satisfies p must also satisfy F. But no world satisfies F. Thus every world satisfies p.
19 Natural Deduction rule: -e 1 p q 2 p 3 p q p r q r r 4 r 5 q 6 7 r 8 r -e, 1, 2 4, 3 5
20 Natural Deduction rule: -e 1 p q 2 p 3 p q p r q r r Proposition: -e is sound. 4 r 5 q 6 7 r 8 r -e, 1, 2 4, 3 5
21 Natural Deduction rule: -e 1 p q 2 p 3 4 r 5 q 6 7 r p q p r q r r Proposition: -e is sound. Proof. Since p r and q r, every world that satisfies p or satisfies q satisfies r. Thus every world that satisfes p q satisfies r. 8 r -e, 1, 2 4, 3 5
22 1 p q 2 p q 3 p 4 p, 2 5 F F-i, 3, 4 6 q 7 q -e, 2 8 F, 6, 7 9 F -e, 1, 3 5, (p q) F-e, 2 9
23 1 p q 2 p q 3 p 4 p, 2 5 F F-i, 3, 4 iclicker 6.2 What is the justification for line 4? A: -i B: -e C: -i D: -e 6 q 7 q -e, 2 8 F, 6, 7 9 F -e, 1, 3 5, (p q) F-e, 2 9
24 1 p q 2 p q 3 p 4 p, 2 5 F F-i, 3, 4 6 q 7 q -e, 2 8 F, 6, 7 9 F -e, 1, 3 5, (p q) F-e, 2 9 iclicker 6.2 What is the justification for line 4? A: -i B: -e C: -i D: -e iclicker 6.3 What is the justification for line 8? A: -i B: -e C: F-i D: F-e
25 Natural Deduction R6: Our PropCalc proof rules are slightly different from Epp s proof rules. F introduction p q p q p q p q p q p q p q p p F p p elimination p q p p q q p q p r q r r p q p p q q q p p F p p F p p p
26 Propositional Equivalence PropCalc formulas p and q are equivalent (p q) iff they agree on every row of their truth tables.
27 Propositional Equivalence PropCalc formulas p and q are equivalent (p q) iff they agree on every row of their truth tables. Observation: p q iff p q is a tautology.
28 Propositional Equivalence PropCalc formulas p and q are equivalent (p q) iff they agree on every row of their truth tables. Observation: p q iff p q is a tautology. Some Important Equivalences (worth memorizing):
29 Propositional Equivalence PropCalc formulas p and q are equivalent (p q) iff they agree on every row of their truth tables. Observation: p q iff p q is a tautology. Some Important Equivalences (worth memorizing): double negation p p
30 Propositional Equivalence PropCalc formulas p and q are equivalent (p q) iff they agree on every row of their truth tables. Observation: p q iff p q is a tautology. Some Important Equivalences (worth memorizing): double negation p p de Morgan (p q) p q de Morgan (p q) p q
31 Propositional Equivalence PropCalc formulas p and q are equivalent (p q) iff they agree on every row of their truth tables. Observation: p q iff p q is a tautology. Some Important Equivalences (worth memorizing): double negation p p de Morgan (p q) p q de Morgan (p q) p q def. of implication p q p q
32 Propositional Equivalence PropCalc formulas p and q are equivalent (p q) iff they agree on every row of their truth tables. Observation: p q iff p q is a tautology. Some Important Equivalences (worth memorizing): double negation p p de Morgan (p q) p q de Morgan (p q) p q def. of implication p q p q contrapositive p q q p
33 Propositional Equivalence PropCalc formulas p and q are equivalent (p q) iff they agree on every row of their truth tables. Observation: p q iff p q is a tautology. Some Important Equivalences (worth memorizing): double negation p p de Morgan (p q) p q de Morgan (p q) p q def. of implication p q p q contrapositive p q q p def. of iff p q (p q) (q p)
34 More Important Equivalences (worth memorizing): commutative p q q p commutative p q q p
35 More Important Equivalences (worth memorizing): commutative p q q p commutative p q q p associative p (q r) (p q) r associative p (q r) (p q) r
36 More Important Equivalences (worth memorizing): commutative p q q p commutative p q q p associative p (q r) (p q) r associative p (q r) (p q) r distributive p (q r) (p q) (p r) distributive p (q r) (p q) (p r)
37 More Important Equivalences (worth memorizing): commutative p q q p commutative p q q p associative p (q r) (p q) r associative p (q r) (p q) r distributive p (q r) (p q) (p r) distributive p (q r) (p q) (p r) excluded middle p p T
38 SAT is an important class. If formula a has n PVARs, p 1,..., p n, it would seem to require about 2 n time in the worst case to test if a SAT.
39 SAT is an important class. If formula a has n PVARs, p 1,..., p n, it would seem to require about 2 n time in the worst case to test if a SAT. But, if we are given a satisfying world W for a, we can check immediately that W = a.
40 SAT is an important class. If formula a has n PVARs, p 1,..., p n, it would seem to require about 2 n time in the worst case to test if a SAT. But, if we are given a satisfying world W for a, we can check immediately that W = a. This sort of search problem: exponentially many possibilities, each one easy to verify, corresponds to Nondeterministic Polynomial Time (NP).
41 SAT is an important class. If formula a has n PVARs, p 1,..., p n, it would seem to require about 2 n time in the worst case to test if a SAT. But, if we are given a satisfying world W for a, we can check immediately that W = a. This sort of search problem: exponentially many possibilities, each one easy to verify, corresponds to Nondeterministic Polynomial Time (NP). In fact, SAT is a hardest such problem: NP complete. You will learn much more about this in CMPSCI 311.
42 Knights and Knaves [Smullyan, What Is the Name of This Book?] Knights always truthful; Knaves always lie; A, B {Kt,Kv} B : A&B opposite types
43 Knights and Knaves [Smullyan, What Is the Name of This Book?] Knights always truthful; Knaves always lie; A, B {Kt,Kv} A : B is Kt B : A&B opposite types
44 Knights and Knaves [Smullyan, What Is the Name of This Book?] Knights always truthful; Knaves always lie; A, B {Kt,Kv} T 1 A : B is Kt B : A&B opposite types def = B is a Kt T 2 def = A&B opposite types
45 Knights and Knaves [Smullyan, What Is the Name of This Book?] Knights always truthful; Knaves always lie; A, B {Kt,Kv} S 1 T 1 def def = A : B is Kt S 2 = B : A&B opposite types def = B is a Kt T 2 def = A&B opposite types
46 Knights and Knaves [Smullyan, What Is the Name of This Book?] Knights always truthful; Knaves always lie; A, B {Kt,Kv} S 1 T 1 def def = A : B is Kt S 2 = B : A&B opposite types def def = B is a Kt T 2 = A&B opposite types S 1 = T 1 A is Kt S 2 = T 2 B is Kt
47 Knights and Knaves [Smullyan, What Is the Name of This Book?] Knights always truthful; Knaves always lie; A, B {Kt,Kv} S 1 T 1 def def = A : B is Kt S 2 = B : A&B opposite types def def = B is a Kt T 2 = A&B opposite types S 1 = T 1 A is Kt S 2 = T 2 B is Kt W A is Kt B is Kt T 1 T 2 T 1 A is Kt T 2 B is Kt W W W W
48 Knights and Knaves [Smullyan, What Is the Name of This Book?] Knights always truthful; Knaves always lie; A, B {Kt,Kv} S 1 T 1 def def = A : B is Kt S 2 = B : A&B opposite types def def = B is a Kt T 2 = A&B opposite types S 1 = T 1 A is Kt S 2 = T 2 B is Kt W A is Kt B is Kt T 1 T 2 T 1 A is Kt T 2 B is Kt W W W W W 0 is only world satisfying S 1 S 2.
49 Knights and Knaves [Smullyan, What Is the Name of This Book?] Knights always truthful; Knaves always lie; A, B {Kt,Kv} S 1 T 1 def def = A : B is Kt S 2 = B : A&B opposite types def def = B is a Kt T 2 = A&B opposite types S 1 = T 1 A is Kt S 2 = T 2 B is Kt W A is Kt B is Kt T 1 T 2 T 1 A is Kt T 2 B is Kt W W W W W 0 is only world satisfying S 1 S 2. Thus A and B are both Knaves.
50 R6 Quiz Answers: Match the Epp Proof Rules to the equivalent Natural Deduction Rules. 1. Modus Ponens: -e 2. Modus Tollens: -e 3. Generalization: -i 4. Specialization: -e 5. Conjunction: -i 6, 7. In the following proof, identify the natural deduction rules used in lines 2 and 3. 1 p q 2 q -e, 1 3 ( r q) -i, 2
51 8. Is the following a sound proof rule? p q p In answering this, you may consider the worlds shown in this truth table: W p q W W W W Not valid: reasoning from the converse fails in world W In Smullyan s Island of Knights and Knaves, two natives C and D approach you but only C speaks. C says: Both of us are knaves. What are C and D? C is a Knave and D is a Knight. 10. In Smullyan s Island of Knights and Knaves, you encounter natives E and F. E says: F is a knave. F says: E is a knave. How many knaves are there? 1 q
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