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Proposi'onal Logic Proofs Proof Method #1: Truth Table Example

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Proposi'onal Logic Proofs Proof Method #1: Truth Table Example 9/9/14 Proposi'onal Logic Proofs • An argument is a sequence of proposi'ons: Inference Rules ² Premises (Axioms) are the first n proposi'ons (Rosen, Section 1.5) ² Conclusion is the final proposi'on. p p ... p q TOPICS • An argument is valid if is a ( 1 ∧ 2 ∧ ∧ n ) → tautology, given that pi are the premises • Logic Proofs (axioms) and q is the conclusion. ² via Truth Tables € ² via Inference Rules 2 Proof Method #1: Truth Table Example Proof By Truth Table s = ((p v q) ∧ (¬p v r)) → (q v r) n If the conclusion is true in the truth table whenever the premises are true, it is p q r ¬p p v q ¬p v r q v r (p v q)∧ (¬p v r) s proved 0 0 0 1 0 1 0 0 1 n Warning: when the premises are false, the 0 0 1 1 0 1 1 0 1 conclusion my be true or false 0 1 0 1 1 1 1 1 1 n Problem: given n propositions, the truth 0 1 1 1 1 1 1 1 1 n table has 2 rows 1 0 0 0 1 0 0 0 1 n Proof by truth table quickly becomes 1 0 1 0 1 1 1 1 1 infeasible 1 1 0 0 1 0 1 0 1 1 1 1 0 1 1 1 1 1 3 4 1 9/9/14 Proof Method #2: Rules of Inference Inference proper'es n A rule of inference is a pre-proved relaon: n Inference rules are truth preserving any 'me the leJ hand side (LHS) is true, the n If the LHS is true, so is the RHS right hand side (RHS) is also true. n Applied to true statements n Axioms or statements proved from axioms n Therefore, if we can match a premise to the n Inference is syntac'c LHS (By suBs'tu'ng proposi'ons), we can n SuBs'tute proposi'ons assert the (suBs'tuted) RHS n if p replaces q once, it replaces q everywhere n If p replaces q, it only replaces q n Apply rule 5 6 Example Rule of Inference Rules of Inference Modus Ponens p p → q (p∧(p → q)) → q ∴ q p q p → q p∧(p → q) (p∧(p → q)) → q € 0 0 1 0 1 0 1 1 € 0 1 € € € 1 0€ 0 € 0 1 1 1 1 1 1 7 8 2 9/9/14 Logical Equivalences Modus Ponens n If p, and p implies q, then q Example: p = it is sunny, q = it is hot p → q, it is hot whenever it is sunny “Given the above, if it is sunny, it must be hot”. 9 10 Modus Tollens Hypothe'cal Syllogism n If not q and p implies q, then not p n If p implies q, and q implies r, then Example: p implies r p = it is sunny, q = it is hot Example: p → q, it is hot whenever it is sunny p = it is sunny, q = it is hot, r = it is dry “Given the above, if it is not hot, it p → q, it is hot when it is sunny cannot be sunny.” q → r, it is dry when it is hot “Given the above, it must be dry when it is sunny” 11 12 3 9/9/14 Disjuncve Syllogism Resolu'on n If p or q, and not p, then q n If p or q, and not p or r, then q or r Example: Example: p = it is sunny, q = it is hot p = it is sunny, q = it is hot, r = it is dry p ∨ q, it is hot or sunny p ∨ q, it is sunny or hot “Given the above, if it not sunny, but it ¬p ∨ r, it is not hot or dry is hot or sunny, then it is hot” “Given the above, if it is sunny or hot, but not sunny or dry, it must be hot or dry” Not obvious! 13 14 Addi'on Simplificaon n If p then p or q n If p and q, then p Example: Example: p = it is sunny, q = it is hot p = it is sunny, q = it is hot p ∨ q, it is hot or sunny p ∧ q, it is hot and sunny “Given the above, if it is sunny, it must “Given the above, if it is hot and sunny, be hot or sunny” it must be hot” Of course! Of course! 15 16 4 9/9/14 A Simple Proof Conjuncon Given X, X→Y, Y →Z, ¬Z ∨ W, prove W n If p and q, then p and q Example: Step Reason p = it is sunny, q = it is hot 1. x → y Premise 2. y → z Premise p ∧ q, it is hot and sunny 3. x → z Hypothetical Syllogism (1, 2) “Given the above, if it is sunny and it is 4. x Premise hot, it must be hot and sunny” 5. z Modus Ponens (3, 4) Of course! 6. ¬z∨w Premise 7. w Disjunctive Syllogism (5, 6) 17 18 A Simple Proof Setup the proof “In order to sign up for CS161, I must complete STEP 2) Extract axioms and conclusion. CS160 and either M155 or M160. I have not completed M155 But I have completed CS161. n Axioms: Prove that I have completed M160.” n A → B ∧ (C ∨ D) STEP 1) Assign proposi'ons to each statement. n A n A : CS161 n ¬C n B : CS160 n Conclusion: n C : M155 n D n D : M160 19 20 5 9/9/14 Now do the Proof Another Example STEP 3) Use inference rules to prove conclusion. Step Reason Given: Conclude: 1. A → B ∧ (C ∨ D) Premise p → q ¬q → s 2. A Premise ¬p → r 3. B ∧ (C ∨ D) Modus Ponens (1, 2) r → s 4. C ∨ D Simplification 5. ¬C Premise 6. D Disjunctive Syllogism (4, 5) 21 22 Proof using Rules of Inference and Proof of Another Example Logical Equivalences Step Reason Prove: ¬(p∨(¬p∧q)) ≡ (¬p∧¬q) 1. p → q Premise ¬(p∨(¬p∧q)) ≡ ¬p ∧ ¬(¬p∧q) n By 2nd DeMorgan’s 2. ¬q → ¬p Implication law (1) ≡ ¬p ∧(¬(¬p)∨¬q) n By 1st DeMorgan’s ≡ ¬p∧(p∨¬q) n By double negation 3. p r Premise ¬ → ≡ (¬p∧p) ∨ (¬p∧¬q) n By 2nd distributive 4. ¬q → r Hypothetical syllogism (2, 3) ≡ F ∨ (¬p∧¬q) n By definition of ∧ ≡ (¬p∧¬q) ∨ F n By commutative law 5. Premise r → s ≡ (¬p∧¬q) n By definition of ∨ 6. ¬q → s Hypothetical syllogism (4, 5) 23 24 6 9/9/14 Example of a Fallacy Example of a fallacy q n If q, and p implies q, then p (q∧(p → q)) → p p → q Example: ∴ p p = it is sunny, q = it is hot p q p → q q∧(p → q) (q∧(p → q)) → p p → q, if it is sunny, then it is hot 0 0 1 0 1 “Given the above, just because it is 0 1 1 1 0 hot, does NOT necessarily mean it is € € € 1 0 0 0 1 sunny. 1 1 1 1 1 This is not a tautology, therefore the argument is not valid 25 26 7 .
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