PHIL 12A Reductio Ad Absurdum: Philosophy problems

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PHIL 12A – Spring 2019 Problem Set 7 65 points Reductio Ad Absurdum 1. (10 points) Recall that π is an irrational number. Outline a natural deduction proof that formalizes the reductio step in the following proof. Theorem 1. There is some digit among 1, 2, . . . , 9 that occurs infinitely many times in the decimal expansion of π. Proof. Suppose it is not the case that there is some digit among 1, 2, . . . , 9 that occurs infinitely many times in the decimal expansion of π. Hence each occurs only finitely many times. Then since there are only finitely many digits among 1, 2, . . . , 9, and each occurs only finitely many times in the decimal expansion of π, it follows that the decimal expansion of π is a finite sequence of those digits (followed by only 0’s). This contradicts the fact that π is irrational. Thus, we conclude that there is some digit among 1, 2, . . . , 9 that occurs infinitely many times in the decimal expansion of π. However, as of 2019, no one knows which digit(s) occur infinitely many times in the decimal expansion of π. 2. (10 points) Give natural deduction proofs of the following formulas from the given assumptions: (a) p from assumption ¬(p → q); (b) ¬¬p → p from no assumptions. 3. (5 points) Extra credit. Give a natural deduction proof of ¬p ↔ q from assumption ¬(p ↔ q). 4. (10 points) Prove that any application of EFQ can be replaced by applications of Reiteration and RAA. Thus, EFQ becomes redundant when we add RAA. Disjunction Introduction 5. (10 points) Give natural deduction proofs of the following formulas from the given assumptions: (a) ¬p ∧ ¬q from assumption ¬(p ∨ q); (b) ¬p ∨ ¬q from assumption ¬(p ∧ q) (hint: use RAA). 1 Disjunction Elimination 6. (10 points) A prime number is a positive integer greater than 1 that is divisible without remainder only by itself and 1. A set is infinite if and only if it is not finite. Outline a natural deduction proof that formalizes the negation introduction step and the proof by cases in the proof of Theorem 3 below. Lemma 1. Let a, b, and c be integers. If a divides b without remainder and a divides c without remain, then a divides (b − c) without remainder. Theorem 2 (Fundamental Theorem of Arithmetic). Every integer greater than 1 is a prime number or a product of prime numbers.1 Theorem 3 (Euclid’s Theorem). The set of prime numbers is infinite. Proof. Suppose that the set of prime numbers is finite. List all the primes as p1 , . . . , pn and let p = p1 × · · · × pn q = p + 1. Either q is prime or it is not prime, so we can reason by cases. Case 1: suppose q is prime. Then since q is greater than each of p1 , . . . , pn , we have a contradiction with our assumption that p1 , . . . , pn are all the primes. Case 2: suppose q is not prime. Then by Theorem 2, q is divisible without remainder by some prime—hence by some pi , as p1 , . . . , pn are all the primes. Since pi divides q without remainder and pi clearly divides p without remainder, it follows by Lemma 1 that pi divides q − p = 1 without remainder. But since pi is prime, it is greater than 1, so it does not divide 1 without remainder. This is a contradiction. In either case, we obtained a contradiction, so the set of prime numbers is infinite. 7. (15 points) Give natural deduction proofs of the following formulas: (a) p ∨ q from assumption q ∨ p; (b) r from assumptions p ∨ q, p → r, and q → r; (c) q from assumptions p ∨ q and ¬p. 1 Moreover, its representation as a product of prime numbers is unique modulo the order of the factors. 2 Natural Deduction X: Disjunction Introduction PHIL 12A – Introduction to Logic with Wes Holliday University of California, Berkeley Proving a Disjunction One way to prove a disjunction j _ y is to prove j. Another way to prove a disjunction j _ y is to prove y. Proving a Disjunction An integer n is even if n = 2k for some integer k. An integer n is odd if n = 2k + 1 for some integer k. Proving a Disjunction An integer n is even if n = 2k for some integer k. An integer n is odd if n = 2k + 1 for some integer k. Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proving a Disjunction An integer n is even if n = 2k for some integer k. An integer n is odd if n = 2k + 1 for some integer k. Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. Proving a Disjunction An integer n is even if n = 2k for some integer k. An integer n is odd if n = 2k + 1 for some integer k. Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. Proving a Disjunction An integer n is even if n = 2k for some integer k. An integer n is odd if n = 2k + 1 for some integer k. Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. For the inductive step, . . . to be continued . . . Proving a Disjunction An integer n is even if n = 2k for some integer k. An integer n is odd if n = 2k + 1 for some integer k. Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. For the inductive step, . . . to be continued . . . We can represent the logical .. . move in the base case as follows: .. . i 0 is even i +1 0 is even _ 0 is odd _I, i Disjunction Introduction _ Introduction states that if you have a correct partial proof in which j appears or y appears, then you may correctly add j _ y on a new line in the same column (of the same subproof): .. . .. . .. . .. . i .. . j .. . i .. . y .. . j j_y j j_y _I, i _I, i Proving Disjunctions Nonconstructively Proving a disjunct is not the only way to prove a disjunction. Proving Disjunctions Nonconstructively Proving a disjunct is not the only way to prove a disjunction. Reductio Ad Absurdum allows one to prove a disjunction by assuming its negation and then deriving a contradiction. Proving Disjunctions Nonconstructively Proving a disjunct is not the only way to prove a disjunction. Reductio Ad Absurdum allows one to prove a disjunction by assuming its negation and then deriving a contradiction. In this way, one can prove disjunctions without have any idea which disjunct holds—just as RAA lets one prove that there are irrational a, b such that ab is rational without proving that any particular a, b work. Ex. 1 n p_ p Ex. 1 n´1 n (p _ p ) j^ p_ p j RAA, 1–n ´ 1 Ex. 1 (p _ p ) n´1 p^ n p_ p p RAA, 1–n ´ 1 Ex. 1 (p _ p ) i p n´2 p n´1 p^ n p_ p p ^I, i, n ´ 2 RAA, 1–n ´ 1 Ex. 1 (p _ p ) 2 p i ´1 j^ i I, 2–i ´ 1 p n´2 p n´1 p^ n j p_ p p ^I, i, n ´ 2 RAA, 1–n ´ 1 Ex. 1 (p _ p ) 2 p 3 p_ p i ´1 j^ i j I, 2–i ´ 1 p n´2 p n´1 p^ n p_ p _I, 2 p ^I, i, n ´ 2 RAA, 1–n ´ 1 Ex. 1 (p _ p ) 2 p 3 p_ p (p _ p ) 4 (p _ p ) ^ (p _ p ) 5 6 p p n´1 p^ p_ p R, 1 ^I, 3, 4 I, 2–5 n´2 n _I, 2 p ^I, 6, n ´ 2 RAA, 1–n ´ 1 Ex. 1 (p _ p ) 2 p 3 p_ p (p _ p ) 4 (p _ p ) ^ (p _ p ) 5 6 p R, 1 ^I, 3, 4 I, 2–5 7 p 8 analogous to 3–5 9 p 10 p^ 11 _I, 2 p_ p I, 7–8 p ^I, 6, 9 RAA, 1–10 Disjunction Property If we drop RAA from our proof system, then the resulting proof system has the DISJUNCTION PROPERTY: Disjunction Property If we drop RAA from our proof system, then the resulting proof system has the DISJUNCTION PROPERTY: § if there is a proof of j _ y (from no assumptions), then there is a proof of j or there is a proof of y. Disjunction Property If we drop RAA from our proof system, then the resulting proof system has the DISJUNCTION PROPERTY: § if there is a proof of j _ y (from no assumptions), then there is a proof of j or there is a proof of y. We lose this property when we add RAA. Disjunction Property If we drop RAA from our proof system, then the resulting proof system has the DISJUNCTION PROPERTY: § if there is a proof of j _ y (from no assumptions), then there is a proof of j or there is a proof of y. We lose this property when we add RAA. As we’ve seen, p _ p is provable. Disjunction Property If we drop RAA from our proof system, then the resulting proof system has the DISJUNCTION PROPERTY: § if there is a proof of j _ y (from no assumptions), then there is a proof of j or there is a proof of y. We lose this property when we add RAA. As we’ve seen, p _ p is provable. But neither p nor p is provable. Disjunction Property If we drop RAA from our proof system, then the resulting proof system has the DISJUNCTION PROPERTY: § if there is a proof of j _ y (from no assumptions), then there is a proof of j or there is a proof of y. We lose this property when we add RAA. As we’ve seen, p _ p is provable. But neither p nor p is provable. How do we know that? Disjunction Property If we drop RAA from our proof system, then the resulting proof system has the DISJUNCTION PROPERTY: § if there is a proof of j _ y (from no assumptions), then there is a proof of j or there is a proof of y. We lose this property when we add RAA. As we’ve seen, p _ p is provable. But neither p nor p is provable. How do we know that? Because our system is SOUND in the sense that if there is a proof of j, then j is semantically valid; Disjunction Property If we drop RAA from our proof system, then the resulting proof system has the DISJUNCTION PROPERTY: § if there is a proof of j _ y (from no assumptions), then there is a proof of j or there is a proof of y. We lose this property when we add RAA. As we’ve seen, p _ p is provable. But neither p nor p is provable. How do we know that? Because our system is SOUND in the sense that if there is a proof of j, then j is semantically valid; and neither p nor p is semantically valid (true under all valuations), so it follows that neither p nor p is provable. Next Up Next, we will conclude our series on natural deduction for propositional logic with an important rule: _ Elimination or “proof by cases.” Natural Deduction XI: Disjunction Elimination PHIL 12A – Introduction to Logic with Wes Holliday University of California, Berkeley Using Disjunctions Suppose you’ve established a disjunction a _ b at some stage of a proof. Using Disjunctions Suppose you’ve established a disjunction a _ b at some stage of a proof. How can you use this to prove some further j? Using Disjunctions Suppose you’ve established a disjunction a _ b at some stage of a proof. How can you use this to prove some further j? Show that if a holds, then j holds, and also if b holds, then again j holds. Using Disjunctions Suppose you’ve established a disjunction a _ b at some stage of a proof. How can you use this to prove some further j? Show that if a holds, then j holds, and also if b holds, then again j holds. Since in either case, j holds, you can conclude that j holds. Using Disjunctions Suppose you’ve established a disjunction a _ b at some stage of a proof. How can you use this to prove some further j? Show that if a holds, then j holds, and also if b holds, then again j holds. Since in either case, j holds, you can conclude that j holds. This important method of proof is called “proof by cases.” Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd. Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd. Case 1: n is even. Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd. Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd. Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the remainder when 4k 2 is divided by 4 is 0. Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd. Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the remainder when 4k 2 is divided by 4 is 0. Case 2: n is odd. Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd. Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the remainder when 4k 2 is divided by 4 is 0. Case 2: n is odd. Then n = 2k + 1 for some k, so n2 = (2k + 1)2 = 4k 2 + 4k + 1, Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd. Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the remainder when 4k 2 is divided by 4 is 0. Case 2: n is odd. Then n = 2k + 1 for some k, so n2 = (2k + 1)2 = 4k 2 + 4k + 1, and the remainder when 4k 2 + 4k + 1 is divided by 4 is 1. Using and Proving Disjunctions Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is divided by 4 is 0 or 1. Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd. Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the remainder when 4k 2 is divided by 4 is 0. Case 2: n is odd. Then n = 2k + 1 for some k, so n2 = (2k + 1)2 = 4k 2 + 4k + 1, and the remainder when 4k 2 + 4k + 1 is divided by 4 is 1. Since in either case, the remainder when n2 is divided by 4 is 0 or 1, we are done. We can represent the preceding proof as follows: 1 n is even _ n is odd 2 .. . n is even .. . i remainder of n2 /4 is 0 i +1 remainder of n2 /4 is 0 _ remainder of n2 /4 is 1 j .. . n is odd .. . k remainder of n2 /4 is 1 k +1 remainder of n2 /4 is 0 _ remainder of n2 /4 is 1 k +2 remainder of n2 /4 is 0 _ remainder of n2 /4 is 1 _I, i _I, k _E, 1, 2–i + 1, j–k + 1 _ Elimination _ Elimination states that if you have a correct partial proof containing a _ b, followed by a subproof of j from a, followed by a subproof of j from b, then you can correctly add j on the next line in the same column as a _ b: i a_b i +1 .. . a .. . j j j +1 .. . b .. . k j n j _E, i, i + 1–j, j + 1–k Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. For the inductive step, suppose that n is even or odd. Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. For the inductive step, suppose that n is even or odd. Case 1: n is even. Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. For the inductive step, suppose that n is even or odd. Case 1: n is even. Then n = 2k for some k, so n + 1 = 2k + 1 and hence n + 1 is odd, so n + 1 has the property. Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. For the inductive step, suppose that n is even or odd. Case 1: n is even. Then n = 2k for some k, so n + 1 = 2k + 1 and hence n + 1 is odd, so n + 1 has the property. Case 2: n is odd. Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. For the inductive step, suppose that n is even or odd. Case 1: n is even. Then n = 2k for some k, so n + 1 = 2k + 1 and hence n + 1 is odd, so n + 1 has the property. Case 2: n is odd. Then n = 2k + 1 for some k, so n + 1 = 2k + 2 = 2(k + 1) and hence n + 1 is even, so n + 1 has the property. Using and Proving Disjunctions Lemma 1. Let n be a nonnegative integer. Then n is even or odd. Proof. We prove by induction that every n has the property that it is even or odd. For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property. For the inductive step, suppose that n is even or odd. Case 1: n is even. Then n = 2k for some k, so n + 1 = 2k + 1 and hence n + 1 is odd, so n + 1 has the property. Case 2: n is odd. Then n = 2k + 1 for some k, so n + 1 = 2k + 2 = 2(k + 1) and hence n + 1 is even, so n + 1 has the property. We conclude that n + 1 has the property, which completes the proof. We can represent the inductive step of the preceding proof as follows: 1 n is even _ n is odd 2 .. . n is even .. . i n + 1 is odd i +1 n + 1 is even _ n + 1 is odd j .. . n is odd .. . k n + 1 is even k +1 n + 1 is even _ n + 1 is odd k +2 k +3 n + 1 is even _ n + 1 is odd (n is even _ n is odd) Ñ (n + 1 is even _ n + 1 is odd) _I, i _I, k _E, 1, 2–i + 1, j–k + 1 ÑI, 1–k + 2 Ex. 1 n pÑq p_q Ex. 1 .. . .. . pÑq 2 p_ p n p_q Ex. 1 .. . 2 .. . pÑq p_ p 3 .. . p .. . i p_q i +1 .. . .. . n´1 p_q n p p_q _E, 2, 3–i, i + 1–n ´ 1 Ex. 1 .. . 2 .. . pÑq p_ p 3 p 4 p_q 5 .. . p .. . n´1 p_q n p_q _I, 3 _E, 2, 3–4, 5–n ´ 1 Ex. 1 .. . 2 .. . pÑq p_ p 3 p 4 p_q 5 p 6 .. . .. . pÑq n´1 p_q n p_q _I, 3 R, 1 _E, 2, 3–4, 5–n ´ 1 Ex. 1 .. . 2 .. . pÑq p_ p 3 p 4 p_q 5 p 6 pÑq _I, 3 R, 1 7 q ÑE, 5–6 8 p_q _I, 7 9 p_q _E, 2, 3–4, 5–8 Putting it All Together We now have a formal proof system for our full language of proposition logic: § § § § § § § Reiteration; Ñ Introduction, Ñ Elimination; ^ Introduction, ^ Elimination; Ø Introduction, Ø Introduction; Introduction, Elimination; Reductio Ad Absurdum; _ Introduction, _ Elimination. Putting it All Together We now have a formal proof system for our full language of proposition logic: § § § § § § § Reiteration; Ñ Introduction, Ñ Elimination; ^ Introduction, ^ Elimination; Ø Introduction, Ø Introduction; Introduction, Elimination; Reductio Ad Absurdum; _ Introduction, _ Elimination. It is (classically) SOUND and COMPLETE: y is a semantic consequence of j1 , . . . , jn if and only if there is a proof of y from assumptions j1 , . . . , jn . Computer-Checkable Proofs It is possible to write a computer algorithm that will check, given any purported natural deduction proof, whether it is in fact a correct natural deduction proof. Computer-Checkable Proofs It is possible to write a computer algorithm that will check, given any purported natural deduction proof, whether it is in fact a correct natural deduction proof. The algorithm simply checks whether each rule has been correctly applied. Computer-Checkable Proofs It is possible to write a computer algorithm that will check, given any purported natural deduction proof, whether it is in fact a correct natural deduction proof. The algorithm simply checks whether each rule has been correctly applied. This is an essential property of a formal proof system: whether a purported proof is in fact a correct proof must be computer-checkable. Syllogistic Logic PHIL 12A – Introduction to Logic with Wes Holliday University of California, Berkeley Goal of this Part This part serves as a transition from our study of propositional logic to our study of predicate logic (first-order logic) in the rest of the course. Goal of this Part This part serves as a transition from our study of propositional logic to our study of predicate logic (first-order logic) in the rest of the course. We’ll discuss one of the earliest topics of study in logic: syllogisms. A Valid Syllogism Here is an instance of a valid form of syllogism: A Valid Syllogism Here is an instance of a valid form of syllogism: § All sophomores are students. A Valid Syllogism Here is an instance of a valid form of syllogism: § § All sophomores are students. All students are invited to the game. A Valid Syllogism Here is an instance of a valid form of syllogism: § § § All sophomores are students. All students are invited to the game. All sophomores are invited to the game. A Valid Syllogism Here is an instance of a valid form of syllogism: § § § All sophomores are students. All students are invited to the game. All sophomores are invited to the game. Abstracting from particular predicates (‘sophomore’, ‘student’, etc.), we have: A Valid Syllogism Here is an instance of a valid form of syllogism: § § § All sophomores are students. All students are invited to the game. All sophomores are invited to the game. Abstracting from particular predicates (‘sophomore’, ‘student’, etc.), we have: § All A are B. A Valid Syllogism Here is an instance of a valid form of syllogism: § § § All sophomores are students. All students are invited to the game. All sophomores are invited to the game. Abstracting from particular predicates (‘sophomore’, ‘student’, etc.), we have: § § All A are B. All B are C . A Valid Syllogism Here is an instance of a valid form of syllogism: § § § All sophomores are students. All students are invited to the game. All sophomores are invited to the game. Abstracting from particular predicates (‘sophomore’, ‘student’, etc.), we have: § § § All A are B. All B are C . All A are C . Another Valid Syllogism Here is another instance of a valid form of syllogism: Another Valid Syllogism Here is another instance of a valid form of syllogism: § All those invited to the game are invited to the after party. Another Valid Syllogism Here is another instance of a valid form of syllogism: § § All those invited to the game are invited to the after party. No faculty are invited to the after party. Another Valid Syllogism Here is another instance of a valid form of syllogism: § § § All those invited to the game are invited to the after party. No faculty are invited to the after party. No faculty are invited to the game. Another Valid Syllogism Here is another instance of a valid form of syllogism: § § § All those invited to the game are invited to the after party. No faculty are invited to the after party. No faculty are invited to the game. Abstracting from particular predicates, we have: § All A are B. Another Valid Syllogism Here is another instance of a valid form of syllogism: § § § All those invited to the game are invited to the after party. No faculty are invited to the after party. No faculty are invited to the game. Abstracting from particular predicates, we have: § § All A are B. No C are B. Another Valid Syllogism Here is another instance of a valid form of syllogism: § § § All those invited to the game are invited to the after party. No faculty are invited to the after party. No faculty are invited to the game. Abstracting from particular predicates, we have: § § § All A are B. No C are B. No C are A. An Invalid Syllogism Here is an example of an invalid form of syllogism: An Invalid Syllogism Here is an example of an invalid form of syllogism: § All students are invited to the game. An Invalid Syllogism Here is an example of an invalid form of syllogism: § § All students are invited to the game. No faculty are students. An Invalid Syllogism Here is an example of an invalid form of syllogism: § § § All students are invited to the game. No faculty are students. No faculty are invited to the game. An Invalid Syllogism Here is an example of an invalid form of syllogism: § § § All students are invited to the game. No faculty are students. No faculty are invited to the game. Abstracting from particular predicates, we have: § All A are B. An Invalid Syllogism Here is an example of an invalid form of syllogism: § § § All students are invited to the game. No faculty are students. No faculty are invited to the game. Abstracting from particular predicates, we have: § § All A are B. No C are A. An Invalid Syllogism Here is an example of an invalid form of syllogism: § § § All students are invited to the game. No faculty are students. No faculty are invited to the game. Abstracting from particular predicates, we have: § § § All A are B. No C are A. No C are B. Formal Language Our first step in formalizing syllogistic logic is to define a formal language. Formal Language Our first step in formalizing syllogistic logic is to define a formal language. Given a set Pred of predicates, the formulas of the basic syllogistic language are the following for any A, B, C in Pred: Formal Language Our first step in formalizing syllogistic logic is to define a formal language. Given a set Pred of predicates, the formulas of the basic syllogistic language are the following for any A, B, C in Pred: § § § § All A are B; Some A are B; No A are B; Not all A are B. Toward a Formal Semantics Our second step in formalizing syllogistic logic is to define a formal semantics. Toward a Formal Semantics Our second step in formalizing syllogistic logic is to define a formal semantics. A predicate like ‘sophomore’ is not a proposition that has a truth value, so we won’t define our semantics using a simple truth valuation as before. Toward a Formal Semantics Our second step in formalizing syllogistic logic is to define a formal semantics. A predicate like ‘sophomore’ is not a proposition that has a truth value, so we won’t define our semantics using a simple truth valuation as before. Rather, a predicate like ‘sophomore’ has an extension: the set of things that are sophomores. Toward a Formal Semantics Our second step in formalizing syllogistic logic is to define a formal semantics. A predicate like ‘sophomore’ is not a proposition that has a truth value, so we won’t define our semantics using a simple truth valuation as before. Rather, a predicate like ‘sophomore’ has an extension: the set of things that are sophomores. To evaluate whether something like All A are B is true or false, we need to know the extension of A and the extension of B. Toward a Formal Semantics Our second step in formalizing syllogistic logic is to define a formal semantics. A predicate like ‘sophomore’ is not a proposition that has a truth value, so we won’t define our semantics using a simple truth valuation as before. Rather, a predicate like ‘sophomore’ has an extension: the set of things that are sophomores. To evaluate whether something like All A are B is true or false, we need to know the extension of A and the extension of B. This data will be provided by a model. . . Models for the Syllogistic Language A model M = (D, I ) for the syllogistic language consists of a nonempty set D and a function I that sends each predicate A to a subset I (A) of D. Models for the Syllogistic Language A model M = (D, I ) for the syllogistic language consists of a nonempty set D and a function I that sends each predicate A to a subset I (A) of D. We call I (A) the extension or interpretation of A in M. Models for the Syllogistic Language A model M = (D, I ) for the syllogistic language consists of a nonempty set D and a function I that sends each predicate A to a subset I (A) of D. We call I (A) the extension or interpretation of A in M. For example, here is a simple model for the syllogistic language with Pred = tsophomores, students, faculty, invitees to the gameu: Models for the Syllogistic Language A model M = (D, I ) for the syllogistic language consists of a nonempty set D and a function I that sends each predicate A to a subset I (A) of D. We call I (A) the extension or interpretation of A in M. For example, here is a simple model for the syllogistic language with Pred = tsophomores, students, faculty, invitees to the gameu: § D = t1, 2, 3, 4u; Models for the Syllogistic Language A model M = (D, I ) for the syllogistic language consists of a nonempty set D and a function I that sends each predicate A to a subset I (A) of D. We call I (A) the extension or interpretation of A in M. For example, here is a simple model for the syllogistic language with Pred = tsophomores, students, faculty, invitees to the gameu: § D = t1, 2, 3, 4u; § I (sophomores) = t1, 3u; Models for the Syllogistic Language A model M = (D, I ) for the syllogistic language consists of a nonempty set D and a function I that sends each predicate A to a subset I (A) of D. We call I (A) the extension or interpretation of A in M. For example, here is a simple model for the syllogistic language with Pred = tsophomores, students, faculty, invitees to the gameu: § D = t1, 2, 3, 4u; § I (sophomores) = t1, 3u; § I (students) = t1, 2, 3u; Models for the Syllogistic Language A model M = (D, I ) for the syllogistic language consists of a nonempty set D and a function I that sends each predicate A to a subset I (A) of D. We call I (A) the extension or interpretation of A in M. For example, here is a simple model for the syllogistic language with Pred = tsophomores, students, faculty, invitees to the gameu: § D = t1, 2, 3, 4u; § I (sophomores) = t1, 3u; § I (students) = t1, 2, 3u; § I (faculty) = t4u; Models for the Syllogistic Language A model M = (D, I ) for the syllogistic language consists of a nonempty set D and a function I that sends each predicate A to a subset I (A) of D. We call I (A) the extension or interpretation of A in M. For example, here is a simple model for the syllogistic language with Pred = tsophomores, students, faculty, invitees to the gameu: § D = t1, 2, 3, 4u; § I (sophomores) = t1, 3u; § I (students) = t1, 2, 3u; § I (faculty) = t4u; § I (invitees to the game) = t1, 2, 3, 4u. Visualizing a Model The model just considered can be visualized as follows: Language: sophomores students faculty invitees I I I 1 I 3 Model: 2 4 Truth in a Model Where ϕ is a formula of the syllogistic language and M is a model, we will define when ϕ is true in M: M ( ϕ. Truth in a Model Where ϕ is a formula of the syllogistic language and M is a model, we will define when ϕ is true in M: M ( ϕ. § M ( All A are B if and only if I (A) Ď I (B ); Truth in a Model Where ϕ is a formula of the syllogistic language and M is a model, we will define when ϕ is true in M: M ( ϕ. § M ( All A are B if and only if I (A) Ď I (B ); § M ( Some A are B if and only if I (A) X I (B ) ‰ ∅; Truth in a Model Where ϕ is a formula of the syllogistic language and M is a model, we will define when ϕ is true in M: M ( ϕ. § M ( All A are B if and only if I (A) Ď I (B ); § M ( Some A are B if and only if I (A) X I (B ) ‰ ∅; § M ( No A are B if and only if I (A) X I (B ) = ∅; Truth in a Model Where ϕ is a formula of the syllogistic language and M is a model, we will define when ϕ is true in M: M ( ϕ. § M ( All A are B if and only if I (A) Ď I (B ); § M ( Some A are B if and only if I (A) X I (B ) ‰ ∅; § M ( No A are B if and only if I (A) X I (B ) = ∅; § M ( Not all A are B if and only if I (A) Ę I (B ). Example Recall our model from before: § § § § § D = t1, 2, 3, 4u; I (sophomores) = t1, 3u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. Example Recall our model from before: § § § § § D = t1, 2, 3, 4u; I (sophomores) = t1, 3u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. For this model, we have, for example: Example Recall our model from before: § § § § § D = t1, 2, 3, 4u; I (sophomores) = t1, 3u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. For this model, we have, for example: § M ( All sophomores are students; Example Recall our model from before: § § § § § D = t1, 2, 3, 4u; I (sophomores) = t1, 3u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. For this model, we have, for example: § § M ( All sophomores are students; M ( Some invitees to the game are sophomores; Example Recall our model from before: § § § § § D = t1, 2, 3, 4u; I (sophomores) = t1, 3u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. For this model, we have, for example: § § § M ( All sophomores are students; M ( Some invitees to the game are sophomores; M ( No students are faculty; Example Recall our model from before: § § § § § D = t1, 2, 3, 4u; I (sophomores) = t1, 3u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. For this model, we have, for example: § § § § M ( All sophomores are students; M ( Some invitees to the game are sophomores; M ( No students are faculty; M ( Not all invitees to the game are students. Syllogistic Validity Formalized We can now formalize the idea of a syllogistic argument with a set of premises tϕ1 , . . . , ϕn u (allowing more than two) and conclusion ψ being valid: Syllogistic Validity Formalized We can now formalize the idea of a syllogistic argument with a set of premises tϕ1 , . . . , ϕn u (allowing more than two) and conclusion ψ being valid: § for every model M, if M ( ϕi for each ϕi , then M ( ψ. Example of Syllogistic Validity § § § All A are B. All B are C . All A are C . That this is valid according to our formal semantics follows from the fact that if I (A) Ď I (B ) and I (B ) Ď I (C ), then I (A) Ď I (C ) (transitivity of Ď). Example of Syllogistic Invalidity § § § All A are B. No C are A. No C are B. That this is invalid according to our formal semantics can be seen in the model we considered before: Example of Syllogistic Invalidity § § § All A are B. No C are A. No C are B. That this is invalid according to our formal semantics can be seen in the model we considered before: § § § § D = t1, 2, 3, 4u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. Example of Syllogistic Invalidity § § § All A are B. No C are A. No C are B. That this is invalid according to our formal semantics can be seen in the model we considered before: § § § § D = t1, 2, 3, 4u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. Here M ( All students are invitees to the game, Example of Syllogistic Invalidity § § § All A are B. No C are A. No C are B. That this is invalid according to our formal semantics can be seen in the model we considered before: § § § § D = t1, 2, 3, 4u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. Here M ( All students are invitees to the game, M ( No faculty are students, Example of Syllogistic Invalidity § § § All A are B. No C are A. No C are B. That this is invalid according to our formal semantics can be seen in the model we considered before: § § § § D = t1, 2, 3, 4u; I (students) = t1, 2, 3u; I (faculty) = t4u; I (invitees to the game) = t1, 2, 3, 4u. Here M ( All students are invitees to the game, M ( No faculty are students, M * No faculty are invitees to the game. Syllogistic Logic We can now formalize the idea of a syllogistic argument with a set of premises tϕ1 , . . . , ϕn u (allowing more than two) and conclusion ψ being valid: § for every model M, if M ( ϕi for each ϕi , then M ( ψ. Syllogistic Logic We can now formalize the idea of a syllogistic argument with a set of premises tϕ1 , . . . , ϕn u (allowing more than two) and conclusion ψ being valid: § for every model M, if M ( ϕi for each ϕi , then M ( ψ. Key Points: § one can define a sound and complete proof system for the syllogistic language in these slides, allowing one to derive any valid consequence of some premises uses a stock of basic inference rules; Syllogistic Logic We can now formalize the idea of a syllogistic argument with a set of premises tϕ1 , . . . , ϕn u (allowing more than two) and conclusion ψ being valid: § for every model M, if M ( ϕi for each ϕi , then M ( ψ. Key Points: § one can define a sound and complete proof system for the syllogistic language in these slides, allowing one to derive any valid consequence of some premises uses a stock of basic inference rules; § moreover, there is an algorithm for deciding whether an argument is valid as above, described in Section 3.5 of Logic in Action. Combining Syllogistic and Propositional Reasoning Note that we can extend our syllogistic language by allowing one to build up complex formulas from our basic formulas § All A are B, § Some A are B, § No A are B, and § Not all A are B, using our propositional connectives , ^, _, Ñ, and Ø. Combining Syllogistic and Propositional Reasoning Note that we can extend our syllogistic language by allowing one to build up complex formulas from our basic formulas § All A are B, § Some A are B, § No A are B, and § Not all A are B, using our propositional connectives , ^, _, Ñ, and Ø. Then we can have formulas such as (Some A are B _ All C are D ). Combining Syllogistic and Propositional Reasoning Note that we can extend our syllogistic language by allowing one to build up complex formulas from our basic formulas § All A are B, § Some A are B, § No A are B, and § Not all A are B, using our propositional connectives , ^, _, Ñ, and Ø. Then we can have formulas such as (Some A are B _ All C are D ). Equivalently, we can start with the language of propositional logic and then replace our proposition symbols p, q, r , etc., by syllogistic formulas. Combining Syllogistic and Propositional Reasoning We learned in this part how to determine the truth value in a model M of one of the basic syllogistic formulas, Combining Syllogistic and Propositional Reasoning We learned in this part how to determine the truth value in a model M of one of the basic syllogistic formulas, so we can recursively calculate the truth value in a model M of any propositional combination of basic formulas: Combining Syllogistic and Propositional Reasoning We learned in this part how to determine the truth value in a model M of one of the basic syllogistic formulas, so we can recursively calculate the truth value in a model M of any propositional combination of basic formulas: § M ( All A are B if and only if I (A) Ď I (B ); § M ( Some A are B if and only if I (A) X I (B ) ‰ ∅; § M ( No A are B if and only if I (A) X I (B ) = ∅; § M ( Not all A are B if and only if I (A) Ę I (B ); Combining Syllogistic and Propositional Reasoning We learned in this part how to determine the truth value in a model M of one of the basic syllogistic formulas, so we can recursively calculate the truth value in a model M of any propositional combination of basic formulas: § M ( All A are B if and only if I (A) Ď I (B ); § M ( Some A are B if and only if I (A) X I (B ) ‰ ∅; § M ( No A are B if and only if I (A) X I (B ) = ∅; § M ( Not all A are B if and only if I (A) Ę I (B ); § M ( ϕ ^ ψ if and only if M ( ϕ and M ( ψ; § M ( ϕ _ ψ if and only if M ( ϕ or M ( ψ; § M ( ϕ if and only if M * ϕ; Combining Syllogistic and Propositional Reasoning We learned in this part how to determine the truth value in a model M of one of the basic syllogistic formulas, so we can recursively calculate the truth value in a model M of any propositional combination of basic formulas: § M ( All A are B if and only if I (A) Ď I (B ); § M ( Some A are B if and only if I (A) X I (B ) ‰ ∅; § M ( No A are B if and only if I (A) X I (B ) = ∅; § M ( Not all A are B if and only if I (A) Ę I (B ); § M ( ϕ ^ ψ if and only if M ( ϕ and M ( ψ; § M ( ϕ _ ψ if and only if M ( ϕ or M ( ψ; § M ( ϕ if and only if M * ϕ; etc. Combining Syllogistic and Propositional Reasoning We learned in this part how to determine the truth value in a model M of one of the basic syllogistic formulas, so we can recursively calculate the truth value in a model M of any propositional combination of basic formulas: § M ( All A are B if and only if I (A) Ď I (B ); § M ( Some A are B if and only if I (A) X I (B ) ‰ ∅; § M ( No A are B if and only if I (A) X I (B ) = ∅; § M ( Not all A are B if and only if I (A) Ę I (B ); § M ( ϕ ^ ψ if and only if M ( ϕ and M ( ψ; § M ( ϕ _ ψ if and only if M ( ϕ or M ( ψ; § M ( ϕ if and only if M * ϕ; etc. We could write the propositional clauses in terms truth values and min, max, and 1 ´ x, but we assume you now understand the truth-functional connectives. Next Up Next, we will take further steps toward the most expressive logical language we will consider in this course: the language of first-order logic.
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Explanation & Answer

This is the version of my answer where all of the screenshots from Excel were deleted.

Question 1:
A natural deduction proof that formularizes the reduction step is shown below:
1
All of the given information or premises.
2
 (There is some digits among 1,2,…,9 occurring infinitely many times)
3
Each of 1,2,…,9 occurs finitely many times.
4
There are only finitely many digits among 1,2,…,9.
5
Decimal expansion of  is a finite sequence of digits.
6
 is irrational.
7 There is some digits among 1,2,…,9 occurring infinitely many times.

I , 2
R, 1
 I , 3, 4
R, 1
RAA, 1-6

Question 2:
a) The natural deduction proof p from assumption  ( p → q ) is shown below:

( p → q )

1
2
3
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5
6...


Anonymous
I was struggling with this subject, and this helped me a ton!

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