Steeple #1 of Rationalistic Genius: Gödel s Completeness Theorem
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1 Steeple #1 of Rationalistic Genius: Gödel s Completeness Theorem Selmer Bringsjord Are Humans Rational? v of 11/3/17 RPI Troy NY USA
2 Gödels Great Theorems (OUP) by Selmer Bringsjord Introduction ( The Wager ) Brief Preliminaries (e.g. the propositional calculus) The Completeness Theorem The First Incompleteness Theorem The Second Incompleteness Theorem The Speedup Theorem The Continuum-Hypothesis Theorem The Time-Travel Theorem Gödel s God Theorem Could a Machine Match Gödel s Genius?
3 Gödels Great Theorems (OUP) by Selmer Bringsjord Introduction ( The Wager ) Brief Preliminaries (e.g. the propositional calculus) The Completeness Theorem The First Incompleteness Theorem The Second Incompleteness Theorem The Speedup Theorem The Continuum-Hypothesis Theorem The Time-Travel Theorem Gödel s God Theorem Could a Machine Match Gödel s Genius?
4 Gödels Great Theorems (OUP) by Selmer Bringsjord Introduction ( The Wager ) Brief Preliminaries (e.g. the propositional calculus) The Completeness Theorem The First Incompleteness Theorem The Second Incompleteness Theorem The Speedup Theorem The Continuum-Hypothesis Theorem The Time-Travel Theorem Gödel s God Theorem Could a Machine Match Gödel s Genius?
5 Gödels Great Theorems (OUP) by Selmer Bringsjord 3/4 Introduction ( The Wager ) Brief Preliminaries (e.g. the propositional calculus) The Completeness Theorem The First Incompleteness Theorem The Second Incompleteness Theorem The Speedup Theorem The Continuum-Hypothesis Theorem The Time-Travel Theorem Gödel s God Theorem Could a Machine Match Gödel s Genius?
6 Some Timeline Points 1978 Princeton NJ USA. 194 Back to USA, for good Hitler comes to power Doctoral Dissertation: Proof of Completeness Theorem Undergrad in seminar by Schlick 1923 Vienna 196 Brünn, Austria-Hungary
7 Some Timeline Points 1978 Princeton NJ USA. 194 Back to USA, for good Hitler comes to power Doctoral Dissertation: Proof of Completeness Theorem Undergrad in seminar by Schlick 1923 Vienna 196 Brünn, Austria-Hungary
8 R&W s Axiomatization of the Propositional Calculus A1 ( _ )! A2! ( _ ) A3 ( _ )! ( _ ) A4 (! )! (( _ )! ( _ ))
9 R&W s Axiomatization of the Propositional Calculus A1 ( _ )! A2! ( _ ) A3 ( _ )! ( _ ) A4 (! )! (( _ )! ( _ )) All instances of these schemata are true no matter what the input (true or false). (Agreed?) And indeed every single formula in the propositional calculus that is true no matter what the permutation (as shown in a truth table), can be proved (somehow) from these four axioms (using the rules of inference given in AHR? 216). This, Gödel knew, and could use.
10 Exercise 1: Verify that these are true-no-matter what in a truth table; then prove using our rules for the prop. calc. ( ^ )! ( _ )! (! )
11 Exercise 1: Verify that these are true-no-matter what in a truth table; then prove using our rules for the prop. calc. ( ^ )! ( _ )
12 Exercise 1: Verify that these are true-no-matter what in a truth table; then prove using our rules for the prop. calc. ( ^ )! ( _ ) Truth Table showing this formula true no matter what the inputs.
13 Exercise 1: Verify that these are true-no-matter what in a truth table; then prove using our rules for the prop. calc. ( ^ )! ( _ ) Truth Table showing this formula true no matter what the inputs. Proof:
14 Exercise 1: Verify that these are true-no-matter what in a truth table; then prove using our rules for the prop. calc. ( ^ )! ( _ ) Truth Table showing this formula true no matter what the inputs. Proof: 1 ^ Supposition 2 1, Simplification 3 _ 2, Addition 4 ( _ )! ( _ ) 1 3, Conditional Intro
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17 But what about the pure predicate calculus and the full predicate calculus?!? Does completeness hold here as well?? Is it true that whatever is true-nomatter-what can also be proved??
18 But what about the pure predicate calculus and the full predicate calculus?!? Does completeness hold here as well?? Is it true that whatever is true-nomatter-what can also be proved?? Gödel: You give me a non-contradictory formula, and I ll show you that there s a Buzz-Lightyear train trip on which that formula is true which* gives you an answer of Yes!.
19 But what about the pure predicate calculus and the full predicate calculus?!? Does completeness hold here as well?? Is it true that whatever is true-nomatter-what can also be proved?? Gödel: You give me a non-contradictory formula, and I ll show you that there s a Buzz-Lightyear train trip on which that formula is true which* gives you an answer of Yes!. *For reasons we won t go into.
20 From Languge-Learning Slides: The Grammar of the Pure Predicate Formula ) AtomicFormula (Formula Connective Formula) Formula AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant Connective ) ^ _! $ Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3...
21 Recall the Examples We Cited Formula ) AtomicFormula (Formula Connective Formula) Formula Sally likes Bill. (Likes sally bill) AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant Connective ) ^ _! $ Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3... Lexicon Sally likes Bill and Bill likes Sally. Sally likes Bill s mother. Sally likes Bill only if Bill s mother is tall. Matilda is Bill s super-smart mother. 5 plus 5 equals the number 1. Did you make sure you can simulate a machine that says Yes that sentence is okay! whenever it s conforms to this grammar?
22 Recall the Examples We Cited Formula ) AtomicFormula (Formula Connective Formula) Formula Sally likes Bill. (Likes sally bill) AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant Connective ) ^ _! $ Likes Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3... Lexicon Sally likes Bill and Bill likes Sally. Sally likes Bill s mother. Sally likes Bill only if Bill s mother is tall. Matilda is Bill s super-smart mother. 5 plus 5 equals the number 1. Did you make sure you can simulate a machine that says Yes that sentence is okay! whenever it s conforms to this grammar?
23 But Last Time We Noted These Examples Formula ) AtomicFormula (Formula Connective Formula) Formula AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant Connective ) ^ _! $ Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3...
24 But Last Time We Noted These Examples Formula ) AtomicFormula (Formula Connective Formula) Formula AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) If Sally likes Bill then Sally likes Bill. Term ) (Function Term 1... Term k ) Constant Connective ) ^ _! $ Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3...
25 But Last Time We Noted These Examples Formula ) AtomicFormula (Formula Connective Formula) Formula AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant If Sally likes Bill then Sally likes Bill. Sally likes Bill s mother, or not. Connective ) ^ _! $ Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3...
26 But Last Time We Noted These Examples Formula ) AtomicFormula (Formula Connective Formula) Formula AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant If Sally likes Bill then Sally likes Bill. Sally likes Bill s mother, or not. Sally likes Bill and Bill likes Jane, only if Bill likes Jane. Connective ) ^ _! $ Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3...
27 But Last Time We Noted These Examples Formula ) AtomicFormula (Formula Connective Formula) Formula AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant If Sally likes Bill then Sally likes Bill. Sally likes Bill s mother, or not. Sally likes Bill and Bill likes Jane, only if Bill likes Jane. Bill s smart mother is a mother. Connective ) ^ _! $ Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3...
28 But Last Time We Noted These Examples Formula ) AtomicFormula (Formula Connective Formula) Formula AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant Connective ) ^ _! $ If Sally likes Bill then Sally likes Bill. Sally likes Bill s mother, or not. Sally likes Bill and Bill likes Jane, only if Bill likes Jane. Bill s smart mother is a mother. Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3...
29 But Last Time We Noted These Examples Formula ) AtomicFormula (Formula Connective Formula) Formula AtomicFormula ) (Predicate Term 1...Term k ) (Term = Term) Term ) (Function Term 1... Term k ) Constant Connective ) ^ _! $ If Sally likes Bill then Sally likes Bill. Sally likes Bill s mother, or not. Sally likes Bill and Bill likes Jane, only if Bill likes Jane. Bill s smart mother is a mother. Predicate ) P 1 P 2 P 3... Constant ) c 1 c 2 c 3... Function ) f 1 f 2 f 3... These are all true, yes; but can they be proved?!
30 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter
31 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter there exists at least one thing x such that
32 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter there exists at least one thing x such that for all x, it s the case that
33 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that for all x, it s the case that
34 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that
35 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that 8 ( >!9 ( > ^8x(d(x, a) <! d(f(x),b) < )))
36 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that
37 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that Every natural number is greater than or equal to zero.
38 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that Every natural number is greater than or equal to zero. 8x(x )
39 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that Every natural number is greater than or equal to zero. 8x(x ) There s a positive integer greater than any positive integer.
40 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that Every natural number is greater than or equal to zero. 8x(x ) There s a positive integer greater than any positive integer. 9x8y(y <x)
41 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that Every natural number is greater than or equal to zero. 8x(x ) 9x8y(y <x)
42 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that Every natural number is greater than or equal to zero. 8x(x )
43 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that Every natural number is greater than or equal to zero. 8x(x ) Every positive integer x is less-than-or-equal-to a positive integer y.
44 Add the Final Addition/Deeper Challenge: Add Two Quantifiers to the Pure Predicate Calculus, Which Yields the Predicate Calculus simpliciter 9x... there exists at least one thing x such that 8x... for all x, it s the case that Every natural number is greater than or equal to zero. 8x(x ) Every positive integer x is less-than-or-equal-to a positive integer y. 8x9y(x apple y) 8x9y(apple (x, y))
45 For every positive integer x is less-than-or-equal-to a positive integer y. 8x9y(x apple y) 8x9y(apple (x, y))
46 For every positive integer x is less-than-or-equal-to a positive integer y. 8x9y(x apple y) 8x9y(apple (x, y)) Gödel: You give me a non-contradictory formula, and I ll show you that there s a Buzz-Lightyear train trip on which that formula is true which gives you an answer of Yes! Equivalently, the predicate calculus is complete: every formula in it that must-be-true, can in fact be proved!
47 For every positive integer x is less-than-or-equal-to a positive integer y. 8x9y(x apple y) 8x9y(apple (x, y)) First, I ll break this down into a list of conjunctions and drop the quantifiers.
48 For every positive integer x is less-than-or-equal-to a positive integer y. 8x9y(x apple y) 8x9y(apple (x, y)) First, I ll break this down into a list of conjunctions and drop the quantifiers. apple x 1 x 2 apple x 1 x 2^applex 2 x 3 apple x 1 x 2^applex 2 x 3^applex 3 x 4 apple x 1 x 2^applex 2 x 3^applex 3 x 4^applex 4 x 5. apple x 1 x 2...apple x 99 x 1
49 For every positive integer x is less-than-or-equal-to a positive integer y. 8x9y(x apple y) 8x9y(apple (x, y)) First, I ll break this down into a list of conjunctions and drop the quantifiers. apple x 1 x 2 apple x 1 x 2^applex 2 x 3 apple x 1 x 2^applex 2 x 3^applex 3 x 4 apple x 1 x 2^applex 2 x 3^applex 3 x 4^applex 4 x 5. apple x 1 x 2...apple x 99 x 1 Next, I ll take you on a Buzz-Lightyear train trip on which this entire list is true
50 Now, recall König s Lemma
51 Toward König s Lemma as Train Travel
52 To infinity and beyond!
53 König s Lemma (train-travel version) In a one-way train-travel map with finitely many options leading from each station, if there are partial paths forward of every finite length, there is an infinite path (= a path to infinity ).
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63 Exercise 2: Is there an algorithm for traveling this way?
64 Exercise 2: Is there an algorithm for traveling this way? No. This strategy for travel is beyond the reach of standard computation.
65 Exercise 2: Is there an algorithm for traveling this way? No. This strategy for travel is beyond the reach of standard computation. (To anticipate our final class & Steeple #3: Does it not then follow, assuming that humans can find and use a provably correct strategy for this travel, that humans can t be fundamentally computing machines?)
66 Proving the Lemma (that there is an infinite branch) Proof: We are seeking to prove that there is an infinite path (= that you can keep going forward forever = that the number of your stops forward are the size of Z + ). To begin, assume the antecedent of the theorem (i.e. that, (1), there are finitely many options leading from each station, and that, (2), in the map there are partial paths forward of every finite size). Now, you are standing at Penn Station (S1), facing k options. At least one of these options must lead to partial paths of arbitrary size (the size of any m in Z + ). (Sub- Proof: Suppose otherwise for indirect proof. Then there is some positive integer n that places a ceiling on the size of partial paths that can be reached. But this violates (2) contradiction.) Proceed to choose one of these options that lead to partial paths of arbitrary size. You are now standing at a new station (S2), one stop after Penn Station. At least one of these options must lead to partial parts of arbitrary size (the size of any m in Z + ). (Sub-Proof: Suppose otherwise for indirect proof ) Since you can iterate this forever, you ll be on an infinite trip to infinity! Buzz will be happy.
67 Simple Buzz-Lightyear-Like Branch
68 Gödel s Buzz-Lightyear Branch
69 For Further Reading
70 For Further Reading
71 For Further Reading
72 For Further Reading
73 For Further Reading
74 For Further Reading
75 Next Time (Final Content-Delivery Class Mtg) Steeple #2: The Immortal Incompleteness Result Steeple #3: Could a machine ever be this rational and creative?
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77 slutten
78 Appendices
79 So what would be a specific g*? A truth of arithmetic that you can t move from the axioms of arithmetic?!?
80 So what would be a specific g*? A truth of arithmetic that you can t move from the axioms of arithmetic?!? Here you go:
81 So what would be a specific g*? A truth of arithmetic that you can t move from the axioms of arithmetic?!? Here you go: That the Goodstein Sequence eventually reaches zero!
82 Goodstein Sequence; Goodstein s Theorem...
83 Pure base n representation of a number r Represent r as only sum of powers of n in which the exponents are also powers of n etc
84 Grow Function
85 Example of Grow
86 Goodstein Sequence For any natural number m
87 Sample Values
88 Sample Values
89 m Sample Values
90 Sample Values m
91 Sample Values m
92 Sample Values m (96th term)...
93 Sample Values m (96th term) ~1 13 ~1 155 ~ ~
94 Yet, The Theorems!!
95 Yet, The Theorems!! Theorem 1 (Goodstein s Theorem). For all natural numbers, the Goodstein sequence reaches zero after a finite number of steps. Theorem 2 (Unprovability of Goodstein s Theorem). Goodstein s theorem is not provable in Peano Arithmetic (PA) (or any equivalent theory of arithmetic).
96 Yet, The Theorems!! Theorem 1 (Goodstein s Theorem). For all natural numbers, the Goodstein sequence reaches zero after a finite number of steps. Theorem 2 (Unprovability of Goodstein s Theorem). Goodstein s theorem is not provable in Peano Arithmetic (PA) (or any equivalent theory of arithmetic). So, Gödel was right, empirically! We have in GT a truth of elementary arithmetic that we can t prove from elementary arithmetic!
97 Could a computing machine get this??...
98 Govindarajulu, N.; Licato, J.; Bringsjord, S Small Steps Toward Hypercomputation via Infinitary Machine Proof Verification and Proof Generation. In Proceedings of UCNC 213. Pdf
99 Needs Understanding of Ordinal Numbers
100 Needs Understanding of Ordinal Numbers 19 1 = <!!!! +!! +!
101 Needs Understanding of Ordinal Numbers 19 1 = <!!!! +!! +! 19 2 = <!!!! +!! +! 1
102 Needs Understanding of Ordinal Numbers 19 1 = <!!!! +!! +! 19 2 = = <!!!! <!!!! +!! +! 1 +!! 1
103 Needs Understanding of Ordinal Numbers 19 1 = <!!!! +!! +! 19 2 = = = <!!!! <!!!! <!!!! +!! +! 1 +!! 1 +! +! 1
104 Needs Understanding of Ordinal Numbers 19 1 = <!!!! +!! +! 19 2 = = = <!!!! <!!!! <!!!! 19 5 = <!!!! +!! +! 1 +!! 1 +! +! 1 +!
105 Needs Understanding of Ordinal Numbers 19 1 = <!!!! +!! +! 19 2 = = = <!!!! <!!!! <!!!! 19 5 = <!!!! +!! +! 1 +!! 1 +! +! 1 +!...
106 Needs Understanding of Ordinal Numbers 19 1 = <!!!! +!! +! 19 2 = = = <!!!! <!!!! <!!!! 19 5 = <!!!! +!! +! 1 +!! 1 +! +! 1 +!...
107 Needs Understanding of Ordinal Numbers 19 1 = <!!!! +!! +! 19 2 = = = <!!!! <!!!! <!!!! 19 5 = <!!!! +!! +! 1 +!! 1 +! +! 1 +!... strictly decreasing
108 19 1 = <!!!! +!! +! 19 2 = = = <!!!! <!!!! <!!!! 19 5 = <!!!! +!! +! 1 +!! 1 +! +! 1 +!... strictly decreasing
109 Ordinal Numbers
110 Yet, Conjecture (C) (see Isaacson s Conjecture )
111 Yet, Conjecture (C) (see Isaacson s Conjecture ) In order to produce a rationally compelling proof of any true sentence S formed from the symbol set of the language of arithmetic, but independent of PA, it s necessary to deploy concepts and structures of an irreducibly infinitary nature.
112 Yet, Conjecture (C) (see Isaacson s Conjecture ) In order to produce a rationally compelling proof of any true sentence S formed from the symbol set of the language of arithmetic, but independent of PA, it s necessary to deploy concepts and structures of an irreducibly infinitary nature. If this is right, and computing machines can t use irreducibly infinitary techniques, they re in trouble or: there won t be a Singularity.
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