IST 4 Information and Logic

Transcription

1 IST 4 Information and Logic

2 HW2 will be returned today Average is 53/6~=88%

3 T = today x= hw#x out x= hw#x due mon tue wed thr fri 3 M 6 oh M oh 3 oh oh 2M2M 2 oh oh 2 Mx= MQx out 27 oh M2 oh oh = office hours 4 3 oh 3 4 T oh oh midterms oh Mx= MQx due 8 oh oh oh 5 oh oh oh

4 MQ2 Memory Due today by pm Please PDF lastname-firstname.pdf to

5 Languages evolution

6 Building Blocks Separation finite number of building blocks infinitely many descriptions Separation A Separation B between syntax and between what is represented semantics and reality, feasibility, time, space,... Separation C between algorithms and implementation

7 The appearance of life is the first Information Megamorphosis DNA ~3.7 Billion ya The appearance of the human brain is the second Information Megamorphosis Spoken languages g ~6Kya

8 Babylonians Written languages ~5,ya The language of numbers positional number systems mathematics

9 Formal languages Greeks ~2,5ya Pythagoras BC Axioms Theorems Euclid,3BC Proofs

10 Formal languages Greeks Logic ~2,5ya Aristotle BC Syllogism Inference... our logical sense is 3 People that are wise are Babylonians Leibniz was wise Leibniz was a Babylonian

11 Algorizms Algorizmi 78-85AD ~ya Fibonacci 7-25AD Algorizms for everything!! Gottfried Leibniz ~3ya

12 Formal languages g for ideas Gottfried Leibniz Let us calculate l without t further ado, to see who is right" Let s Google it!! Let s Leibniz it!!

13 Algorizms and syntax boxes Gottfried Leibniz instead of progression by tens, I have for many years used the most simple of all, which goes by two...

14 The Binary Gottfried Leibniz d d2 majority c 2 symbol adder c a b m s parity

15 Gottfried Leibniz we need a language for logic... We need a language for syntax boxes...

16 ~2 years after Aristotle... George Boole Calculus for logic and for Syntax boxes... Boole was a Babylonian...

17 Boolean Algebra Huntington 94; concise set of axioms Edward Huntington April 26, Undergrad and Masters at Harvard PhD at the U of Strasbourg, Germany (9) Professor at Harvard until 94 Huntington was a Greek...

18 The Algebra (Boolean Calculus) Algebraic system: set of elements B, two binary operations + and B has at least two elements ( and ) If the following axioms are true then it is a Boolean Algebra: A. identity A2. complement A3. commutative A4. distributive

19 Properties of an Axiomatic System consistent No contradictions in the math theory complete Every true statement in the math theory can be derived d using the axioms independent d No redundancy d in the set of axioms

20 Complete: Every true statement in the math theory can be derived using the axioms Theorems and Algorizms Can we prove / compute EVERYTHING?

21 Consistent: No contradictions in the math theory Complete: Every true statement in the math theory can be derived using the axioms Kurt Gödel April 28, : For any axiomatic system that is powerful enough to describe the arithmetic of the natural numbers: If the system is consistent, it cannot be complete In a consistent system there are statements that are not provable... The key: represent the axiomatic system using numbers, a proof is a calculation algorizm...

22 Will be posted on the class web site Kurt Gödel April 28, Olga Taussky-Todd Caltech professor starting 957

23 A simple example Can we prove / compute EVERYTHING?

24 6, 3,, 5, 6, 8, 4, 2,, 34, 7, 52, 26, 3, 4, 2,, 5, 6, 8, 4, 2, Source: wikipedia

25 Does it always reach? 6, 3,, 5, 6, 8, 4, 2,, 34, 7, 52, 26, 3, 4, 2,, 5, 6, 8, 4, 2, n even n odd Source: wikipedia

26 Does it always reach? Which number in -5 has the longest sequence to reach? 9, 28, 4, 7, 22,, 34, 7, 52, 26, 3, 4, 2,, 5, 6, 8, 4, 2, The number 9, a sequence with 2 numbers Source: wikipedia

27 Lothar Collatz 9-99 The Collatz conjecture (937): For every starting ti value m, the sequence always reaches Empirical evidence: Verified up to some large number: True False Prove that it is impossible to decide if the conjecture true or false

28 Lothar Collatz 9-99 The Collatz conjecture (937): For every starting ti value m, the sequence always reaches A generalization:

29 The Collatz conjecture (937): For every starting value m, the sequence always reaches It is likely undecidable... Open problem... This generalization is undecidable, J. Conway, 972 Undecidable: Given a number n, there is no algorithm that can decide if the Collatz sequence reaches (Collatz conjecture) We can prove that it is impossible to decide if true or false

30 Euclid,3BC There are theorems that cannot be proved Algorizmi 78-85AD There are problems that cannot be solved by an algorizm

31 consistent No contradictions complete Every true statement can be derived using the axioms independent No redundancy in the set of axioms

32 Boolean Algebra Proving theorems If you would not see a giraffe you will not believe it exists

33 If I satisfy the axioms then I am a GIRAFFE You do not need to see it to believe it exists!

34 Boolean Algebra - algebra You can see this one

35 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND xy OR(x,y) xy AND(x,y) iff both x and y are iff both x and y are Is it a Boolean Algebra?

36 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND The following axioms are obviously true:??? A. identity A2. complement A3. commutative A4. distributive

37 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A. identity a + = a a x = a xy OR(x,y) xy AND(x,y)

38 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A. identity a + = a a x = a xy OR(x,y) + = + = xy AND(x,y)

39 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A. identity a + = a a x = a xy OR(x,y) x = x = xy AND(x,y)

40 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A2. complement a + a = a x a = xy OR(x,y) xy AND(x,y)

41 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A2. complement a + a = a x a = + = x = + = x = xy OR(x,y) xy AND(x,y) a complement of a complement of

42 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A3. commutative ti a + b = b + a a x b = b x a xy OR(x,y) xy AND(x,y)

43 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A3. commutative ti a + b = b + a a x b = b x a xy OR(x,y) + = + + = + x = x x = x + = + x = x + = + x = x xy AND(x,y)

44 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A4. distributive a + (b x c) = (a + b) x (a + c) a x (b + c) = (a x b) + (a x c) xy OR(x,y) xy AND(x,y)

45 Two-valued Boolean Algebra Boolean Algebra: set of elements B={,}, two binary operations OR and AND A4. distributive a + (b x c) = (a + b) x (a + c) + ( x ) = ( + ) x ( + ) xy OR(x,y) + ( x ) = ( + ) x ( + ) We can check all the cases... xy AND(x,y)

46 Now, to our Boolean

47 Self Absorption ME-MYSELF&I Lemma : Proof: Two-valued Boolean Algebra: set of elements B={,}, two binary operations OR and AND xy OR(x,y) Is the lemma true? xy AND(x,y)

48 Self Absorption ME-MYSELF&I Lemma : Proof: A A2 A4 A2 A Q

49 Self Absorption ME-MYSELF&I Lemma : Proof: We only proved that Need to prove Ideas?

50 Boolean Algebra Duality

51 Duality Theorem : Theorem : Any identity that is true in a Boolean algebra, is also true if + and are interchanged, and and are interchanged.

52 ME-MYSELF&I Lemma : Proof: if + and. are interchanged,, and and are interchanged A A2 A4 A2 A

53 Duality Theorem : Any identity that is true in a Boolean algebra, is also true if + and are interchanged, and and are interchanged. Proof:????

54 Theorem : Any identity that is true algebra, is also true if + and. are interchanged, and and are interchanged. It is a syntax machine: It is true for the axioms!

55 Back to the Axioms Q: Is the complement unique / well defined?

56 Boolean Algebra One way to say NO

57 One Way to Say No! Theorem : Each element of a Boolean Algebra has exactly one complement. Proof: Warm-up: First we will prove that an element is not self-complement m Assume that: By Lemma :?? However by A2:

58 One Way to Say No! Theorem : Each element of a Boolean Algebra has exactly one complement. Proof: Warm-up: First we will prove that an element is not self-complement m Assume that: By duality: By Lemma : However by A2: Contradiction! and are distinct Q

59 One Way to Say No! Theorem : Each element of a Boolean Algebra has exactly one complement. Proof: We proved that an element is not self-complement Next will prove that the complement is unique

60 One Way to Say No! Proof: Need to prove that the complement m is unique By contradiction: Assume an element has two distinct complements A A2 A4 A3 A2

61 One Way to Say No! Proof: Need to prove that the complement m is unique By contradiction: Assume an element has two distinct complements A A2 A4 A3 A2 A4 A2 A

62 One Way to Say No! Proof: Need to prove that the complement m is unique By contradiction: Assume an element has two distinct complements A A2 A4 A3 A2 A2 A3 A4 A2 A Contradiction! Q

63 So far True for any Boolean Algebra T: duality principle T: one complement per element L: Self Absorption

64 So far True for any Boolean Algebra T: one complement m per element We also proved that an element is not self-complement Q: Is the complement of the complement the original element? True for two elements? Four elements?

65 Saying No Twice? uniqueness of complement (T) complement (T) NONO Property: Proof: Need to prove that t a is the complement of Definition of complement: A2 A3 Q

66 Boolean Algebra How many elements?

67 Is there a Boolean algebra with 3 elements? Idea: use the properties p of the complement NO!! Must have an even number of elements!? Question resolved only in