1. can’t be c can’t be b Could be cthis is bcould be c

4. Names  Constants are used to name existing objects a, b, c, d, e, f max, claire, carl  No constant can name more than one object  An object can have more than one name or no name at all

5. Tiberius Sempronius Gracchus Gaius Sempronius Gracchus Examples Leonard Euler Honest Abe Lincoln

7. Predicates  A (determinate) property possessed by an object Shape Size  A (determinate) relationship among objects Shape relationship Size relationship Positional relationship Equality =

8. Atomic Sentences  A sentence formed by a single predicate followed by one or more names Max is tall Tall(max) e is larger than b Larger(e,b) e is identical to a e = a  A sentence expresses a claim that is either true or false

9. Atomic Sentences in FOL  Predicate(arg 1, arg 2,…, arg n ) Predicates have names beginning with an uppercase letter or are represented by an operator symbol The number of arguments is called the predicate’s arity The order of the arguments is important Larger(e,c) – e is larger than c Larger(c,e) – c is larger than e Between(a,b,e) – a is between b and e Between(b,a,e) – b is between a and e  =(a,b) a and b are identical Usually, written in infix form a = b

10. Function Symbols  A function is used to express complex names (a reference to an individual without using a name) father(b) – b’s father password(c) – c’s password  A function may be nested Max’s father’s father  father(father(max))  A function is never a predicate Can’t nest predicates  Tall(Tall(max)) A predicate forms a sentence, while a function names an individual

11. Functions in FOL  function(arg 1, arg 2,…, arg n ) Function names begin with a lowercase letter or are expressed with a symbol father(max) Max’s father father(mother(max)) Max’s mother’s father youngestChild(max,ann) Max and Ann’s youngest child *(5,+(2,4)) 30 starship(son(dr_crusher)) Dr_Crusher’s son’s starship

12. Connectives  Not   And, Or ,   Material Conditional   Biconditional 

13. Examples   Larger(e,c)  Cube(b)  Large(b)  SameRow(e,c)  BackOf(e,b) e is not larger than c b is a cube or b is large e and c are in the same row and e is in back of b

14. First Order Logic  Names  Predicates  Functions  Connectives Are there more? Atomic Sentences

15. Example FOL

16. Translation  Brando is Nancy’s favorite actor.  brando = favoriteActor(nancy)  BetterActor(favoriteActor(nancy), favoriteActor(max))  Nancy’s favorite actor is better than Max’s favorite actor.  sean = favoriteActor(sean)  Sean is his own favorite actor.  Brando is someone’s favorite actor.   x(brando = favoriteActor(x))

17. Quantifiers and Variables  For every x  x  There exists y  y

18. First Order Logic  Names  Predicates  Functions  Connectives  Quantifiers and variables Revised List

19. Translation using functions  c is the front-most block in b’s column.  c is in the same row as the front-most block in b’s column.  The right-most block in the same row as the front-most block in b’s column is small. c = fm(b) SameRow(c,fm(b)) Small(rm(fm(b)))

20. First-order Arithmetic  Names Zero 0 One 1  Predicates Equality = Less than <  Functions Addition + Multiplication  1)0 and 1 are terms. 2)If t 1 and t 2 are terms then so are (t 1 + t 2 ) and (t 1  t 2 ). 3) Nothing is a term unless formed from the above rules.

21. What is an argument?  A series of statements in which one (called the conclusion) is meant to follow from or be supported by the others (called the premises).

22. Fitch-style Argument  P 1 P 2... P n Q premises conclusion

23. Valid Argument  A valid argument is one that guarantees the truth of its conclusion on the assumption that the premises are true.  A valid argument ensures the conclusion is true provided the premises are true.  A valid argument does not depend on any world for its validity

24. Valid Argument  Large(b) v Cube(b)  Cube(b) Large(b) premisesconclusion 

25. Invalid Argument  Large(b) v Cube(b) Cube(b) Large(b) premisesconclusion

26. Sound Argument  If an argument is valid and its premises are true, then the argument is said to be sound.  The soundness or unsoundness of an argument is determined with respect to some world

27. Sound Argument

28. Argument is not sound

29. Methods of Proof  Formal We will use a Fitch-style proof employed in the text and software of the same name. “Formal” proof evokes images of being rigorous. In fact, it has to do creating a proof with strict syntax rules.  Informal This style of proof, used by mathematicians, is just as rigorous. It consists of sentences describing the situation at hand, the inferences being made, and the justification of each inference.

30. What constitutes a proof?  A proof that sentence Q follows from the premises P 1, P 2, …, P n is a step-by-step demonstration that shows Q must be true in any circumstances in which the premises are all true.

31. Types of Proof  Direct  Indirect  Proof by cases  Proof by contradiction  Proof by induction  Proof by counterexample

32. Fitch-style Proof  P 1 P 2 … P n S 1 S 2 … S n Q Premises Deductions & Justifications may contain sub-proofs Conclusion Fitch Bar

33. Rules/Axioms  = Elimination If b = c and P(b) then P(c).  = Introduction a = a  Symmetry of Identity If a = b then b = a.  Transitivity of Identity If a = b and b = c then a = c These follow from above

34. = Elimination  P(n) n = m P(m)

35. = Introduction  n = n

36. Symmetry of Identity  a = b 1) 2) 3) a = a = Introduction b = a = Elimination 1, 2

38. Example Formal Proof  Smaller(a,b) c = b Larger(b,a)Ana Con 1 c = c = Introduction b = c = Elim 2, 4 Larger(c,a) = Elim 5, 3 1) 2) 3) 4) 5) 6)

39. Example Informal Proof Prove: If a is smaller than b and c is identical to b then c is larger than a. Since a is smaller than b, it follows that b must be larger than a. Moreover, since c is identical to b, it follows that c must be larger than a. QED

40. Consequence Rules  There are three consequence “rules” in Fitch Tautological Consequence (Taut Con) First-order Consequence (FO Con) Analytic Consequence (Ana Con) Cons rules are proof seekers that work behind the scenes. Success is indicated by the beloved blue check mark. Failure is indicated by the dreaded red x.

41. Taut Con  Weakest of the three Cons “rules”  Attempts to prove if the current step follows from the cited statements by virtue of the truth-functional connectives. E.g., p  q can be replaced by  p  q.

42. FO Con  More powerful than Taut Con but weaker than Ana Con  Attempts to prove if the current step follows from the cited steps by virtue of the truth- functional connectives, the quantifiers, and the identity predicate. E.g., a=b  b=c can be replaced by a=c.

43. Ana Con  Most powerful Cons “rule”  Attempts to prove if the current step follows from the cited steps by virtue of the truth- functional connectives, the quantifiers, the the identity predicate and the meanings of each predicate of Tarski’s World. E.g., Larger(a,b) can be replaced by Smaller(b,a).

44. Showing Non-consequence  To show Q is not a consequence of premises P 1, P 2, …, P n, create a world where the premises are simultaneously true and the conclusion Q is false.  This shows the argument below is invalid P 1 P 2 … P n Q

45. Invalid Argument  SameRow(b,c) SameRow(a,d) SameRow(d,f) LeftOf(a,b) LeftOf(f,c) bcad ff

47. Boolean Connectives  Negation   Conjunction   Disjunction  It is not the case that And, but, moreover Or

48. Negation P PP TF FT

49. Negation Facts   P is translated as It is not the case that P.   (a = b) is equivalent to a  b    P is equivalent to P P PP  P P TFT FTF

50. The Game  Used to understand the truth value of a complex sentence  Strategy: Given a sentence of the form    P that you believe to be True (False) implies   P is False (True) implies  P is True (False) implies P is False (True)

51. Assessment is incorrect

52. Conjunction and Disjunction PQ P  QP  Q TTTT TFFT FTFT FFFF

53. Game Revisited  If you believe P  Q is true you will be asked to select the disjunct that is true.  If you believe P  Q is false Tarski’s World will attempt to find a disjunct that is true.  If you believe P  Q is true Tarski’s World will attempt to find a conjunct that is false.  If you believe P  Q is false you will be asked to select a conjunct that is false.

54. Translation  Both A and B  Either A or B  Neither A nor B A  B A  B  (A  B) Both c and e are cubes. Cube(c)  Cube(e) Either c or e is a cube. Cube(c)  Cube(e) Neither c nor e is a cube.  (Cube(c)  Cube(e))

55. Avoid Ambiguity A  B  C(A  B)  C Both A or B and C A  (B  C) Either A or both B and C A  B  C(A  B)  C Either both A and B or C A  (B  C) Both A and either B or C Either both Max is at home and Claire is tall or Carl is happy. [Home(Max)  Tall(Claire)]  Happy(Carl)

56. Distributive Rules  A  (B  C) is tautologically equivalent to (A  B)  (A  C)  A  (B  C) is tautologically equivalent to (A  B)  (A  C)

57. Famous Equivalences  DeMorgan’s Laws  (A  B)   A   B  (A  B)   A   B  Idempotent   A  A

58. Negation Normal Form NNF  A sentence  S is in negation normal form if the  is moved as far inside S as possible.  [(A   B)  C]  (A   B)   C (  A    B)   C (  A  B)   C

59. Conjunctive Normal Form CNF  A sentence S of the form A 1  A 2  …  A n where each A i, 1  i  n is of the form B 1  B 2  …  B m(i) where each B j is a literal or the negation of a literal, 1  j  m(i).

60. Disjunctive Normal Form DNF  A sentence S of the form A 1  A 2  …  A n where each A i, 1  i  n is of the form B 1  B 2  …  B m(i) where each B j is a literal or the negation of a literal, 1  j  m(i).

61. Example   [(A   B)  C]  (A   B)   C (  A    B)   C (  A  B)   C NNF, CNF (  A   C)  (B   C) NNF, DNF

62. Reiteration  P P

63. Conjunctive Elimination  P  Q... P  Elimination Q

64. Conjunctive Introduction  P... Q... P  Q  Introduction

65. Proof  Cube(b)  Tet(d) Large(d) Tet(d)  Elim 1 Large(d) Reit 2 Tet(d)  Large(d)  Intro 3, 4 2 1 3 4

66. Disjunction Introduction  P... P  Q  Introduction

67. A  B C (B  C)  D 1. A  B 2. C 3. 4. 5. B  Elim: 1 B  C  Intro: 3, 2 (B  C)  D  Intro: 4 Prove:

68. Disjunctive Elimination Proof by cases  P  Q... P … S Q … S S  Elimination Same conclusion

69. A  B B  C A  C 1. A  B 2. B  C 3. A 4. A  C  Intro: 3 5. B 6. C  Elim: 2 7. A  C  Intro: 6 8. A  C  Elim: 1, 3-4, 5-7 Prove:

70. Tautology  A sentence S is a tautology if and only if every row of its truth table assigns true to S.

71. Example

72. Logical Possibility  A sentence S is a logical possibility if there is some logically possible circumstance in which S is true.  A sentence S is a TW-possible if there is a world in which S is true.

73. Examples  Cube(b)  Large(b) b is a large cube.   (Tet(c)  Cube(c)  Dodec(c)) It is not the case that c is a tet or a cube or a dodec.  e  e e is not identical to iteself. Logically possibleTW-possible Not TW-possibleLogically possible Not Logically possibleNot TW-possible

74. Spurious Rows  A spurious row in a truth table is a row whose reference columns describe a situation or circumstance that is impossible to realize on logical grounds.

75. Example Reference Columns

76. Logical Necessity  A sentence S is a logical necessity if and only if S is true in every logical circumstance.  A sentence S is a logical necessity if and only if S is true in every non-spurious row of its truth table. Logical-Necessity TW-Necessity

77. Example Logical NecessityTW-Necessity Reference ColumnsS

78. Example Not a TW-NecessityNot a Logical Necessity Reference Columns S Can you find an example which is a TW-necessity but not a logical necessity?

79. Tet(b)   Tet(b) a=a Tet(b)  Cube(b)  Dodec(b) Cube(a)  Small(a)

80. Tautological Equivalence  Two sentences S and S’ are tautologically equivalent if and only if every row of their joint truth table assigns the same values to S and S’.

81. Example Reference ColumnsSS’

82. Logical Equivalence  Two sentences S and S’ are logically equivalent if and only if every non-spurious row of their joint truth table assigns the same values to S and S’.

83. Example Not Tautologically EquivalentLogically Equivalent Reference ColumnsSS’

84. Tautological Consequence  Sentence Q is a tautological consequence of P 1, P 2, …, P n if and only if every row that assigns true to all of the premises also assigns true to Q.

85. Example Reference ColumnsPremisesConclusion

86. Logical Consequence  Sentence Q is a logical consequence of P 1, P 2, …, P n if and only if every non-spurious row that assigns true to all of the premises also assigns true to Q.

87. Example Reference ColumnsPremisesConclusion

88. Summary  Every tautological consequence of a set of premises is a logical consequence of these premises.  Not every logical consequence of a set of premises is a tautological consequence of these premises.

89. Summary  Every tautological equivalence is a logical equivalence.  Not every logical equivalence is a tautological equivalence.

90. Summary  Every tautology is a logical necessity.  Not every logical necessity is a tautology.

91. Negative Elimination    P... P  Elimination

92. Negative Introduction Proof by contradiction   P...    P  Introduction

93. Negative Introduction Proof by contradiction  P...   P  Introduction

94. Bottom Elimination  ... P  Elimination Anything you wish follows from a contradiction

95. A  B  B A 1. A  B 2.  B 3.  A 4. A 5.   Intro: 4, 3 6. B 7.   Intro: 6, 2 8.   Elim: 1, 4-5, 6-7 9.  A  Intro: 3-8 10. A  Elim: 9 Prove:

96. Material Conditional  P  Q P is called the antecedent and Q is called the consequent.

97. Definition of 

98. English Translations  If P then Q  P implies Q  P only if Q  P is sufficient for Q  Q provided that P  Q is necessary for P  Q if P P Q Large Cube

100.  P  Q  If not P then Q Unless P, Q Q, unless P Unless Max is at home, Claire won’t get the message. b is cube, unless it is large.  Home(max)   GetsMessage(claire)  Large(b)  Cube(b)

101. Equivalencies  P  Q   P  Q   Q   P

102. Biconditional  P  Q

103. English Translations  P if and only if Q  P just in case Q  P is a necessary and sufficient condition for Q.

104. Definition of 

105. Equivalencies  P  Q  (P  Q)  (Q  P)  (P  Q)  (  P   Q) P and Q are sufficient and necessary for each other.

106. Well-formed Formula (wff)  Any atomic sentence is a wff.  If A are B are wffs then so are   A  A  B  A  B  A  B  A  B

107.  Elimination  P  Q … P … Q  Elim

108. Inference Patterns  Modus Tollens P  Q  Q  P

109. Modus Tollens

110.  Introduction  P … Q P  Q  Intro

111. Example  Prove this argument is valid. P  (Q  P)

112.  Elimination  P  Q … P … Q  Elim

113.  Introduction  P … Q Q … P P  Q  Intro

114. Example  Deduce A  C from A  B and B  C. Called Hypothetical Syllogism

115. Law of Excluded Middle  P   P A Tautology Weakest form of Taut Con Prove  from P   P (may use Law of Excluded Middle)

116. Example

117. Valid Argument  P 1 P 2 … P n Q  Q is a tautological (logical) consequence of P 1, P 2, …, P n  (P 1  P 2  …  P n )  Q is a tautology (logical necessity). Valid Argument

118. Example  Show  P is a tautological consequence of  (P  Q).  Methods of attack: Boole  Show  P is a tautological consequence of  (P  Q).  Show  (P  Q)   P is a tautology. Fitch  Show  (P  Q) is a valid argument PP

119. Tautological Consequence

120. Tautology

121. Using Fitch

122. Example  Show  P is not a tautological consequence of  (P  Q).  Method of attack: Boole  Show  P is not a tautological consequence of  (P  Q).  Show  (P  Q)   P is not a tautology. Build a world  Show  (P  Q) is an invalid argument PP

123. Not a Tautological Consequence

124. Not a Tautology

125. Build a World  Let P be assigned true and Q false.  (P  Q) is true while  P is false. premises conclusion

126. Example  Show the following argument is valid. Cube(b)  (Cube(c)  Cube(b))  Cube(c)

127. Logical Consequence

128. Logical Necessity Every non-spurious row is true! In fact, every row is true, so we have a tautology!!

129. Fitch

130. Non-consequence  Show the following argument is invalid. Cube(a)  Cube(b)  (Cube(c)  Cube(b))  Cube(c)

131. Counterexample

132. Tautological Consequence

133. Tautology

134. Clausal Form  Given a sentence S written in CNF S = ( )  ( ) ...  ( )  Convert each ( ) into a clause - a set consisting of each of the literals in the disjunction ( ). Example: S = (A)  (  B  C  D)  (  A  D) Clauses are: {A}, {  B, C, D} and {  A, D}

135. Satisfying a Set of Clauses  Assigning truth-values to each of the literals so that at least one of the literals in each clause is assigned True. Example: {{A,  C, D}, {C}, {  A}, {A, D}} is satisfied by A = False D =True C = True

136. Empty Clause  { } denoted by .  The empty clause is not satisfied by any truth-assignment. Example: Consider the fallacy A   A represented by a set of clauses {{A}, {  A}}. No truth-assignment satisfies these clauses.

137. Resolvent  A clause R is said to be a resolvent of clauses C 1 and C 2 if there is a literal in C 1 whose negation is in clause C 2 and R consists of all the remaining literals in either clause. Example: {A, B,  D } is a resolvent of the clauses {A, C,  D} and {B,  C}.

138. Resolution Theorem  For any set of clauses  that are not satisfiable, it is possible to arrive at the empty clause by applying a succession of resolution operations on Z. Example: T he fallacy (A)  (  A), whose set of clauses Z = {{A}, {  A}} is not satisfiable since, {A} {  A} 

139. Example  Show  A  (B  C)  (  C   D)  (A  D)  (  B   D) is not satisfiable. Clauses: Z = { {  A}, {B, C}, {  C,  D}, {A, D}, {  B,  D}}. {  A} {A,D} {B,C} {  C,  D} {D} {B,  D} {  B,  D} {  D} 

140. Determine Validity of Arguments using Resolution S T = S  A sentence Q is a logical consequence of a set of sentences S if and only if T = S  {  Q} is not satisfiable. Q P1P2PnP1P2Pn S

141. Logical Truths and Resolution  A sentence S, in conjunctive normal form, is a logical truth if and only if  S is not satisfiable. Example: S = P   P is a logical truth.  S =  P  P = {{  P}, {P}} {  P} {P} this implies  S is not satisfiable 

142. Example  Modus Pones P  Q or  P  Q or {  P,Q} P or {P} Q negate to get {  Q} Apply resolution: {  P,Q} {P} {Q} {  Q} 

143. PROLOG  Programs = facts + rules + queries B. Fact B:-A. Rule :-A. Query

144. Example Prolog Program  grandfather(X,Y) :- father(X,Z), father(Z,Y). grandFather(X,Y) :- father(X,Z), mother(Z,Y). mother(ann,bill). father(carl,ed). father(nick,ann). father(ed,sam).

145. What is a Horn sentence?  A positive literal is any literal that is not preceded with a . For example, Cube(b) and P are positive literals and  Tet(c) and  Q are negative literals.  A sentence S is a Horn sentence if and only if it is in conjunctive normal form and every conjunct is composed of at most one positive literal.

146. Examples  (A   B   C)  (  A   B)  (A   B  C)  (D)  (A  B)  (  C   D) Horn sentence Not Horn sentence (A   C)  (B   C)  (A   D)  (B   D) Horn sentence

147. Why are Horn sentences important?  Unlike sentences in general, Horn sentences can easily be determined to be or not to be satisfiable. Consider a sentence with 50 distinct literals. Its truth table has 2 50 rows. Attack problem with a truth table?

148. Satisfaction Algorithm for Horn sentences  (1) No conjunct in S consists of a single literal  (2) Some conjunct in S consists of a single literal S= L 1  L 2  …  L t  ( )  …  ( ) (May or may not be Satisfiable!!!) Every conjunct in S has a negative literal Take one conjunct at a time, assign a negative literal the truth-value of False. Since (  False  anything) is True this conjunct may be removed. Continue substituting this literal’s value into the remaining conjuncts. Repeat this process for the next conjunct. Do this until no literals are left to assign. (ALWAYS Satisfiable!!!)

149. Done = False; W=S; Repeat Assign L i ’s a truth-value of True (False) if positive (negative). Also replace their values inside ( )s. If all conjuncts in S are True Done = True; W is satisfiable Else if there exists a conjunct in S whose value is False Done = False; W is not satisfiable Else Remove any conjunct in S which has at least one True. What remains are ( )s with disjunction of unassigned literals. Call this sentence S’. If S’ is of form (1) Done = True; W is satisfiable Else W = S’ Until Done

150. Examples  S = (P   Q)  (  P   Q)  (  P  Q)  S = (C   A   B)  (  A  B)  (  C   B)  A  S = (  A  B)  C   B

151. Alternate Form of Horn Conjunct   A 1   A 2 ...   A n  B  (A 1  A 2 ...  A n )  B (A 1  A 2 ...  A n )  B Conditional Form of Horn sentence

152. Special Cases  No positive literal  A 1   A 2 ...   A n  No negative literals B (A 1  A 2 ...  A n )  False True  B

153. Logical Consequence Revisited SS  Theorem: A sentence Q is a logical consequence of a set of sentences S if and only if S  {  Q} is not satisfiable.

154. CNF to PROLOG   A  B   C =  (A  C)  B = (A  C)  B or A C B B :- A, C. A. C. Query: :- B.

155. Variables  Cube(b)  Cube(d) The truth of this sentence is determined by the truth values of P and Q.  Cube(x)  Cube(d) Cube(x) is neither true nor false, since x is a placeholder for the name of an object. x is said to be free or unbound. Wff and a sentence Wff but not a sentence

156. wff  If P is wff then so is  P.  If P 1, P 2, …, P n are wffs then so is (P 1  P 2  …  P n ).  If P 1, P 2, …, P n are wffs then so is (P 1  P 2  …  P n ).  If P and Q are wffs, so is (P  Q).  If P and Q are wffs, so is (P  Q).  If P is a wff and v is a variable, then  v P is a wff. Every occurrence of v is said to be bound.  If P is a wff and v is a variable, then  v P is a wff. Every occurrence of v is said to be bound.

157. Sentences  A sentence is a well-formed formula with no unbound (free) variables.  The scope of a quantifier is defined as those variables that fall under the quantifier’s influence as indicated by the enclosing parentheses.

158. Unbounded Variables   x LeftOf(x, y)   x Cube(x)  Large(x)   x Small(x)  Tet(y)   x (Cube(c)  SameRow(x, c))

159. Satisfaction of a wff  Let S(x) be a wff with free variable x. An object b is said to satisfy S(x) if and only if S(b) is a true sentence.

160. Semantics of Quantifiers   x S(x) is true if and only if there is at least one object that satisfies S(x).   x S(x) is true if and only if every object satisfies S(x). The truth-value is determined with respect to a domain of discourse!

162. Aristotelian Forms  All P’s are Q’s  x [P(x)  Q(x)] Q P 

163. Aristotelian Forms  Some P’s are Q’s  x [P(x)  Q(x)] Q P 

164. Aristotelian Forms  No P’s are Q’s  x [P(x)   Q(x)] Q P 

165. Aristotelian Forms  Some P’s are not Q’s  x [P(x)   Q(x)] Q P 

166. Translation   x (Tet(x)  LeftOf(x, c))   x [Cube(x)  BackOf(x, c)] Some Tet is to the left of c. Every cube is in back of c.

167. Quantifiers and Functions  Everyone is taller than Max’s father.  Someone’s father is taller than Max.  x Taller(x, father(max))  x Taller(father(x), max)

168. Quantifiers and Tautology  When is a sentence involving quantifiers a tautology? Example:  x Cube(x)   x  Cube(x)

169. Truth-Functional Form Algorithm  Start at the beginning of sentence S and proceed to the right. If you encounter a quantifier, underline the quantifier and the entire formula that it is applied to. If you encounter an atomic sentence simply underline it. When you come to the end of an underline assign a letter (A, B, C, …). If an underlined sentence is identical to one that occurs previously (i.e., character for character), use the same letter to label it. Determine if the sentences formed in terms of these letters is a tautology. If it is, then the original is a tautology.

170. When is a sentence a tautology?  A quantified sentence S is a tautology if and only if its truth-functional form is a tautology.

171. Example   x Cube(x)   x  Cube(x) A B Since A  B is not a tautology, then neither is the sentence above.

172. Example   x Cube(x)   [  x Cube(x)] A A A Since A   A is a tautology, then so is the sentence above.

173. Example   (Tet(d)   x Small(x))  (  Tet(d)   y Small(y)) A B A C  (A  B)  (  A  C)

175. Tautological Valid?  Is the following argument valid or invalid?  x Cube(x)  x Small(x)  x (Cube(x)  Small(x)) A B C Not tautologically valid!!! In fact, it is not valid.

176. Tautological Valid?  Is the following argument valid or invalid?  x Cube(x)   x Small(x)  x Cube(x)  x Small(x) A  B A B Is tautologically valid!!!

177. Tautological Valid?  Is the following argument valid or invalid?  x [ Cube(x)  Small(x)]  x Cube(x)  x Small(x) A B C Is not tautologically valid. It is logically valid!

178. Propositional Logic  Tautology  Tautological Consequence  Tautological Equivalence Based on the truth-functional Connectives

179. First-Order Logic  Takes into consideration all of the truth-functional connectives (      ), the identity symbol (=), and the quantifiers (  x  y).  Does not take the meanings of the names, functions or any predicates into consideration (other than =) when determining whether or not a sentence is a logical truth or whether or not an argument is valid, or whether or not two sentences are equivalent.

180. First-Order Logic  FO Validity is a sentence that can’t be false.  FO Consequence applies to an argument whose conclusion can’t be made false, while at the same time, all of its premises are simultaneously true.  FO Equivalence applies to a pair of sentences that, in all possible circumstances, are satisfied by the same objects.

181. Facts  All tautologies are FO Validities.  All FO Validities are logical truths. Tet(b)   Tet(b) Tautologies FO Validities Logical truths  x Cube(x)  Cube(a)   x LeftOf(x, x)

182. Using Boole spurious

183. Using Fitch to Prove

184.  S is not a tautology, since  A is not a tautology  Even though we know that no object can be to the left of itself. Consider replacing LeftOf with a meaningless predicate say R. Then it is not obvious that   x R(x,x) is still a true statement.  In fact, if we let R be the = predicate,   x (x = x) is false when the domain = integers. S =   x LeftOf(x, x)

185. Using Fitch to Disprove

186. Facts  All tautological consequences are FO Consequences.  All tautological equivalencies are FO Equivalencies.

187. FO Consequence   x [P(x)  Q(x)]  Q(b)  P(b) Q P b  x [Tet(x)  Large(x)]  Large(b)  Tet(b) A B CA B C C is not a tautological consequence of A and  B

188. Using Fitch

189. Not an FO Consequence   x R(x,a)  x R(b,x) R(c,d) R(a,b) Interpret R(x,y) as x loves y. The first line says, nobody loves a (Scrooge) including a (Scrooge). The second line says, b (Moriarty) loves nobody including b (Moriarty). The third line says, c (Romeo) loves d (Juliet). The conclusion says a (Scrooge) loves b (Moriarity). MoriartyScroogeRomeoJuliet

190. Replacement Method  This method is used to determine if a sentence is an FO Validity or if an argument is an FO Consequence.

191. Replacement Method  Replace all predicates in the sentence or in the argument with symbolic ones making sure that if a predicate appears more than once it is replaced with the same symbolic name.|  See if you can describe a circumstance where the sentence is false, if this is impossible then the sentence is a FO Validity.  See if you can describe a circumstance where the conclusion is false and the premises are all true. If this is impossible, then the conclusion is an FO Consequence of its premises.

192. DeMorgan’s Laws for Quantifiers   x P(x)   x [  P(x)] Nobody is P. Everyone is not P.   x P(x)   x [  P(x)] It is not the case that everyone is P. Somebody is not P. P PP

193. Aristotelian Forms Revisited  Negate: All P’s are Q’s.  x[P(x)  Q(x)]   x [  (P(x)  Q(x)) ]   x [  (  P(x)  Q(x))]   x [P(x)   Q(x)] Some P’s are not Q’s

194. A Special Form and its Equivalent  Only the Q’s are P’s  All P’s are Q’s P Q  x [P(x)  Q(x)]

196. Other Equivalences   x [P(x)  Q(x)]   x P(x)   x Q(x)   x [P(x)  Q(x)]   x P(x)   x Q(x)   x [P(x)  Q(x)]   x P(x)   x Q(x)   x [P(x)  Q(x)]   x P(x)   x Q(x)

197. Other Equivalences   x P  P, where x is not free in P   x P  P, where x is not free in P   x [P  Q(x)]  P   x Q(x)   x [P  Q(x)]  P   x Q(x)   x P(x)   y P(y)   x P(x)   y P(y)

198. Proofs Involving Quantifiers  Universal Elimination  x S(x) … S(c)  Elim

199. Example  Prove  x Cube(x)  x Large(x) Large(d)  Cube(d)]

201. Proofs Involving Quantifiers  Universal Introduction c … S(c)  x S(x)  Intro Assume c is an arbitrary element in the domain of discourse.

202. Example  Prove  x Cube(x)  x Large(x)  x [Large(x)  Cube(x)]

204. Proofs Involving Quantifiers  Existential Introduction S(c) …  x S(x)  Intro

205. Example  Prove Cube(e) Large(e)  LeftOf(e,a)  x [Cube(x)  LeftOf(x,a)]

207. Proofs Involving Quantifiers  Existential Elimination  x S(x) c S(c) … Q Q  Elim Since there exists an x such that S(x), let c designate this object. Symbol c cannot appear outside this subproof!

208. Example  Prove  x Large(x)  x Cube(x)  x [Large(x)  Cube(x)]

210. General Conditional Proof  Universal Introduction c P(c) … Q(c)  x [P(x)  Q(x)]  Intro Assume c is an arbitrary element in the domain of Discourse and assume P(c)

211. Example  Prove  x [P(x)  Q(x)]  z [Q(z)  R(z)]  x [P(x)  R(x)]

213. Order of Quantifiers vs. Meaning   x  y P(x,y) is logically equivalent to  y  x P(x,y)   x  y P(x,y) is logically equivalent to  y  x P(x,y)   x  y P(x,y) is not logically equivalent to  y  x P(x,y)

214. Careful   x  y[(Cube(x)  Cube(y))  (LeftOf(x,y)  RightOf(x,y))]   x  y[(Cube(x)  Cube(y)  x  y)  (LeftOf(x,y)  RightOf(x,y))]

215. Translation   x  y P(x,y) For each x there is a y x such that P(x,y).

216. Translation   x  y P(x,y) There us a special x such that for all y, P(x,y).

217. Example Let Q(x,y) means x + y = x – y. Determine if  y  x Q(x,y) is true or false using the integers as the domain of discourse. Let y = 0 (special element). Consider the reduced sentence,  x Q(x,0) or  x (x = x) is obviously true. Determine if  y  x Q(x,y) is true or false using the integers as the domain of discourse. Let c represent an arbitrary but fixed y, i.e., y = c. Consider the reduced sentence,  x Q(x,c) or  x (x+c = x-c) or  x (c = -c) which is false.

218. Translation  All cubes are to the left of something large.  x [Cube(x)  x is to the left of something large]  x [Cube(x)   y (Large(y)  LeftOf(x, y))]  Some cube is to the left of everything large.  x [Cube(x)  x is to the left of everything large]  x [Cube(x)   y [Large(y)  LeftOf(x,y)]]

219. Translation Worksheet Handout

220. At Least One   x P(x) There exists an x such that P(x) There is at least one x such that P(x)

221. At Least Two   x  y [P(x)  P(y)  x  y] There exists a pair (x,y) such that P(x) and P(y) and x is not equal to y. There are at least two objects that satisfy P(x).

222. At Least Three   x  y  z [P(x)  P(y)  P(z)  x  y  x  z  y  z] There are at least three objects that satisfy P(x).

223. At Least n   x 1  x 2 …  x n [P(x 1 )  P(x 2 )  …  P(x n )  x 1  x 2  x 1  x 3  …  x 1  x n  x 2  x 3  x 2  x 4  …  x 2  x n  x 3  x 4  x 3  x 5  …  x 3  x n …  x n-1  x n ] There are at least n objects that satisfy P(x). n existential quantifiers and Combinations(n,2) = n(n-1)/2 not equal to symbols

224. Exactly One   x [P(x)   y (P(y)  y  x)] There exists an x such that P(x) and for any y that satisfies P(y) that y must equal x. Exactly one x satisfies P(x).

225. Exactly Two   x  y [P(x)  P(y)  x  y   z (P(z)  (z  x  z  y)] There exists an x and a y where x  y, such that P(x) and P(y) and for any z that satisfies P(z) that z must equal either x or y. Exactly two objects satisfies P(x).

226. Exactly n   x 1  x 2 …  x n [P(x 1 )  P(x 2 )  …  P(x n )  x 1  x 2  x 1  x 3  …  x 1  x n  x 2  x 3  x 2  x 4  …  x 2  x n  x 3  x 4  x 3  x 5  …  x 3  x n …  x n-1  x n   z (P(z)  (z  x 1  z  x 2  …  z  x n ))] Exactly n objects satisfies P(x).

227. At Most One  “at most one” means either none or exactly one.  x P(x) says “nobody satisfies P(x)”  x [P(x)   y (P(y)  y  x)] says “exactly one” Combining gives  x P(x)   x [P(x)   y (P(y)  y  x)]

228. At Most One  “at most one” means  (“at least two”)  (  x  y [P(x)  P(y)  x  y])  x  y [  P(x)   P(y)  x=y]  x  y [(P(x)  P(y))  x=y] At most one objects satisfy P(x).

229. At Most Two  “at most two” means  (“at least three”)  (  x  y  z [P(x)  P(y)  P(z)  x  y  x  z  y  z])  x  y  z [  P(x)   P(y)   P(z)  x=y  x=z  y=z]  x  y  z [(P(x)  P(y)  P(z))  (x=y  x=z  y=z)] At most two objects satisfy P(x).

230. At Most n  “at most n” means  (“at least n+1”)  (  x 1  x 2 …  x n+1 [P(x 1 )  P(x 2 )  …  P(x n+1 )  x 1  x 2  x 1  x 3  …  x 1  x n+1  x 2  x 3  x 2  x 4  …  x 2  x n+1  x 3  x 4  x 3  x 5  …  x 3  x n+1 …  x n  x n+1 ]) At most n objects satisfy P(x).

231. At Most n   x 1  x 2 …  x n+1 [  P(x 1 )   P(x 2 )  …   P(x n+1 )  x 1 = x 2  x 1 = x 3  …  x 1 = x n+1  x 2 = x 3  x 2 = x 4  …  x 2 = x n+1  x 3 = x 4  x 3 = x 5  …  x 3 = x n+1 …  x n = x n+1 ])

232. At Most n   x 1  x 2 …  x n+1 [(P(x 1 )  P(x 2 )  …  P(x n+1 ))  ( x 1 = x 2  x 1 = x 3  …  x 1 = x n+1  x 2 = x 3  x 2 = x 4  …  x 2 = x n+1  x 3 = x 4  x 3 = x 5  …  x 3 = x n+1 …  x n = x n+1 )]

233. Models of Computation  Function (table look-up)  Finite-state Automata

234. XOR Table  Combine two n-bit binary numbers using exclusive or  Table size = 2 n x 2 n rows  If n = 32 then table size  1.8 x 10 19

235. XOR Automata 01 00, 11 01, 10 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1 0 1 0 0 1 0 1 0 1

236. Mathematical Preliminaries  Set – a group of objects represented as a unit. {a, b, c}= {a, b, c, b} = {b, c, a}  Sequence – a list of objects in some order. (a, b, c)  (b, c, a) (a, b, c, b)  (a, b, c)

237. Mathematical Preliminaries  Cartesian Product A x B = {(x,y) | x  A and y  B}  Generalized Cartesian Product A 1 x A 2 x … x A n = {(a 1, a 2, …, a n ) | a i  A i for each i}

238. Mathematical Preliminaries  Function f: D  R f associates each element d  D with some element r  R. This association is denoted f(d) = r. Example: f(x) = (x + 1) mod 3 D = {1, 2, 3, 4} R = {0, 1, 2}

239. Mathematical Preliminaries  Example: f(x,y) = max(x,y) f: A x A  A where A = {1, 2, 3} (1,1)  1 (1,2)  2 (1,3)  3 (2,1)  2 (2,2)  2 (2,3)  3 (3,1)  3 (3,2)  3 (3,3)  3

240. Mathematical Preliminaries  Alphabet – a finite set of symbols called .  A string s over  is a finite sequence of symbols from . The length of s is the number of symbols it contains, designated |s|.   is used to represent the empty string.  Let s and t be strings over , st is the string obtained by appending string t to the end of string s.

241. Mathematical Preliminaries  A language is a set of strings over .  Lexicographic ordering is dictionary order of strings in which shorter strings precede longer strings. Example: { 0, 1, 00, 01, 10, 11, 000, 001} – lexicographic order { 0, 00, 000, 001, 01, 1, 10, 11} – dictionary order

242. Finite Automaton  5 - tuple M = (Q, , , q 0, F) Q is a finite set of states  is an alphabet  :Q x   Q is the transition function q 0  Q is the starting state F  Q is the set of accept states.

243. JFLAP software  www.jflap.org www.jflap.org

244. Example M 1 q0q0 q1q1 q2q2 0 1 1 0 0, 1 q1q1 Q = {q 0, q 1, q 2 }  = {0, 1} F = {q 1 }

245. Language Recognized  A string w is accepted by M if and only if after processing each symbol of w M finds itself in a state belonging to F (otherwise, we say w is rejected by M).  If A is the set of all strings accepted by M, we say A is the language of machine M, denoted by L(M) = A. M is said to recognize A.

246. More Formally  Let w = w 1 w 2 …w n be a string accepted by M then there exists states r 0, r 1, …, r n in Q satisfying 1) r 0 = q 0 2)  (r i, w i+1 ) = r i+1, where i = 0, …, n-1 3) r n in F

247. Regular Language  A language is called a regular language if some finite automaton recognizes it.

248. Creating a Automaton  Given a language L over an alphabet , design a deterministic finite automaton (DFA) M such that L(M) = L.

249. Example 1  L 1 = {w | w is a string over  ={0,1} that contains an even number of 0s and an odd number of 1s }  Method: Define nodes to represent when a) both an even number of 0s and 1s have been seen in the input b) both an odd number of 0s and 1s have been seen in the input c) an even number of 0s and an odd number of 1s have been seen in the input d) an even number of 1s and an odd number of 0s have been seen in the input

250. q oe q oo q ee 1 0 1 0 0 0 1 1 q eo Example 1

252. Example 2 (complement)  L 2 = {w | w is a string over  ={0,1} that does not contain an even number of 0s and an odd number of 1s } = L 1

253. q eo 1 0 1 0 0 0 1 1 Example 2 q ee q oe q oo

254. Example 3  L 3 = { w | w is a string over {0, 1} such that |w|  3} = { , 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111} Starting state must be final

255. Example 3 q0q0 q1q1 q2q2 0, 1 q3q3 q4q4

256. Example 4  L 4 = { w | w is a string over  ={0,1} such that w contains the substring 11 } = { w | w = x11y, where x and y are strings over  ={0, 1}}

257. Example 4 q0q0 q1q1 0 1 0 1 q2q2 0, 1 f s

258. Machine M accepts string w  If there exists a sequence of states r 0, r 1, …, r n in Q such that 1) r 0 = q 0 2)  (r i, w i+1 ) = r i+1, for i=0,…,n-1 3) r n in F Note: w = w 1 w 2 …w n

259. Regular Languages  Machine M recognizes language A if A = {w| M accepts w}  A language is called regular if some finite automaton recognizes it.

260. Regular Languages  Machine M recognizes language A if A = {w| M accepts w}  A language is called regular if some finite automaton recognizes it.

261. Regular Operations  Let A and B be languages. Union A  B = { x | x in A or x in B} Concatenation A  B = {xy | x in A and y in B} Star A* = {x 1 x 2 …x k | k  0 and each x j in A} Note:  is always a member of A*.

262. Regular languages are closed under union  Let A 1 and A 2 be regular languages. We want to show A 1  A 2 is a regular language. Since A 1 and A 2 are regular languages there exists a finite automaton M 1 and there exists a finite automaton M 2 such that M 1 recognizes A 1 and M 2 recognizes A 2. Assume M 1 = (Q 1, ,  1, q 1, F 1 ) and M 2 = (Q 2, ,  2, q 2, F 2 ) It suffices to create a finite automaton M that recognizes A 1  A 2.

263. Continue …  Let a be a symbol in  and states r 1 in Q 1 and r 2 in Q 2. Define M = (Q, , , q 0, F) where Q = Q 1 x Q 2 states  ((r 1, r 2 ), a) = (  1 (r 1, a),  2 (r 2, a)) transition function q 0 = (q 1, q 2 ) start state F = (F 1 x Q 2 )  (Q 1 x F 2 ) final states

264. Example u v 0,1 0 1 M1M1 x z 0 M2M2 y 1 1 0 M = (Q, , , q 0, F) Q = {(u,x), (v,x), (u,y), (v,y), (u,z), (v,z)} states q0 = (u, x) start state F = {(v,x), (v,y), (v,z)}  {(u,z), (v,z)} = {(v,x), (v,y), (v,z), (u,z)} final states  01 (u,x) (   (u),  2 (x)) = (v,z)(   (u),  2 (x)) = (v,y) (u,y) (   (u),  2 (y)) = (v,y)(   (u),  2 (y)) = (v,z) (u,z) (   (u),  2 (z)) = (v,y) (v,x) (   (v),  2 (x)) = (v,z)(   (v),  2 (x)) = (u,y) (v,y) (   (v),  2 (y)) = (v,y)(   (v),  2 (y)) = (u,z) (v,z) (   (v),  2 (z)) = (v,y)(   (v),  2 (z)) = (u,y)

265. Regular languages are closed under concatenation  Let A 1 and A 2 be regular languages. We want to show A 1  A 2 is a regular language. Since A 1 and A 2 are regular languages there exists a finite automaton M 1 and there exists a finite automaton M 2 such that M 1 recognizes A 1 and M 2 recognizes A 2. Assume M 1 = (Q 1, ,  1, q 1, F 1 ) and M 2 = (Q 2, ,  2, q 2, F 2 ) It suffices to create a finite automaton M that recognizes A 1  A 2. There is a problem since M doesn’t know where to subdivide the input string into the part accepted by M 1 and the remaining part that will be accepted by M 2. We will return to this later.

266. Regular Expressions  R is a regular expression if 1) x for some x in  (note: regular expression x represents language {x}) 2)  (empty string) (note: regular expression  represents language {  }) 3)  (empty set) 4) (R 1  R 2 ) where R 1 and R 2 are regular expressions 5) (R 1  R 2 ) where R 1 and R 2 are regular expressions 6) (R 1 *) where R 1 is a regular expression If R is a regular expression then L(R) is the language of R.

267. Non-Deterministic Automaton  NFAs generalize DFAs. In a DFA, each state has exactly one transition for each symbol in the alphabet. In an NFA, at any state there may be zero or more transitions for a symbol in the alphabet. In a DFA, a label on a transition arrow is a symbol in the alphabet. In an NFA, a label on a transition arrow is a symbol in the alphabet or .

268. Example q1q1 q2q2 q3q3 0, 1 1 0,  1 0, 1 q4q4

269. q1q1 q1q1 q1q1 q2q2 q1q1 q3q3 q2q2 q4q4 q3q3 q3q3 q1q1 q4q4 q4q4 q3q3 q2q2 q1q1 q3q3 q1q1 q4q4 q4q4 0 111 0 0 1 1 1 1 00 00 1 1 1 11 Input: 010110

270. Non-Deterministic Finite Automaton  N = (Q, , , q 0, F) Q is a finite set of states  is a finite alphabet  : Q x (   {  } )   (Q) F  Q is a set of accept states  (Q) is the power set of Q = {X| X  Q}

271. Machine N accepts string w  If there exists a sequence of states r 0, r 1, …, r n in Q such that 1) r 0 = q 0 2) r i+1 in  (r i, w i+1 ) for i=0,…,n-1 3) r n in F Note: w = w 1 w 2 …w n  (r i, w i+1 ) is a set of states

272. Are NFAs more powerful than DFAs?  Every deterministic finite automaton has an equivalent non-deterministic finite automaton. (see next slide)  Every non-deterministic finite automaton has an equivalent deterministic finite automaton.

273. Non-deterministic? q0q0 q1q1 0 1 0 1 q2q2 0, 1 Non-deterministic interpretationDeterministic interpretation

274. Deterministic Equivalent? 1 2 3 b a, b a  a with  edgeswithout  edges 12 0 3 0 1 NN Next four slides color coded with respect to these examples  = {0, 1}  = {a, b}

275. DFA from NFA Construction  Assume no  edges. Let N = (Q, , , q 0, F) be an NFA that recognizes language A. We construct a DFA called M = (Q’, ,  ’, q 0 ’, F’) 1) Q’ =  (Q) 2) For R in Q’ and a in  let  ’(R,a) = {q in Q| q in  (r,a) for some r in R} =   (r,a) r in R Q’ = {{}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}  ’({1,2},b) =  (1,b)   (2,b) = {2}  {3} = {2,3} Q’ = {{}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}  ’({2, 3},0) =  (2,0)   (3,0) = {}  {} = 

276. Continued … 3) q 0 ’ = { q 0 } 4) F’ = {R in Q’| R contains an accept state of N} (do blue example) Assume  edges, then we need these modifications. Let R be a state of M. Define E(R) = {q in Q| q can be reached from R traveling along 0 or more  edges} Modify  ’(R,a) = {q in Q| q in E(  (r,a)) for some r in R} =  E(  (r,a)) transition function r in R  ’({1,2},b) = E(  (1,b))  E(  (2,b)) = E({2})  E({3}) = {2,3}  ’({3},a) = E(  (3,a)) = E({1}) = {1,3} q 0 ’ = E({q 0 }) start state (do red example)

277. Find Deterministic Equivalent Deterministic Equivalent Start state q 0 ’= {1} Final states F’ = { {2,3} } without  edges 12 0 3 0 1 N

278. Solution {1} { } {2,3} 0, 1 0 1 0 1 M

279. Find Deterministic Equivalent 1 2 3 b a, b a  a Deterministic Equivalent Start state q 0 ’= E({1}) = {1,3} Final states F’ = { {1}, {1,2}, {1,3}, {1,2,3} } N

280. Final Solution a {1,3} {2} b {3} a { } b b {2,3} a b a, b a b a {1,2,3} b {1,2} a, b {1} a M

281. Regular languages are closed under union  Let A 1 and A 2 be regular languages. We want to show A 1  A 2 is a regular language. Since A 1 and A 2 are regular languages there exists an NFA N 1 and there exists an NFA N 2 such that N 1 recognizes A 1 and N 2 recognizes A 2. Assume N 1 = (Q 1, ,  1, q 1, F 1 ) and N 2 = (Q 2, ,  2, q 2, F 2 ) It suffices to create a NFA N that recognizes A 1  A 2.

282. Construction of NFA N1N1 N2N2 q1q1 q2q2 q0q0   N = (Q, , , q 0, F) Q = {q 0 }  Q 1  Q 2 F = F 1  F 2  1 (q,a) q in Q 1  (q,a) =  2 (q,a) q in Q 2 {q 1, q 2 } q = q 0 and a=  { } q = q 0 and a  N q1q1 q2q2

283. Example Union u v 0,1 0 1 M1M1 x z 0 M2M2 y 1 1 0 q0q0  See slide 260 (union) for original machines

284. Regular languages are closed under concatenation  Let A 1 and A 2 be regular languages. We want to show A 1  A 2 is a regular language. Since A 1 and A 2 are regular languages there exists an NFA N 1 and there exists an NFA N 2 such that N 1 recognizes A 1 and N 2 recognizes A 2. Assume N 1 = (Q 1, ,  1, q 1, F 1 ) and N 2 = (Q 2, ,  2, q 2, F 2 ) It suffices to create a NFA N that recognizes A 1  A 2.

285. Construction of NFA N1N1 N2N2   N = (Q, , , q 1, F 2 ) Q = Q 1  Q 2 and q 1 (start state) and F = F 2  1 (q,a) q in Q 1 and q not in F 1  (q,a) =  2 (q,a) q in Q 2  1 (q,a)  {q 2 } q in F 1 and a=  (add  edges to q 2 keep  edges from final states in N 1 )  1 (q,a) q in F 1 and a   (keep non-  edges from final states in N1) q1q1 q2q2 q1q1 q2q2 N   Not 

286. Example Concatenation u v 0,1 0 1 M1M1 x z 0 M2M2 y 1 1 0 

287. Regular languages are closed under the star operation  Let A be a regular language. We want to show A* is a regular language. Since A is regular language there exists an NFA N 1 such that N 1 recognizes A. Assume N 1 = (Q 1, ,  1, q 1, F 1 ) It suffices to create a NFA N that recognizes A*.

288. Construct NFA N1N1   N N = (Q, , , q 0, F) Q = {q 0 }  Q 1 F = F 1  {q 0 } and q 0 the start state  1 (q,a) q in Q 1 and q not in F 1  (q,a) =  1 (q,a) q in F 1 and a   1 (q,a)  {q 1 } q in F 1 and a=  {q 1 } q = q 0 and a=  { } q= q 0 and a  q0q0 q1q1  q1q1

289. Find Star of M 2 x z 0 0,1 M2M2 y 1 1 0 x z 0 M2*M2* y 1 1 0 q0q0  

290. Regular Expressions  R is a regular expression if 1) x for some x in  (note: regular expression x represents language {x}) 2)  (empty string) (note: regular expression  represents language {  }) 3)  (empty set) 4) (R 1  R 2 ) where R 1 and R 2 are regular expressions 5) (R 1  R 2 ) where R 1 and R 2 are regular expressions 6) (R 1 *) where R 1 is a regular expression If R is a regular expression then L(R) is the language of R.

291. Examples  0*0 {w| w contains at least one zero}   * = {  }  11  00 = {11, 00}  0  *1 = {w| w begins with a 0 and ends in a 1}  (01)* = { , 01, 0101, 010101, 01010101, …}  1*0 = {w| w contains any number of 1s followed by exactly one 0}

292. Using Regular Expressions

293. Beginning or End?

294. Regular Expressions vs. Regular Languages  A language is regular if and only if some regular expression describes it. Part a) If a regular expression describes a language then it is regular. Part b) If a language is regular then a regular expression describes it.

295. x  NFA that recognizes {x} x

296.   NFA that recognizes {  }

297.   NFA that recognizes 

298. R 1  R 2, R 1  R 2, or R 1 *  Construct a machine the same way we did to show regular languages are closed under , , or *.

299. NFA to recognize (0  11)* 01 11  0 11   

300. 0 11    0 11      

301. Part b) If a language is regular then a regular expression describes it.  Properties of GNFA 1) The start state has transition arrows going to every other state but no arrows coming in from any other state. 2) There is one accept state, and it has arrows coming in from every other state but no arrows going to any other state. The start state is not the same as the final state. 3) Except for the start and accept states, one arrow goes from every state to every other state and also from each state to itself. 4) The labels on each edge is a regular expression.

302. Example GNFA start accept ab*  b ab  ba b* ab aa (aa)* a*