CS375: Logic and Theory of Computing Fuhua (Frank) Cheng Department of Computer Science University of Kentucky 9/26/2014
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Table of Contents: Week 1: Preliminaries (set algebra, relations, functions) (read Chapters 1-4)
Weeks 2&3: Propositional Logic (Chapter 6) Weeks 4&5: Predicate Logic (Chapter 7) Weeks 6&7: Computing with Logic (Chapter 9) Week 8: Algebraic Structures (Chapter 10) Weeks 9&10: Regular Languages, Finite
Antomata (Chapter 11) 9/26/2014
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Table of Contents (conti): Weeks 11&12: Context-Free Languages,
Pushdown Automata (Chapters 12) Weeks 13&14: Turing Machines (Chapter 13)
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3. Predicate Logic
quantifiers
, , ...
Goal: use the notions and notations of first-order predicate calculus to analyze and symbolize a wider variety of statements and arguments that can be done with propositional logic. cannot have other predicates or functions as arguments 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Predicate calculus studies internal structure of sentences where subjects are applied to predicates existentially or universally. A predicate describes a property of the subject (s) of a sentence. If p is a predicate that describes a property of x, then we write p(x). Example. x is prime: p(x) Example. x is a parent of y: q(x, y) 9/26/2014
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Indicates disjunction
3. Predicate Logic – First-order Predicate Calculus Existential Quantifier “there exists an x such that p(x)” : ∃x p(x) Example. What does p(1) ∨ p(2) ∨ p(3) mean? ∃x p(x), where x ∈ {1, 2, 3} Quiz. p(y, a) ∨ p(y, b) ∨ p(y, c) = ? Answer: ∃x p(y, x), where x ∈ {a, b, c}. Indicates conjunction
Universal Quantifier “for every x, p(x)” : ∀x p(x) Example. What does p(1) ∧ p(2) ∧ p(3) mean? ∀x p(x), where x ∈ {1, 2, 3} 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Quiz. p(a, x) ∧ p(b, x) ∧ p(c, x) = ? Answer: ∀y p(y, x), where y ∈ {a, b, c} Quiz. If x, y ∈ {0, 1}, then ∃x ∀y p(x, y, z) = ? Answer 1: ∀y p(0, y, z) ∨ ∀y p(1, y, z) = (p(0, 0, z) ∧ p(0, 1, z)) ∨ (p(1, 0, z) ∧ p(1, 1, z)) Answer 2: ∃x (p(x, 0, z) ∧ p(x, 1, z)) = (p(0, 0, z) ∧ p(0, 1, z)) ∨ (p(1, 0, z) ∧ p(1, 1, z)) 9/26/2014
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non-logical things
3. Predicate Logic – First-order Predicate Calculus Syntax of wffs in first-order predicate calculus Terms: constants a, b, c, …, variables x, y, z, …. and function symbols applied to constants, variables or even other functions ƒ(a), g(x, ƒ(y)), h(c, z), … . Atoms: predicate symbols applied to terms p(x), q(a, ƒ(x)), ... . Wffs: either atoms or if U and V are wffs then the following expressions are also wffs: ¬ U, U ∧ V, U ∨ V, U → V, ∃x U, ∀x U, and (U). 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Inductive definition of a wff: 1. any atom is a wff 2. if U and V are wffs and x is a variable, then the following expressions are also wffs: ¬ U, U ∧ V, U ∨ V, U → V, ∃x U, ∀x U, and (U). Hierarchy in the absence of parentheses: ¬, ∃x, ∀x (highest, group rightmost operator with smallest wff to its right) ∧ ∨ → (lowest, and it is left associative) 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Example. Right most
Right most
∀x ¬ ∃y p(x, y) → ∀x q(x) = ? = ∀x ¬ (∃y p(x, y)) → (∀x q(x)) =∀x (¬ (∃y p(x, y))) → (∀x q(x))
= (∀x (¬ (∃y p(x, y)))) → (∀x q(x))
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3. Predicate Logic – First-order Predicate Calculus Scope, Bound, and Free Scope of ∃x in ∃x W is W. Scope of ∀x in ∀x W is W. An occurrence of x is bound if it occurs in either ∃x or ∀x or in their scope. Otherwise the occurrence of x is free. Example.
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Bound occurrences of variables are not assigned values. Why?
3. Predicate Logic – First-order Predicate Calculus
Interpretation for a first-order wff consists of a domain D (nonempty set) and an assignment of the symbols of the wff as follows: 1. Predicate letters are assigned to relations over D. 2. Function letters are assigned to functions over D. 3. Constants are assigned to elements of D. 4. Free occurrences of variables are assigned to elements of D. Example. W = p(x). An interpretation can be defined by letting D = N, p(x) means x is odd, and x = 4. 9/26/2014
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Also write W(d)=W(x/d)
3. Predicate Logic – First-order Predicate Calculus Notation: x is free in W, d ∈ D, then W(x/d): wff obtained by replacing all free occurrences of x by d. After the substitution, we get a proposition
Example. Let W = ∀y (p(x, y) → q(x)).
Then W(x/d) =?
∀y (p(d, y) → q(d)) 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Truth Value of a Wff The truth value of a wff w.r.t. an interpretation with domain D is obtained by recursively applying the following rules: 1. An atom (e.g., p(x), q(a, f(x))) has the truth value of the proposition obtained from its interpretation. 2. Truth values for ¬ U, U ∧ V, U ∨ V, U → V are obtained by applying truth tables for ¬, ∧, ∨, → to the truth values for U and V. 3. ∀x W is true iff W(x/d) is true for all d ∈ D. 4. ∃x W is true iff W(x/d) is true for some d ∈ D. 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus If a wff is true with respect to an interpretation I, we say the wff is true for I. Example. (same domain, different assignments) Let W = p(x). We can define an interpretation I by letting D = N, p(x) means x is odd, and x = 4. W is false for I If we let J be the same as I except that we assign x = 3, then W is true for J 9/26/2014
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called a model in this case
3. Predicate Logic – First-order Predicate Calculus Example. W = ∀x (p(x) → q(x, y)). Two interpretations: 1. I : D = {0, 1}, p(0) = True, p(1) = False, q(0, 1) = True, otherwise q(x, y) is false, and y = 1. Then W is true for I because ∀x (p(x) → q(x, 1)) = (p(0) → q(0, 1)) ∧ (p(1) → q(1, 1)) = (True → True) ∧ (False → False) ≡ True ∧ True ≡ True. 2. J : any domain D, p is true, q is false, and y any value of D, then W is false for J because ∀x (p(x) → q(x, 1)) = (True → False) ∧ … ≡ False ∧ … ≡ False. University of Kentucky 16 9/26/2014
3. Predicate Logic – First-order Predicate Calculus Example. W = ∃x (p(x) ∧ q(x)). Two interpretations : 1. I : D = N, p(x) : x is prime, q(x) : x is odd. Then W is true for I because, for example, p(3) ∧ q(3) = True ∧ True ≡ True. counter model
2. I : D = N, p(x) : x is even, q(x) : x is odd. Then W is false for I because, for example, p(3) ∧ q(3) = False ∧ True ≡ False. 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Example. W = ∀x (g(x, c) → ∃y (p(y) ∧ d(y, x))). Here are two interpretations of W: 1. I : D = {a}, p(a) = False, d(a, a) = True, g(a, a) = True, and c = a. Then W is false for I because W = g(a, a) → p(a) ∧ d(a, a) ≡ True → False ∧ True ≡ False. 2. I : D = N, g(x, c) means x > c, p(y) means y is prime, d(y, x) means y divides x, and c = 1. This gives the sentence, “every natural number greater than 1 has a prime divisor,” which is known to be true. (I is a model) 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Models and Countermodels An interpretation that makes a wff true is called a model. Otherwise, called a countermodel. Example. Let W be the following wff. ∀x (r(x, a) ∧ ¬ p(x) → ∃y (r(x, y) ∧ r(y, a) ∧ d(y, x))). Let I be the interpretation defined by D = N, r(x, y) means x > y, p(x) means x is prime, d(y, x) means y divides x, and a = 1. Is I a model or a countermodel for W? Answer: We get the statement, “every natural number x > 1 that is not a prime has a divisor y between 1 and x,” which is known to be true. So I is aUniversity model of W. of Kentucky 19 9/26/2014
3. Predicate Logic – First-order Predicate Calculus Validity A wff is valid if every interpretation is a model. Otherwise the wff is invalid. A wff is unsatisfiable if every interpretation is a countermodel. Otherwise the wff is satisfiable. Quiz. Every wff has exactly two of these four properties. What are the possible pairs of properties that a wff can have? Answer: Satisfiable valid Unsatisfiable 9/26/2014
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We cannot check every interpretation (there are too many)
3. Predicate Logic – First-order Predicate Calculus Proofs of Validity or Unsatisfiability
Example. Prove the following wff is valid. ∃x (A(x) ∧ B(x)) → ∃x A(x) ∧ ∃x B(x). Proof: Let I be an interpretation with domain D for the wff and assume that the antecedent is true for I. Then A(d) ∧ B(d) is true for I for some d ∈ D. So both A(d) and B(d) are true for I. Therefore both ∃x A(x) and ∃x B(x) are true for I. So the consequent is true for I. Thus I is a model for the wff. Since I was an arbitrary interpretation for the wff, every interpretation for the wff is a model. Therefore, the wff is valid. QED. University of Kentucky 21 9/26/2014
3. Predicate Logic – First-order Predicate Calculus Quiz. Is ∃x A(x) ∧ ∃x B(x) → ∃x (A(x) ∧ B(x)) valid? Answer: No. e.g., let D = N, A(x) mean x is even, and B(x) mean x is odd. Some Valid Conditionals Whose Converses Are Invalid (a) ∀x A(x) → ∃x A(x) . (b) ∃x (A(x) ∧ B(x)) → ∃x A(x) ∧ ∃x B(x). (c) ∀x A(x) ∨ ∀x B(x) → ∀x (A(x) ∨ B(x)). (d) ∀x (A(x) → B(x)) → (∀x A(x) → ∀x B(x)). (e) ∃x ∀y P(x, y) → ∀y ∃x P(x, y). 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Quiz. Find a countermodel for each converse of the above wffs.
(a) ∃x A(x) → ∀x A(x) A(x): x is prime
(b) ∃x A(x) ∧ ∃x B(x) → ∃x (A(x) ∧ B(x)) A(x): x is even; B(x): x is odd
(c) ∀x (A(x) ∨ B(x)) → ∀x A(x) ∨ ∀x B(x) A(x): x is even; B(x): x is odd 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Quiz. Find a countermodel for each converse of the above wffs.
(d) (∀x A(x) → ∀x B(x)) → ∀x (A(x) → B(x)) D={a, b}; A(a)=True, A(b)=False; B(a)=False, B(b)=True
(e) ∀y ∃x P(x, y) → ∃x ∀y P(x, y) D=N ; P(x,y): x=y+1
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3. Predicate Logic – First-order Predicate Calculus Closures (Observation: there are two interesting transformations that we can apply to any wff containing free variables, one is to universally quantify each free variable, and the other one is to existentially quantify each free variable.) When we quantify free variables of a wff, what properties of the original wff would be preserved?
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3. Predicate Logic – First-order Predicate Calculus Closures (Observation: there are two interesting transformations that we can apply to any wff containing free variables, one is to universally quantify each free variable, and the other one is to existentially quantify each free variable.) When we quantify free variables of a wff, what properties of the original wff would be preserved? satisfiability 9/26/2014
validity
unsatisfiability
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invalidity
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3. Predicate Logic – First-order Predicate Calculus Closures (Observation: there are two interesting transformations that we can apply to any wff containing free variables, one is to universally quantify each free variable, and the other one is to existentially quantify each free variable. When we quantify free variables of a wff, what properties of the original wff would be preserved? satisfiability
unsatisfiability 9/26/2014
validity
invalidity University of Kentucky
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3. Predicate Logic – First-order Predicate Calculus Closures (Observation: there are two interesting transformations that we can apply to any wff containing free variables, one is to universally quantify each free variable, and the other one is to existentially quantify each free variable. When we quantify free variables of a wff, what properties of the original wff would be preserved? - validity is preserved if we universally quantify the free variables, and - unsatisfiability is preserved if we existentially quantify the free variables ) 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Closures Let W be a wff with free variables x1, …, xn. Then we have the following definitions: ∀x1 … ∀xn W is the universal closure of W. ∃x1 … ∃xn W is the existential closure of W. ( Why do we want to define these closures? Note that ∀x1 … ∀xn W has no free variables, hence, can no longer do any new assignment, or the status of W is fixed, you can not set a new goal for W any more. ) 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Sometimes a wff and one of its closures have the same properties and sometimes they don’t, as we can see in the following examples. Example. The wff p(x) ∨ ¬ p(y) is satisfiable and invalid. The universal closure ∀x ∀y (p(x) ∨ ¬ p(y)) is satisfiable and
invalid. The existential closure ∃x ∃y (p(x) ∨ ¬ p(y)) is valid.
Example. The wff p(x) ∧ ¬ p(y) is satisfiable and invalid. The universal closure ∀x ∀y (p(x) ∧ ¬ p(y)) is unsatisfiable. The existential closure ∃x ∃y (p(x) ∧ ¬ p(y)) is satisfiable and invalid. 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Closure Properties 1. A wff is valid if and only if its universal closure is valid. 2. A wff is unsatisfiable if and only if its existential closure is
unsatisfiable . Example. The wff p(x) → ∃y p(y) is valid and its universal closure ∀x (p(x) → ∃y p(y)) is also valid. Example. The wff p(x) ∧ ∀y ¬ p(y) is unsatisfiable and its existential closure ∃x (p(x) ∧ ∀y ¬ p(y) )) is also unsatisfiable. 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Closure Properties 1. A wff is valid if and only if its universal closure is valid.
Proof. First, prove that if x is a free variable in a wff W, then W is valid iff x W is valid. (->) Let I be an interpretation with domain D for x W . If I is not a model for x W , then there exist d in D such that W(x/d) is false for I. This being the case, we can define an interpretation J for W by letting J be I, with the free variable x assigned the element d in D. Since W is valid, it follows that W(x/d) is true for J. But W(x/d) w.r.t. J is the same as W(x/d) w.r.t. I, which is false. This contradiction shows that I is a model for x W . So x W is valid. 9/26/2014
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3. Predicate Logic – First-order Predicate Calculus Closure Properties 1. A wff is valid if and only if its universal closure is valid.
Proof (conti). (
