PHIL 12A Syntax and Semantics Philosophy Problems Set 8 and 9

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Please complete all questions in problem set 9. You may also need problem set 8, lecture slides, and the textbook called Logic in Action. I will provide all of them in attachments.

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PHIL 12A – Spring 2019 Problem Set 9 100 points 1 Syntax and Semantics of Monadic Predicate Logic 1.1 1.1.1 Identity and Substitution The Identity Predicate 1. 4 points Exercise 4.36 of Logic in Action. Write out your formula with no abbreviations. 2. 4 points Exercise 4.38(2) of Logic in Action. Demonstrate the difference in meaning by providing a model in which the two formulas have different truth values. 1.1.2 Substitution 3. 10 points total; 2.5 points each part Give a recursive definition of the set of terms t that are substitutable for a variable x in a formula ϕ. That is, fill in the question marks in the following template (note that we do not assume that the metavariables ‘x’ and ‘y’ refer to distinct variables, so you need to analyze the case where x = y and the case where x 6= y in part (d)): (a) t is always substitutable for x in an atomic formula P (u) or s = ˙ t (because there are no quantifiers in an atomic formula to capture a variable in t); (b) t is substitutable for x in ¬ϕ iff ? (c) for # ∈ {∧, ∨, →, ↔}, t is substitutable for x in (ϕ # ψ) iff ? (d) for Q ∈ {∀, ∃}, t is substitutable for x in Qyϕ iff ? 2 Syntax and Semantics of Predicate Logic 4. These problems concern translating English sentences into the language of predicate logic: (a) 3 points Assume the domain of discourse is all dogs. Translate the following sentences into predicate logic: (Use f for Fido, r for Rover, and L for loves) • Fido is loved by everyone. • Rover loves everyone who loves Fido. • Fido and Rover love each other. 1 (b) 5 points For each of the following, specify an appropriate domain of discourse, specify a translation key, and translate into predicate logic. (Note: you do not have to understand what a sentence means before you can attempt to translate it.) • Cats that love dogs don’t taunt dogs. • There is a greatest natural number. • Friend’s of Fido’s friends are Rover’s friends. • There is no largest even number. • Every apple is tastier than every orange. (c) 4 points Translate the following sentences into predicate logical formulas. Assume the domain of discourse is cats and dogs. • Some cat doesn’t love all dogs. • Every dog who loves all cats doesn’t love every dog. • Every cat who loves a dog is purred at by some cat. • No cat loves a cat who loves a dog, except for the cats who don’t love dogs. 5. These problems concern describing situations using the language of predicate logic: (a) 4 points Exercise 4.18 of Logic in Action. (b) 5 points Exercise 4.22 of Logic in Action. 6. These problems concern describing binary relations: (a) 3 points The formula ∀x∀y∀z((R(x, y)∧R(y, z)) → R(x, z)) expresses the transitivity of the relation R. Which of the following relations are transitive? • Being a cousin of ... on the set of human beings. • Being a sibling of ... on the set of human beings. • The divides relation on the set of natural numbers. (Recall that a natural number n divides a natural number m if there is some natural k such that n × k = m) (b) 5 points Exercise 4.24 of Logic in Action. (c) 5 points Exercise 4.25 of Logic in Action. 7. 12 points The following natural language sentences are ambiguous: they can be interpreted in at least two different ways. For each one of them, provide two predicate logic formulas that correspond to two different readings (i.e., four formulas in total). (a) There is a dog who likes only cats who like only dogs. (b) Fido and Rover are liked by some cat. Decide for each of the four predicate logic formulas that you have given whether they are true or false in the model in Figure 1. Explain your answers. 2 : dog •: cat • → : • likes  f :Fido, r:Rover, t:Tibbles r f t Figure 1: Model for Problem 7. Figure 2: Model for Problem 8. 8. 10 points Consider the model in Figure 2. Let the set of solid dots be the interpretation of the unary predicate symbol P . Let the edge relation of the graph be the interpretation of the binary predicate symbol R. What does ∀x(P (x) → (∃y(¬P (y) ∧ ∀z(R(y, z) → P (z))))) mean? Is this true in the model? 9. 8 points Consider the four models in Figure 3. Let R be the binary predicate interpreted by the arrow in the diagram. Give for each model a predicate logic formula that is true in that model but not in the other three. 3 Figure 3: Models for Problem 9. 10. These problems concern validity and consequence in predicate logic: (a) 8 points Which of the following statements are true? (i)  ∃xF (x) ∨ ∀x¬F (x) (ii)  ∀xF (x) ∨ ∀x¬F (x) (iii)  ∀xF (x) ∨ ∃x¬¬F (x) (iv)  ∀x∀y∀z((R(x, y) ∧ R(y, z)) → R(x, z)) (v)  ∀x∀yR(x, y) → ∀xR(x, x) (vi)  ∀x∀yR(x, y) → ∀zR(z, z) (vii)  ∀xR(x, x) → ∀x∃yR(y, x) (viii)  ∃xR(x, x) → ∀x∃yR(y, y) (b) 6 points Which of the following statements are true? If the statement is false, provide a counterexample. (i) ∀x∀yR(y, x)  ∀y∀xR(y, x) (ii) ∀xR(x, x)  ∀x∃yR(x, y) (iii) ∃x∃yR(x, y)  ∃xR(x, x) (iv) ∀x∃yR(x, y)  ∀xR(x, x) (v) ∃x∀yR(y, x)  ∃x∀yR(x, y) (vi) ∃xR(x, x)  ∀x∃yR(y, y) (c) 4 points Which of the following statements are true? If the statement is false, provide a counterexample. (i) {∀x∀y(R(x, y) → R(y, x)), R(a, b)}  R(a, a) (ii) {∀x∀y(R(x, y) → R(y, x)), R(a, b)}  R(b, b) (iii) {∀x∀y(R(x, y) → R(y, x)), ¬R(b, a)}  ¬R(a, b) (iv) {∀x∀y(R(x, y) → R(y, x)), R(b, c), ¬R(a, c)}  ¬R(a, b) 11. 10 points Extra Credit In an extra credit problem for Problem Set 8, we showed that if a formula of pure monadic predicate logic is satisfiable, then it is satisfiable in a 4 model of size at most 2k where k is the number of predicates occurring in the formula. (We stated this in terms of ‘falsifiable’, but ϕ is satisfiable iff ¬ϕ is falsifiable, so we can state it either way.) Let us now consider a sentence with a binary predicate symbol: ∀x∃yR(x, y) ∧ ∀x¬R(x, x) ∧ ∀x∀y∀z((R(x, y) ∧ R(y, z)) → R(x, z)). Is this sentence satisfiable? If so, are there any finite models that make the sentence true? If so, give an example. If not, explain why not. 12. 8 points Extra Credit A formula of predicate logic is in prenex normal form iff it is of the form Q1 x1 . . . Qn xn ψ where Qi ∈ {∀, ∃} and ψ does not contain quantifiers. The following is an important fact about predicate logic. Proposition 1. Every formula ϕ of predicate logic is equivalent to a formula in prenex normal form. To get a feel for this, find prenex equivalents of the following formulas: (a) ∀x(P (x) → ∀yR(x, y)); (b) ∃x(∀yR(y, x) → P (x)); (c) ∃x(P (x) → ∃yR(x, y)); (d) ∀x(∃yR(y, x) → P (x)). 5 PHIL 12A – Spring 2019 Problem Set 8 60 points 1 Syntax and Semantics of Monadic Predicate Logic 1.1 1.1.1 Pure Monadic Predicate Logic Pure Monadic Predicate Logic I 1. 2 points Exercise 4.6 in Logic in Action. 2. 2 points Exercise 4.7 in Logic in Action. 3. 10 points total; each part 2.5 points Give a recursive definition of the set of free variables of a formula ϕ, i.e., those variables with at least one free occurrence in ϕ, as described in the slides. To do so, fill in the question marks in the following template (note that we do not assume that the metavariables ‘x’ and ‘y’ refer to distinct variables, so you need to analyze the case where x = y and the case where x 6= y): (a) x is a free variable of P (y) iff ? (b) x is a free variable of ¬ϕ iff ? (c) for # ∈ {∧, ∨, →, ↔}, x is a free variable of (ϕ # ψ) iff ? (d) for Q ∈ {∀, ∃}, x is a free variable of Qyϕ iff ? 4. 10 points Given a model M = (D, I), a subset A ⊆ D is said to be definable in the language of pure monadic predicate logic iff there is a pure monadic formula ϕ with exactly one free variable x such that for some variable assignment g, A = {d ∈ D | M g[x:=d] ϕ}, i.e., A is exactly the set of objects d such that ϕ is true under the variable assignment that maps x to d. For example, in the model in Figure 1, the set {2, 4} is defined by the formula ¬Sophomore(x). For every subset of the domain of that model, say whether it is definable by a formula or not. If it is definable, give a formula that defines it. 1.1.2 Pure Monadic Predicate Logic II 5. 3 points Exercise 4.14 of Logic in Action. 6. 3 points Exercise 4.15 of Logic in Action. 1 Language: Sophomore Student Faculty I I Invitee I 1 I 3 Model: 2 4 Figure 1: Model for Problem 4. 7. 12 points Determine whether each of the following formulas are true in the model in Figure 2, where R stands for red, G stands for green, B stands for blue, P stands for purple, S stands for square, and C stands for circle: (a) practice ∃x(R(x) ∧ C(x)); (b) practice ∀x(C(x) ∨ S(x)); (c) ∃xG(x) ∨ ∃xC(x); (d) ∃xR(x) ∧ ∃xC(x); (e) ∀xC(x) ∨ ∀xS(x); (f) ∃x(G(x) ∨ C(x)); (g) ∀x(¬P (x) → S(x)); (h) ∀x(S(x) → (P (x) ∨ R(x))). Figure 2: 8. practice Exercise 4.17 of Logic in Action. 9. 8 points For each of the following sentences, say whether or not it is valid. If it is not valid, present a model in which it is false. (a) practice ∀x(P (x) ∧ Q(x)) → (∀xP (x) ∧ ∀xQ(x)); (b) ∀x(P (x) ∨ Q(x)) → (∀xP (x) ∨ ∀xQ(x)); (c) ∀xP (x) → ∃xP (x); 2 (d) ∃x∃y(P (x) ∧ Q(y)) → ∃z(P (z) ∧ Q(z)); (e) ∃x(P (x) → ∀xP (x)). 10. 5 points An important fact about pure monadic predicate logic is the following. Proposition 1. Each sentence ϕ of pure monadic predicate logic is equivalent to a sentence ϕ? containing the same predicate symbols as ϕ and only one variable. To get a feel for this, explain the following equivalences by appealing to the semantics of monadic predicate logic: (a) ∀x∃y(P (x) ∧ Q(y)) is equivalent to ∀xP (x) ∧ ∃yQ(y); (b) ∀xP (x) ∧ ∃yQ(y) is equivalent to ∀zP (z) ∧ ∃zQ(z). 11. 5 points Extra Credit. In the slides, we mentioned the important lemma that if a formula ϕ of pure monadic predicate logic is not valid, then it is falsified in a model on the domain D = {1, . . . , 2k } where k is the number of predicate symbols appearing in ϕ. In this extra credit problem, we will show that if ϕ is not valid, then it is falsified in a model where D has at most 2k elements (from which the lemma easily follows). Suppose M = (D, I) is a model and g an assignment such that M 2g ϕ. We will shrink M to a model with no more than 2k elements that also falsifies ϕ. Let P red(ϕ) be the set of all unary predicates that occur in ϕ. For example, if ϕ is ∀x(P1 (x) → P3 (x)), then P red(ϕ) = {P1 , P3 }. We assume that P red(ϕ) has k elements. It follows that P red(ϕ) has 2k subsets. For each d ∈ D, let db = {d0 ∈ D | for all P ∈ P red(ϕ) : d ∈ I(P ) iff d0 ∈ I(P )}. That is, db is the set of all objects that are “indistinguishable” from d in M using predicates that appear in ϕ. The idea behind the proof is that such indistinguishable b (For example, in the model in objects should be collapsed into a single object d. Figure 4, the objects 1 and 3 cannot be distinguished from each other by any of the four predicates, so they can be collapsed into a single object b 1=b 3.) There can be at k most 2 distinct sets db because each distinct db defines a distinct subset of P red(ϕ), namely the subset {P ∈ P red(ϕ) | d ∈ I(P )}, and there are only 2k subsets of c = (D, b I): b P red(ϕ). This leads us to the definition of our collapsed model M b = {db | d ∈ D}; • D b ) iff d ∈ I(P ); • for each predicate P ∈ P red(ϕ), db ∈ I(P b ) = ∅. • for each predicate P 6∈ P red(ϕ), I(P c by Given any variable assignment g for M, define the variable assignment gb for M [ gb(xi ) = g(x i ). 3 It follows that for any variable x and d ∈ D, \ b g[x := d] = gb[x := d]. It also follows that for any subformula ψ of ϕ and variable assignment g, c gb ψ iff M g ψ. M Prove this ‘iff’ by induction on ψ. Give only the base case where ψ is an atomic formula of the form P (x) and the inductive step where ψ is of the form ∃xα. Use the equations given above (plus the inductive hypothesis for the ∃ case). c 2gb ϕ, From what you proved and our initial assumption that M 2g ϕ, it follows that M which shows that ϕ is indeed falsified in a model with at most 2k elements! 1.2 1.2.1 Constants and Functions Constants 12. 5 points Let ϕ be a formula in which the constant c appears but the variable x does not appear. Prove that if ϕ is valid, then ∀xϕcx is valid, where ϕcx is the formula obtained from ϕ by replacing all occurrences of c with x. Intuitively: “if ϕ holds of an arbitrary element, then it holds of everything.” In your argument you may assume (without proof) the fact that for any model M = (D, I) and variable assignment g for M: M g ϕ iff M g[x:=JcKgI ] ϕcx . You may also assume that if two models M and N differ only on the interpretation of a constant that does not appear in a formula ψ, then M g ψ iff N g ψ. 1.2.2 Function Symbols 13. 5 points Extra Credit In the slides, we mentioned the fact that there is an algorithm that converts any formula ϕ of monadic predicate logic with unary function symbols into a formula ϕ0 of monadic predicate logic without function symbols such that ϕ is valid iff ϕ0 is valid. To give you a feel for how this could be true, let ψ be a formula of monadic predicate logic, P a unary predicate, and f a unary function symbol not occurring in ψ. Prove that the formula ∀z(ψ(z) → P (f (z))) is valid if and only if the formula ∀x∃y(P (y) ↔ Q(x)) → ∀z(ψ(z) → Q(z)) is valid, where Q is a new predicate symbol that does not appear in ψ. Suggestion: suppose ∀z(ψ(z) → P (f (z))) is not valid, so it is falsified in some model 4 M. Then define a model M0 that differs from M at most in the interpretation of 0 the predicate Q (i.e., just define I M (Q) for us), such that M0 falsifies ∀x∃y(P (y) ↔ Q(x)) → ∀z(ψ(z) → Q(z)) (prove it). In the other direction, suppose ∀x∃y(P (y) ↔ Q(x)) → ∀z(ψ(z) → Q(z)) is not valid, so it is falsified in some model N . Then define a model N 0 that differs from N at most in the interpretation of the function symbol 0 f (i.e., just define I N (f ) for us) such that N 0 falsifies ∀z(ψ(z) → P (f (z))) (prove it). 5 Logic in Action –New Edition, November 23, 2016– Johan van Benthem, Hans van Ditmarsch, Jan van Eijck, Jan Jaspars 0-2 Contents 1 General Introduction 1-1 1.1 Inference, Observation, Communication . . . . . . . . . . . . . . . . . . 1-1 1.2 The Origins of Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-3 1.3 Uses of Inference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-5 1.4 Logic and Other Disciplines . . . . . . . . . . . . . . . . . . . . . . . . 1-9 1.5 Overview of the Course . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-11 Classical Systems 2-1 2 2-1 Propositional Logic 2.1 Reasoning in Daily Life . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1 2.2 Inference Patterns, Validity, and Invalidity . . . . . . . . . . . . . . . . . 2-3 2.3 Classification, Consequence, and Update . . . . . . . . . . . . . . . . . . 2-5 2.4 The Language of Propositional Logic . . . . . . . . . . . . . . . . . . . 2-8 2.5 Semantic Situations, Truth Tables, Binary Arithmetic . . . . . . . . . . . 2-13 2.6 Valid Consequence and Consistency . . . . . . . . . . . . . . . . . . . . 2-18 2.7 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-22 2.8 Information Update . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-24 2.9 Expressiveness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-26 2.10 Outlook — Logic, Mathematics, Computation . . . . . . . . . . . . . . . 2-28 2.11 Outlook — Logic and Practice . . . . . . . . . . . . . . . . . . . . . . . 2-32 2.12 Outlook — Logic and Cognition . . . . . . . . . . . . . . . . . . . . . . 2-34 0-3 0-4 3 4 CONTENTS Syllogistic Reasoning 3-1 3.1 Reasoning About Predicates and Classes . . . . . . . . . . . . . . . . . . 3-1 3.2 The Language of Syllogistics . . . . . . . . . . . . . . . . . . . . . . . . 3-4 3.3 Sets and Operations on Sets . . . . . . . . . . . . . . . . . . . . . . . . . 3-5 3.4 Syllogistic Situations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-10 3.5 Validity Checking for Syllogistic Forms . . . . . . . . . . . . . . . . . . 3-12 3.6 Outlook — Satisfiability and Complexity . . . . . . . . . . . . . . . . . 3-18 3.7 Outlook — The Syllogistic and Actual Reasoning . . . . . . . . . . . . . 3-21 The World According to Predicate Logic 4-1 4.1 Learning the Language by Doing . . . . . . . . . . . . . . . . . . . . . . 4-2 4.2 Practising Translations . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-8 4.3 Reasoning Patterns with Quantifiers . . . . . . . . . . . . . . . . . . . . 4-13 4.4 Formulas, Situations and Pictures . . . . . . . . . . . . . . . . . . . . . . 4-17 4.5 Syntax of Predicate Logic . . . . . . . . . . . . . . . . . . . . . . . . . . 4-25 4.6 Semantics of Predicate Logic . . . . . . . . . . . . . . . . . . . . . . . . 4-30 4.7 Valid Laws and Valid Consequence . . . . . . . . . . . . . . . . . . . . . 4-35 4.8 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-38 4.9 Identity, Function Symbols, Algebraic Reasoning . . . . . . . . . . . . . 4-41 4.10 Outlook — Mathematical Background . . . . . . . . . . . . . . . . . . . 4-46 4.11 Outlook — Computational Connection . . . . . . . . . . . . . . . . . . . 4-49 4.12 Outlook — Predicate Logic and Philosophy . . . . . . . . . . . . . . . . 4-51 Knowledge, Action, Interaction 5 Logic, Information and Knowledge 4-57 5-1 5.1 Logic and Information Flow . . . . . . . . . . . . . . . . . . . . . . . . 5-1 5.2 Information versus Uncertainty . . . . . . . . . . . . . . . . . . . . . . . 5-3 5.3 Modeling Information Change . . . . . . . . . . . . . . . . . . . . . . . 5-10 5.4 The Language of Epistemic Logic . . . . . . . . . . . . . . . . . . . . . 5-12 5.5 Models and Semantics for Epistemic Logic . . . . . . . . . . . . . . . . 5-15 CONTENTS 0-5 5.6 Valid Consequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-21 5.7 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-25 5.8 Information Update . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-30 5.9 The Logic of Public Announcement . . . . . . . . . . . . . . . . . . . . 5-37 5.10 Outlook — Information, Knowledge, and Belief . . . . . . . . . . . . . . 5-42 5.11 Outlook – Social Knowledge . . . . . . . . . . . . . . . . . . . . . . . . 5-44 5.12 Outlook – Secrecy and Security . . . . . . . . . . . . . . . . . . . . . . 5-47 6 Logic and Action 6-1 6.1 Actions in General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-1 6.2 Sequence, Choice, Repetition, Test . . . . . . . . . . . . . . . . . . . . . 6-6 6.3 Viewing Actions as Relations . . . . . . . . . . . . . . . . . . . . . . . . 6-10 6.4 Operations on Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-13 6.5 Combining Propositional Logic and Actions: PDL 6.6 Transition Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-20 6.7 Semantics of PDL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-23 6.8 Axiomatisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-26 6.9 Expressive power: defining programming constructs . . . . . . . . . . . . 6-30 . . . . . . . . . . . . 6-17 6.10 Outlook — Programs and Computation . . . . . . . . . . . . . . . . . . 6-31 6.11 Outlook — Equivalence of Programs and Bisimulation . . ...
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Hadizza8690
School: Carnegie Mellon University

Please find the answer in the attachment below, (i was not able to the finish number 11 since i couldn't understand how it connects with the notes- i hope you understand, there was barely enough time to finish everything credibly). Anyway of you have any questions and clarrificarions please feel free to reach out

Surname 1

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Problem Set 9
1

Syntax and Semantics of Monadic Predicate Logic

1.1

Identity and Substitution

1.1.1 The Identity Predicate

1. 4 points Exercise 4.36 of Logic in Action. Write out your formula with no
abbreviations.

 ∃x∃y∃z (¬x = y ∧ ¬x = z ∧ ¬x = y ∧ ∀v (P v ↔ (v = y ∨ v =
x ∨ v = z)))
2. 4 points Exercise 4.38(2) of Logic in Action. Demonstrate the difference in
meaning by providing a model in which the two formulas have different
truth values.

 In a model with two elements one of which is A but not B and
the other is B but not A the formulas have a different truth
value: ∃!x(Ax ∨ Bx) is false and ∃!x Ax ∨ ∃!x Bx is true.

Surname 2

1.1.2 Substitution
3. 10 points total; 2.5 points each part Give a recursive definition of the set of
terms t that are substitutable for a variable x in a formula ϕ. That is, fill in
the question marks in the following template (note that we do not assume
that the metavariables ‘x’ and ‘y’ refer to distinct variables, so you need to
analyze the case where x = y and the case where x ≠ y in part (d)):
 t is always substitutable for x in an atomic formula P(u) or s
˙= t (because there are no quantifiers in an atomic formula to
capture a variable in t);
Solution
For any terms s and t and variable x, S xt recursive as follows:
Consider the term Biomom (x) which can be substituted to get
biomom (biomom (Alex)
 If c ∈ Const, then cxt = c;
 if y ∈ Var and y ≠ x, then yxt = y;
 Xxt = t;
 if f ∈ Func and u ∈ Term, then f (u) xt = f (u xt ).

 t is substitutable for x in ¬ϕ iff ?
 P(s) xt = P(sxt ), where Sxt is defined as before
 For # ∈ {∧, ∨, →, ↔}, t is substitutable for x in (ϕ # ψ) iff?
 (ϕ) xt = ϕxt and (ϕ # ꝕ) xt = (ϕ xt # ꝕ xt)
 For Q ∈ {∀, ∃}, t is substitutable for x in Qyϕ iff?
 (∀yP(x) ∧ ∃xQ(x)) xalex = (∀yP(x)) xalex ∧ (∃xQ(x)) xalex

2

Syntax and Semantics of Predicate Logic
4. These problems concern translating English sentences into the language of
predicate logic:
(a) 3 points assume the domain of discourse is all dogs (d). Translate the
following sentences into predicate logic: (Use f for Fido, r for Rover,
and L for loves)
 Fido is loved by everyone.
∃x →Lxf
 Rover loves everyone who
love Fido.
∀x (Rdf→LRd)
 Fido and Rover love each
other.
Lfr ∧ Lrf

Surname 3

(b) 5 points for each of the following, specify an appropriate domain of
discourse, specify a translation key, and translate into predicate logic.
(Note: you do not have to understand what a sentence means before you
can attempt to translate it.)

 Cats that love dogs don’t
taunt dogs.
∀x ((Cx → Lyx) ¬ ∀y(Ty)
 There is a greatest natural
number.
∃x∀yG xy
 Friends of Fido’s friends are
Rover’s friends.
∀xy ((Ffx ∧ Fxr) →Ffr)
 There is no largest even number.
∀x(Ex→∃y(Ey ∧ Sxy ))
 Every apple is tastier than every orange.
∀xy(Ax >Oy)

(c) 4 points translate the following sentences into predicate logical
formulas. Assume the domain of discourse is cats and dogs.
 Some cat doesn’t love all dogs.
∃X ∃y (Cx ∧ Dy ∧ ¬ Lxy)
 Every dog who loves all cats doesn’t
love every dog.
∀x((C x ∧ ∃y(Dy ∧ Lxy)) → ∃z(Dz ∧ Lzx))
 Every cat who loves a dog is purred at
by some cat.
∀x((C x ∧ ∀y (Dv→Lxy)) → ∃z(Cz ∧ Lz...

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