COMP 409: Homework 2

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1 COMP 409: Homework 2 Sample Solutions 1 Propositional semantics 1. Implement a truth evaluator eval(φ, τ), which evaluates whether a formula φ holds for a truth assignment τ. Use the evaluator to answer the question. The code is available upon request. The results are: α(i 1 ) = β(i 1 ) = γ(i 1 ) = δ(i 1 ) = 0 ψ(i 1 ) = 1 α(i 2 ) = 0; β(i 2 ) = γ(i 2 ) = δ(i 2 ) = 1 ψ(i 2 ) = 1 2 Validity 2.1 Prove Validity (((p q) p) p) - Peirce s Law Let τ 2 P rop. Case 1: τ = p and τ = q τ = (p q) τ = p and τ = (p q) τ = ((p q) p) τ = p and τ = ((p q) p) τ = (((p q) p) p) Case 2: τ = p and τ q τ (p q) τ (p q) τ = ((p q) p) τ = p and τ = ((p q) p) τ = (((p q) p) p) Case 3: τ p τ = (p q) τ p and τ = (p q) τ ((p q) p) τ ((p q) p) τ = (((p q) p) p) false p - Ex Falso Quolibet Let τ 2 P rop. τ false τ = (false p) 1

2 2.1.3 (( p) p) - Excluded Middle Let τ 2 P rop. If τ = p, then τ = (p ( p)). If τ p, then τ = ( p) = τ = (p ( p)). Therefore = (p ( p)) ((p q) p) - Weak Excluded Middle Let τ 2 P rop. We know that τ = ((p q) p) τ = (p q) or τ = p. Case 1: τ = p, then τ = ((p q) p). Case 2: τ p then τ = (p q). Therefore, τ = ((p q) p). 2.2 Disjunctional Propositional Completeness Suppose α and β were both not valid. This would mean that there exist assignments τ 1 2 AP(α) and τ 2 2 AP(β) s.t τ 1 α and τ 2 β. Note that τ 1 τ 2 =. Take the assignment τ 2 AP(α β) where τ = τ 1 τ 2. Since, because AP(alpha) AP(beta) =, τ AP(α) = τ 1 and τ AP(β) = τ 2. Since = (α β), τ = (α β) = either τ = α or τ = β since it is sufficient to look at the relevant propositions only, = either τ AP(α) = α or τ AP(β) = β = either τ 1 = α or τ 2 = β. We know that this is not possible. Therefore our assumption must be faulty. Hence either = α or = β. 3 Logical implication 1. Analyze the binary connectives,,, to see whether they are commutative or associative. Prove your claims. We will show that,, and are commutative and associative. The basic strategy in all cases is to use the commutativity and the associativity of the set-theoretic operations,, and set equality (=). An alternative approach is to use truth tables. is commutative. Proof: models(θ ψ) = models(θ) models(ψ) = models(ψ) models(θ) = models(ψ θ) (since set intersection is commutative). By definition of == it follows that (θ ψ) == (ψ θ). is associative. Proof: models((φ θ) ψ) = models(φ θ) models(ψ) = models(φ) models(θ) models(ψ) = models(φ) models(θ ψ) = models(φ (θ ψ)). By definition of == it follows that ((φ θ) ψ) == (φ (θ ψ)). 2

3 is commutative. Proof: models(θ ψ) = models(θ) models(ψ) = models(ψ) models(θ) = models(ψ θ) (since set union is commutative). By definition it follows that (θ ψ) == (ψ θ). is associative. Proof: models((φ θ) ψ) = models(φ θ) models(ψ) = models(φ) models(θ) models(ψ) = models(φ) models(θ ψ) = models(φ (θ ψ)) (since set union is associative). By definition it follows that ((φ θ) ψ) == (φ (θ ψ)). is NOT commutative. Proof by counterexample. Take P rop = {p, q, r}. Notice that models(q p) = (2 AP models(q)) models(p) = {, {p}} {{p}, {p, q}} = {, {p}, {p, q}}. Similarly, models(p q) = {, {q}, {p, q}}. Since models(p q) models(q p) then (p q) is NOT logically equivalent to (q p). is NOT associative. Proof by counterexample (derent approach than above). Take P rop = {p, q, r} and a world τ = (i.e. all propositions are false). From the EE view, we have ((p q) r)(τ) = 0 and (p (q r))(τ) = 1. Then, ((p q) r) (p (q r)) is not valid, so by definition of valid ((p q) r) is NOT logically equivalent to (p (q r)). is commutative. Proof: models(θ ψ) = ((2 P rop models(θ)) (2 P rop models(ψ))) (models(θ) models(ψ)) = ((2 P rop models(ψ)) (2 P rop models(θ))) (models(ψ) models(θ)) = models(ψ θ) by set intersection and union commutativity. By definition of == it follows that (θ ψ) == (ψ θ). is associative. Proof: models((φ θ) ψ) = models(φ θ) models(ψ)) ((2 P rop models(φ θ)) (2 P rop models(ψ))). Expanding models(φ θ) = (models(φ) models(θ)) ((2 P rop models(φ)) (2 prop models(θ))) and regrouping using associativity from set theory, we get this is equal to (models(φ) models(θ ψ)) ((2 P rop models(φ)) (2 prop models(θ ψ))) = models(φ (θ ψ)). By definition of == it follows that 2. Prove de Morgan s Laws = (( (p q)) (( p) ( q)) = (( (p q)) (( p) ( q)) ((φ θ) ψ) == (φ (θ ψ)). 3

4 Let φ = (( (p q)) (( p) ( q) ψ = (( (p q)) (( p) ( q)) Proof: By definition, φ is valid models(φ) = 2 P rop models(φ) = 2 AP (φ) (by Relevance Lemma) φ(τ) = 1, τ 2 AP (φ) (by Lemma from class) Similarly, ψ is valid models(ψ) = 2 P rop models(ψ) = 2 AP (ψ) (by Relevance Lemma) ψ(τ) = 1, τ 2 AP (ψ) (by Lemma from class) The following table shows that models(φ) = 2 AP (φ) and models(ψ) = 2 AP (ψ). p q (( (p q)) (( p) ( q) φ (( (p q)) (( p) ( q)) ψ Show that logical and material equivalence coincide, that is, φ == ψ = (φ ψ). By definition, φ == ψ φ = ψ and ψ = φ (by definition of == ) = (φ ψ) and = (φ ψ) (by Lemma 1 in Lecture 8 ) = (φ ψ) (by definition of ) 4

5 4. Show that φ = ψ (φ ψ) is not satisfiable. Proof: By definition, (φ ψ) is not satisfiable (φ ψ) is valid = (φ ψ) = ( φ ψ) (by de Morgan s Law) = (φ ψ) (by definition of ) φ = ψ (by definition of logical implication) 5. Prove that = is reflexive, transitive, but not symmetric. Prove that == is reflexive, transitive, and symmetric. Notice that = is not a relation between 2 Prop and Form, nor between 2 Prop and 2 Form (otherwise neither of the properties would make sense). The only option that is left is = 2 Form 2 Form. Let Φ, Ψ 2 Form = is reflexive. Proof: since models(φ) models(φ) then Φ = Φ by definition. = is transitive. Proof: if Φ = Ψ and Ψ = Θ then models(φ) models(ψ) and models(ψ) models(θ). Since subset is a transitive relation, models(ψ) models(θ), so Φ = Θ. = is NOT symmetric. Proof by counterexample. Let Φ = {(p q)}, Ψ = {p}, P rop = {p, q}. Then, models(φ) = {{p, q}} and models(ψ) = {{p}, {p, q}}. Since models(φ) models(ψ) but models(ψ) models(φ), it follows from the definition of = that Φ = Ψ but Ψ = Φ. In the case of == we also have sets of formulas on both sides. == is reflexive. Proof: Since models(φ) = models(φ) then Φ == Φ by definition. == is transitive. Proof: if Φ == Ψ and Ψ == Θ then models(φ) = models(ψ) and models(ψ) = models(θ), so models(φ) = models(θ) and by definition Φ == Θ. == is symmetric. Proof: Φ == Ψ models(φ) = models(ψ) models(ψ) = models(φ) Ψ == Φ. 6. Show that if φ and ψ are logically equivalent and φ is valid, then ψ is valid. Restating the problem formally, we show that if φ == ψ and = φ, then = ψ (by definition of logical equivalence and logical validity). 5

6 Proof: By definition of ==, φ == ψ models(φ) = models(ψ). By definition of logically valid, = φ models(φ) = 2 P rop. Thus, models(ψ) = 2 P rop and hence = ψ. 7. Which of the formulas implies the other? Let φ = (p (q r)) ψ = ((p (q r)) (( p) ( q) ( r))) Then, models(φ) = {{p}, {p, q, r}, {q}, {r}} and models(ψ) = {{p, q, r}, }. Since neither models(φ) models(ψ) nor models(ψ) models(φ) then neither φ = ψ nor ψ = φ. 8. Show that Σ {α} = β Σ = (α β). First, we need some help: Lemma 1. If A and B are sets, then A (A B) B. Proof. Suppose that a A. We must show that a A (A B) B. Either a B or a B. In the first case a (A B), in the second, a B. In either case a A (A B) B, which is what we had to prove. Proof of the main problem. Σ {α} = β models(σ {α}) models(β) models(σ) models(α) models(β) (models(σ) models(α)) (2 P rop models(α)) (2 P rop models(α)) models(β) (1) (union the same set to each side) models(σ) (2 P rop models(α)) models(β) (by Lemma) (2) models(σ) models(α β) (by defn of ) Σ = (α β) (by defn of =) Note that the step from (1) to (2) is justified by Lemma 1, but this is not the case when we traverse the proof in the opposite direction. For this we need another lemma, which we state and prove below. Lemma 2. Let Σ, U, A and B be sets. If Σ (U A) B, then (Σ A) (U A) (U A) B. 6

7 Proof. Since (Σ A) Σ and Σ (U A) B, we can conclude that (Σ A) (U A) B. Clearly (U A) (U A) B, therefore the left-hand side is the union of two sets which are both contained in the right-hand side. The proof of Lemma 2 concludes the proof of the main problem. 9. Prove or refute the following statements. If either Σ = α or Σ = β then Σ = (α β). TRUE. Proof: If either models(σ) models(α) or Then Σ = (α β). models(σ) models(β) then models(σ) models(α) models(β) models(α β) by definition. If Σ = (α β), then either Σ = α or Σ = β. FALSE. Proof by counterexample: Take P rop = {p, q}, Σ = {p}, α = (p q), β = ( (p q)) Then: models(σ) = {{p}, {p, q}} models(α) = {{p, q}} models(β) = {, {p}, {q}} Here, models(σ) models(α) models(β) Σ = (α β) but models(σ) models(α) therefore Σ = α and models(σ) models(β) therefore Σ = β Which completes the proof. 10. A set Σ of formulas is independent if Σ {σ} = σ, σ Σ. Let Σ = {σ 1, σ 2,..., σ n } be a finite set of formulas, if Σ is independent already then we have the final result we wanted. If Σ is not independent, then it means that there is at least one σ i, such that σ 1 σ 2... σ i 1 σ i+1... σ n = σ i. That is σ i is implied by the conjunction all of the formulas from Σ except σ i itself. In order to make Σ an independent set of formulas, we can eliminate σ i from the set. The set Σ {σ i } will still be equivalent to the original set. We can continue to do this until there are no more dependent formulas in 7

8 the resulting set. On the other hand, if Σ is an infinite set of formulas (Σ = {σ 1, σ 2,...}), such as: σ 1 = p 1 σ 2 = p 1 p 2.. σ n = p 1 p 2... p n.. In this case, the set itself is not independent. If we take a finite subset of the formulas, it s not going to be equivalent, since an infinite number of formulas can not be generated from the finite subset (any formula with atomic propositions with subscripts greater than the greatest subscript present in the subset). If we take an infinite subset, say by eliminating σ i from the set, we are still able to generate the missing formula from any σ j, j > i. Take σ i+1, for example, if p 1 p 2... p i p i+1 holds then p 1 p 2... p i holds also. This infinite subset is equivalent, but it is not independent since it possesses the same properties as the original set. Any other infinite subset is also equivalent to the original set, due to the same reasoning. 4 Substitutions If φ is valid, then φ[p θ] is valid. 1. If = ϕ, then = ϕ[p θ]. Proof: Suppose that = ϕ. Let τ 2 Prop. We wish to show that τ(ϕ[p θ]) = 1. Let σ 2 Prop be identical to τ except that σ(p) = τ(θ). First we show that τ(ϕ[p θ]) = σ(ϕ). Case: ϕ = p. Then τ(p[p θ]) = τ(θ) = σ(p). Case ϕ = q p. Then τ(q[p θ]) = τ(q) = σ(q). Case ϕ = ( ψ). Then τ(( ψ)[p θ]) = τ( (ψ[p θ])) = (τ(ψ[p θ])) = (σ(ψ)) = σ( (ψ)) = σ( ψ). Case ϕ = (θ ψ). Then τ((θ ψ)[p θ]) = τ(θ[p θ] ψ[p θ]) = (τ(θ[p θ]), τ(ψ[p θ])) = (σ(θ), σ(ψ)) = σ(θ ψ). Now, since every truth assignment makes ϕ true (because it is valid), σ in particular makes it true, and so τ makes ϕ[p θ] true. 8

9 2. If α is equivalent to β, then ϕ[p α] is equivalent to ϕ[p β]. Proof: Let τ be an arbitrary truth assignment. We wish to show ϕ[p α] is equivalent to ϕ[p β]. We proceed by structural induction. Case ϕ = p. Then τ(p[p α]) = τ(α) = τ(β) (because α, β equivalent) = τ(p[p β]). Case ϕ = q p. Then τ(q[p α]) = τ(q) = τ(q[p β]). Case ϕ = ( θ). Then τ(( θ)[p α]) = τ( θ[p α]) = (τ(θ[p α])) = (τ(θ[p β])) (by the inductive hypothesis) = τ( θ[p β]) = τ(( θ)[p β]). Case ϕ = (θ ψ). Then τ((θ ψ)[p α]) = τ(θ[p α] ψ[p α]) = (τ(θ[p α]), τ(ψ[p α])) = (τ(θ[p β]), τ(ψ[p β])) (I.H.) = τ(θ[p β] ψ[p β]) = τ((θ ψ)[p β]). 5 Puzzle Let the assertion Box i has the gold be represented by p i and the assertion The clue on Box i says the truth be represented by q i.. The following facts are given: Only one box have the gold: og = ((p 1 p 2 p 3 ) ( (p 1 p 2 ) (p 1 p 3 ) (p 2 p 3 ))). Box1: b1 = (q 1 p 1 ). Box2: b2 = (q 2 p 2 ). Box3: b3 = (q 3 p 2 ). Only one clue is true: ot = (q 1 q 2 q 3 ) ( (q 1 q 2 ) (q 1 q 3 ) (q 2 q 3 )). We need a satisfying assignment to og b1 b2 b3 ot to solve the puzzle. The only satisfying assignment would be p 1 = 1, p 2 = 0, p 3 = 0, q 1 = 0, q 2 = 1, q 3 = 0, which means box 1 contains the gold. 9