# symbolic logic homework help

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• Q1:Exercise 144. Let R be a binary relation symbol and f a unary function symbol. def 1. Show that the sentence of Vr R(x, f(x)) → Vry R(x, y) is valid. Do it in two different ways: (a) proof-theoretically, p, using natural deduction, and (b) semantically, = 4. def 2. Show that the sentence Valy R(x,y) → Vr R(x, f(x)) is not valid. Note that is just the converse implication of . Hint: Try a semantic approach, i.e., show. You need to define a structure A so that the left-hand side of "" in is true in A but the right-hand side of "" is false in A. 3. Conclude that Vrly R(x, y) and Vr R(x, f(x)) are not equivalent first-order wff's. Remark: Despite the conclusion in part 3, Proposition 142 asserts Vray R(x, y) and Vr R(x, f(x)) are equisatisfiable, i.e., if there is a model for one, then there is a model for the other, and vice- versa. 42 Review the definition of o[a] in Section B.3.See Answer
• Q2:36 The next proof makes explicit reference to notions in topology. As indicated in the preceding alternative proof, a truth assignment can be denoted by a path in the full binary tree Tall, now viewed as an w-sequence in the product space (T,F). (We write for the first infinite ordinal, which is the set of natural numbers listed in their standard order.) We view {T,F) as the underlying space of a product topology ({T,F)", O), thus making every truth assignment a "point" in that topology. O is a family of open sets that define the topology, which are in this case subsets of points in {T, F)~ and satisfy the usual requirements of a topology: • The empty set and the full space {T,F) are in O. • O is closed under arbitrary unions, • O is closed under finite intersections. We can define a subset UC {T,F}" to be open iff there is a finite set of indices IC such that: U = II { Ai | i €w and A; C {T,F}} where A. = {T,F) for every i Ew-I. In words, in the infinite product U = Ao × A₁ × --- × A; x--, for all but finitely many indices i it is the case that A; = {T,F). A set U C {T,F)" is closed iff it is the complement of an open APPENDIX F. ALTERNATIVE COMPACTNESS PROOFS set. In the case of the product topology, every open subset UC (T,F) is also closed, and thus called clopen. Let A be a subset of (T,F). An open covering of A is a family of open sets {U; | i I} CO such that A CU{U; i 1}. And A is said compact if every open covering of A has a finite subcovering; i.e., there exists a finite subfamily U₁₁, UU₁ of (U; | i I} such that A (U₁, UU, U...UU.). By Tychonoff's Theorem in topology, the product topology ({T,F), O) is compact, which means that every open covering of a subset of points AC {T,F) has a finite subcovering. Alternative Proof II of Theorem 2 (Compactness for Propositional Logic). Again here, we only need to consider the non-trivial implication "+": If I is finitely satisfiable, then I is satisfiable. For every propositional wffy, let A, C {T,F}" be the collection of all points/truth assignments that satisfy p. The set A, is a closed (and open) subset of {T,F}" in the topology ({T,F)",0), which follows from the fact that only mentions finitely many propositional variables. It is easy to check that, for every finite subset A of I, if A is satisfiable, then {A | A} is not empty. Hence, the family of closed sets {Apyr} satisfies the finite intersection property. Moreover, the product topology ({T.F),O) is compact, as noted above. Hence, {A, ET) is not empty, which is the desired conclusion. 0/nExercise 147. This exercise is couched in a language which is a little more familiar to computer scientists. Given a set IWFF₁(P), the binary tree T(I) induced by I' is defined in the 35 proof above. In T(I), every finite path is a finite sequence, and every infinite path is an infinite sequence, where all the entries are in {T,F). The set of all such finite sequences is denoted {T,F)", and the set of all such infinite sequences is denoted {T,F). The root node of T(I) is represented by the empty sequence e, every leaf node is connected to the root node by a finite path, and every infinite path is one that can be traversed without ever reaching a leaf node. An arbitrary binary tree can therefore be represented by a subset of {T, F}*U{T,F}", satisfying two conditions: • U is prefix-closed, i.e., for every (possibly infinite) path ₁ € {T,F)* U{T,F) and every finite path #₂ € {T,F)", if ₁ EU and ₂ is a prefix of ₁, then #₂ EU. • For every finite path = {T,F)", it holds that T U iff F EU, i.e., every non-leaf node has two immediate successors. Given an arbitrary IWFF (P), let T(T) be the subset of infinite paths in T(I): T(T) = T(r)n(T.F)". By the preceding proof, T. (T) ‡ Ø iff the set I is satisfiable. Let To, F₁, and I2, be defined as follows: To = {-P₁ ^P₂}, T₁ = {P₁ → (P₂A---A-PE) | A>2}, del 1₂ = {~-P₁ → (P²2 A---A-¬-Pk) |>2}. There are two parts in this exercise: 1. Define the sets of infinite paths T. (To), T.(₁UF₂), and T(Tour₁ UF₂) as subsets of {T,F). In your answers, use the notational conventions of regular languages and regular w-languages that we have used earlier in the presentation of this exercise. 2. For each of the three sets defined in part 1, explain how they indicate whether the corre- sponding set of wff's is satisfiable or not. You will find it useful to consult Figure F.1. 0See Answer
• Q3:In this assignment, you will use what you learn in Logic module to: Create a compound statement that contains at least 2 connectives (negation, conjunction, disjunction, conditional, biconditional). Express the compound statement in the symbolic form. Construct a truth table for the compound statement that you created.See Answer
• Q4:Exercises 1.5 1. Show that a formula is valid iff T = 0, where T is an abbreviation for an instance p V -p of LEM. 2. Which of these formulas are semantically equivalent to p → (q V r)? (a) qv (-p Vr) (b) q^-r-p (c) p^rq * (d)¬q^r-p. 3. An adequate set of connectives for propositional logic is a set such that for every formula of propositional logic there is an equivalent formula with only connectives from that set. For example, the set {-, V} is adequate for propositional logic, because any occurrence of A and → can be removed by using the equivalences → ¬V and o^= -(-V¬). (a) Show that {¬, ^}, {¬,→} and {→,1} are adequate sets of connectives for propositional logic. (In the latter case, we are treating as a nullary con- nective.) (b) Show that, if C C {¬, ^, V, →, 1} is adequate for propositional logic, then ¬EC or LEC. (Hint: suppose C contains neither nor and consider the truth value of a formula o, formed by using only the connectives in C, for a valuation in which every atom is assigned T.) (c) Is {,} adequate? Prove your answer.See Answer
• Q5:Class Activity 6/28/23 MTH 231, done in class Wednesday, 6/28/23 1. Let r be the proposition "It is raining," s the proposition "The sun is shining," and w the proposition "It is windy." Write the following propositions using r, s, and w and logical connectives. (a) It is windy, and either the sun is shining or it's raining. (b) If it's not raining, then the sun is shining but it's not windy. (c) If it's raining or windy, then the sun is not shining.See Answer

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