COS2661 Assignment 3 Summary 2023

COS2661 A3 - summary of entire book Formal Logic II (University of South Africa) lOMoARcPSD| COS2661 Assignment 3 2022 Unique No.: Due date: 31 August 2022 lOMoARcPSD| In this document you will find 1. The complete suggested solutions to COS2661 Assignment 3 2022 2. As a bonus you will have access to download the COS2661 prescribed textbook(Language proof and formal logic 2md edition) completely free of charge by clicking the google drive link on the last page of this document. Please note: This should only be used as a guide for your own assignment. Downloaded by Genial Adventures () lOMoARcPSD| …Preview COS2661 Question 1 1.1. If any bananas are yellow, then some bananas are ripe. ∃x(Banana(x) ' Yellow(x)) = ∃y(Banana(y) ' Ripe(y)) • Explanation ∃x: Denotes some x or there exists x ' : Denotes conjunction symbol 1.2. If all ripe bananas are yellow, some yellow things are ripe. ∀x((Banana(x) ' Ripe(x) = Yellow(x)) = ∃y(Yellow Things(y) ' Ripe(y)) • Explanation: ∀x: Denotes all x ∃y: Denotes some y or there exists y ' : Denotes conjunction symbol Downloaded by Genial Adventures () lOMoARcPSD| ASSIGNMENT 03 100% Marks SUBMISSION: Electronically through myUnisa (as one .pdf file) It will be to your own advantage to check whether the assignment has been registered on the system after a few days. This assignment is submitted electronically and if myUnisa is off-line during that time, you need not contact us, because we will be aware of it. Simply submit it as soon as myUnisa is available again. Due date Extension Tutorial matter Weight of contribution to semester mark Unique number 31 August 2022 No Extension Textbook: Chapter 11: 11.1 – 11.8 Chapter 12: 12.1 – 12.4 Chapter 13: 13.1 - 13.5 Chapter 14: 14.1 – 14.6 Tutorial Letter 102 Chapters 11 to 14 40% Downloaded by Genial Adventures () lOMoARcPSD| QUESTION 1 [20] In this question you have to translate English sentences into sentences of First-Order Logic, using the predicates and names given in Table 1 below. English FOL Names Thabang thabang Logic logic Predicates x is ripe Ripe(x) x is yellow Yellow(x) x is dead Dead(x) x tells y Tells(x, y) x is worth selling WorthSelling(x) x keeps y Keeps(x, y) x are yellow things YellowThings(x) x lays y Lays(x, y) x is a tale Tale(x) x failed y Failed(x, y) x is a banana Banana(x) x is a farmer Farmer(x) x is a hen Hen(x) x is an egg Egg(x) x is a man Man(x) x is a student Student(x) Table 1 Question 1.1 (4) If any bananas are yellow, then some bananas are ripe. Question 1.2 (4) If all ripe bananas are yellow, some yellow things are ripe. Question 1.3 (4) If a farmer keeps only hens, none of them will lay eggs that are worth selling. Question 1.4 (4) Dead man tells no tale. Question 1.5 (4) Only one student failed Logic. 2 Downloaded by Genial Adventures () lOMoARcPSD| COS2661/104/0/2022 QUESTION 2 [20] In this question you have to translate sentences of First-Order Logic into English sentences, using the predicates and names given in Table 1. Question 2.1 (4) ∀x [(Banana(x) ù Yellow(x)] → [∀y((Banana(y) ù Yellow(y)) → Ripe(y)) → Ripe(x))] Question 2.2 (4) x[Hen(x) → y(Egg(y) ù Lays(x, y))] Question 2.3 (4) ¬ x (Farmer(x) ù Keeps(x, Banana(x))) ù x (Farmer(x) ù Keeps(x, Egg(x))) Question 2.4 (4) x(Banana(x) ù Yellow(x) ù ¬Ripe(x)) Question 2.5 (4) x(Man(x) ù y(Man(y) ù Tells(y, x) ù (x  y) → Tale(x)) Question 3 [5] Use De Morgan’s laws and show that: ø (x Jumbo(x) ù y Eggs(y)) is logically equivalent to x y (ø Jumbo (x) ú ø Eggs(y)) Downloaded by Genial Adventures () lOMoARcPSD| QUESTION 4 , [10] Below a Tarski World is given followed by ten sentences. Which of the sentences are true in the given world and which sentences are false in the given world? back left a: C, S right d: T, S e: C, S b: D, L c: T, M front Tarski World: Question 4 Sentences: 1. x Between(x, e, a) ù x Between(x, e, c) 2. x (Cube(x) → y z (Dodec(y) ù Tet(z) ù RightOf(y, x) ù LeftOf(x, z))) 3. ø x (Cube(x) → Small(x)) 4. x (Small(x) ú Medium(x) ú Dodec(x)) 5. x (Dodec(x) ù Large(x)) 6. y x (Medium(y) → (Tet(y) ù (Cube(x) → Larger(y, x)))) 7. y ((Dodec(y) ù Small(y)) ú (Tet(y) ù Medium(y))) 8. x y (Cube(x) ù Tet(y) ù ø SameSize(x, y)) 9. x y SameCol(x, y) 10. Cube(d) → x y SameShape(x, y) QUESTION 5 [10] Convert the following sentences into prenex form. 5.1 xP(x) →x Q(x) (5) 5.2 ∀x(∃yR(x, y) ù ∀y¬S(x, y) → ¬(∃yR(x, y) ù P)) (5) 4 Downloaded by Genial Adventures () lOMoARcPSD| COS2661/104/0/2022 QUESTION 6 [35] In this question, you have to construct formal proofs using the natural deduction rules. The Fitch system makes use of these rules. A summary of the rules of natural deduction is given on pages 573 to 578 of your textbook. Consult this when you do this question. Remember that De Morgan’s laws and other tautologies are not permissible natural deduction rules. You are also not allowed to use Taut Con, Ana Con or FO Con. It is important to number your statements, to indicate subproofs and at each step to give the rule that you are using. Hint: If you have access to a computer, take advantage of the fact and use Fitch. Question 6.1 (11) Using the natural deduction rules, give a formal proof of x (Large(x) ù LeftOf(x, b)) from the premises x (Cube(x) → Large(x)) y (Large(y) → LeftOf(y, b)) x Cube(x) Question 6.2 (14) Using the natural deduction rules, give a formal proof of x ( Smiling(x) ù Happy(x) ) from the premises x [ (AtHome(x) ú InLibrary(x)) → (Happy(x) ù Reading(x))] y [ (Smiling(y) ú Happy(y)) → InLibrary(y) ] x Smiling(x) Question 6.3 (10) Using the natural deduction rules, give a formal proof of: x y ¬SameSize(y, x) from premises. x y Adjoins(x, y) x y (Adjoins (x, y) → ¬SameSize(x, y)