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- please type out so i can copy paste*** Exercise 1.5.4: Logical relationships between the inverse, converse, and contrapositive. Use the laws of propositional logic to prove each of the following assertions. Start by defining a generic conditional statement p → q, and then restate the assertion as the equivalence or non-equivalence of two propositions using p and q. Finally prove that the two propositions are equivalent or non-equivalent. For example, the statement: "A conditional statement is not logically equivalent to its converse" is proven by showing that that p → q is not logically equivalent to q → p. (a) A conditional statement is not logically equivalent to its converse. (b) A conditional statement is not logically equivalent to its inverse. (c) A conditional statement is logically equivalent to its contrapositive. (d) The converse and inverse of a conditional statement are logically equivalent.Let Q(x, y) denote xy< 0 The domain of x is {1,2,3} and the domain of y is {0, -1, -2, -3) Implement the code that outputs the truth values of the following quantifications: 2. VX ay Q(X, y) **Display values of variables x and y if the quantification is true.X Bird(x)=Can- Fly(x) This universal quantifier used here implies that All reptiles cannot fly Oa. Ob. Ob. All that flies is a Bird All Birds can fly Oc. All chicken are birds d.
- Q2: Answer the fallowing Suppose you have the Degree, and then find the Result As fallow: * A C 1 Result Degree 77 89 45 ? 66 6. 95 55 ? 9 max degree 10 ? 1- Find the result depending on Degree as fallow: Degree in [0:49] → "F". Degree in [50:59] → "E" Degree in [60..69] →"D" Degree in [70..79] →"C" Degree in [80..89] →"B" Degree in [100..90 →"A" 2-Find the max degree value Your answer N345 o709e13. p(x): x is intelligent q(x): x reads book r(x): x is a student in your class Find the correct English translation for the following logical expression: ∀x(r(x)→(p(x) Λ q(x)) a. None b. There is an intelligent student in your class who reads book c. Every student in your class is intelligent and reads book d. All student in your class is intelligent and doesn't read bookDefinition. A Pythagorean triple (x, y, z) is a triple of positive integers where x² + y² = z². This can be thought of as describing an x × y rectangle with the property that the diagonal z is also of integer length. A Pythagorean triple (x, y, z) is primitive if x, y, z are coprime (i.e. there is no integer k > 1 which divides all of them). (a) Write a Python function PrimPyth (n) which returns a list of primitive Pythagorean triples (x, y, z) where 0 < x < y < z~(P ↔ Q) –| |– (P ↔ ~Q) Section 8.5. Tautologies and Equivalencies. Can you solve this question for me, because it seems complicated for me to solve this using the advanced method of assumptions, subproofs, and the use of negation rules ("for reductio") for the biconditional proofs? Try to use this method that my class uses: | ~(P ↔ Q) Want: P ↔ Q | And then the other way: | P ↔ ~Q Want: ~(P ↔ Q) |a) Give a recursive definition for the set of all strings of a’s and b’s where n a’s followed by n b’s (where n= 0, 1, 2, 3, ...). (Assume, S is set of all strings of a’s and b’s. Then S = {λ, ab, aabb, aaabbb, ... ) b) Give a recursive definition for the set of all strings of 'a' 's and 'b' 's that ends with an 'ab'. (Assume, S is set of all strings of a’s and b’s. Then S = {ab, aab, bab, abab, baab, aaab, bbab, ... ).1): ¬( p ∨ q ) ≡ ¬p ∧ ¬q The above law is called Group of answer choices De Morgans Law Absorption Law Complement Law Double Negation Law 2): To prove the this identity ¬p → (q → r) ≡ q → (p ∨ r) The following reasoning is valid: ¬p → (q → r)¬¬p ∨ (q → r) Conditional identity p ∨ (q → r) Double negation law p ∨ (¬q ∨ r) Conditional identity (p ∨ ¬q) ∨ r Associative law (¬q ∨p) ∨ r Commutative law ¬q ∨ (p ∨ r) Associative law q → (p ∨ r) Conditional identity Group of answer choices True FalseLet H(x) = "x plays hockey". Let B(x) = "x plays basketball". Which statement below means "No one who plays basketball also plays hockey"? OVx (-H(x) A-B(x)) OVx-(H(x) V B(x)) OVx (B(x)→-H(x)) Ox-(H(x) ^ B(x)) OVx¬(H(x) → B(x)).***ASKED BEFORE BUT NO ANSWER WAS GIVEN*** ***PLEASE ANSWER CORRECTLY*** WILL LIKE IF ALL CORRECT 1. Let S5 be the set of binary strings of length at least 5, i.e. S5 = {x ∈ {0, 1} ∗ : |x| ≥ 5}. (a) Build (design do not describe) a recogniser for S5 which is not a decider for S5. (b) Build an enumerator for S5. Note that we can enumerate or list all strings in lexicographic order s1 = epsilon , s2 = 0, s3 = 1, s4 = 00, . . . where is the empty string “”. In this case we only consider binary strings but this is valid over any alphabet. Let sn be the nth string enumerated in this way. (c) Is S5 decidable? (d) If you had the option to have access to either a decider or a recogniser when solving membership of a set, which would you choose and why? 2. Consider the set RejTM = {<M, w> | M is a TM such that M rejects on input w}. (a) Describe an element in the set and one not in the set. You should provide pseudocode for any TM you discuss. (b) Prove RejTM is recognisable. Hint: Similar…The correct statements are: A set is countable iff it is either finite or countably infinite. The set of natural numbers is countable. The power set of the set of natural numbers is countable. For any Σ Ø, Σ* is countably infinite.SEE MORE QUESTIONS