5) Given the following propositions: p = ́ ́105 is a multiple of 3 ́ ́ q = ́ ́69 is a prime number ́ ́ r = ́ ́-1 is a number less than -2 ́ ́ Write in symbolic form respecting, brackets and keys: ́ ́102 is not a multiple of 3 if and only if -1 is not a number less than -2 and 69 is a prime number ́ ́ 6) Taking into account the propositions p, q and r from the previous exercise, find the truth value of the following compound proposition: [(q↔ r) ˄ ((-q) →p)] Justify the answer of the intervening propositions

Let's first translate the given propositions into symbolic form: p: 105 is a multiple of 3 q: 69 is a prime number r: -1 is a number less than -2 Now, let's write the compound proposition in symbolic form: Compound proposition: [(q ↔ r) ∧ ((-q) → p)] Let's break down the compound proposition step by step: 1. q ↔ r: This means "q if and only if r," which is equivalent to "(q → r) ∧ (r → q)." q → r: If 69 is a prime number, then -1 is a number less than -2. r → q: If -1 is a number less than -2, then 69 is a prime number. 2. (-q) → p: This means "if not q, then p." If 69 is not a prime number, then 105 is a multiple of 3. Now let's evaluate the truth values of these sub-propositions using the given information: q → r: Since 69 is a prime number (q is true), and -1 is a number less than -2 (r is true), this implication is true. r → q: Since -1 is a number less than -2 (r is true), but 69 is not a prime number (q is false), this implication is false. (-q) → p: Since 69 is a prime number (q is true), the negation of q is false, and thus, this implication is true. Now, let's put it all together: q ↔ r: (q → r) ∧ (r → q) = (true ∧ false) = false (-q) → p: true Finally, let's evaluate the entire compound proposition: [(q ↔ r) ∧ ((-q) → p)] = [false ∧ true] = false So, the compound proposition [(q ↔ r) ∧ ((-q) → p)] is false based on the given propositions p, q, and r.