1. We are given the universe $U$ as the set of integers, and predicates: - $P(x)$: $x$ is a prime number - $Q(x)$: $x$ is an even number - $R(x)$: $x$ is an odd number 2. First, consider the statement: $$ (\forall x) (R(x) \land P(x)) $$ This means "for all integers $x$, $x$ is odd and prime." 3. This statement is not a well-formed proposition because $R(x) \land P(x)$ is a predicate, not a truth value by itself. Usually, we consider statements like $(\forall x) (R(x) \land P(x))$ to mean "for all $x$, $x$ is odd and prime," which is false because not all integers are odd primes. 4. If the question is about the truth value of $(\forall x) (R(x) \land P(x))$, it is false because many integers are not odd primes. 5. Next, consider the statement: $$ (\forall x) \left[ (R(x) \land P(x)) \Rightarrow Q(x) \right] $$ This means "for all integers $x$, if $x$ is an odd prime, then $x$ is even." 6. Check if this implication is true or false: - The only even prime number is 2. - 2 is even but not odd, so $R(2)$ is false. - For odd primes like 3, 5, 7, $R(x) \land P(x)$ is true but $Q(x)$ is false. 7. Since there exists an $x$ (e.g., 3) where $R(x) \land P(x)$ is true but $Q(x)$ is false, the implication is false. **Final answers:** - Truth value of $(\forall x) (R(x) \land P(x))$ is **F** (False). - Truth value of $(\forall x) \left[ (R(x) \land P(x)) \Rightarrow Q(x) \right]$ is **F** (False).