Let P(x), Q(x), and R(x) be the following predicates:
P(x):x^2 −7x+10=0 Q(x):x^2 −2x−3=0 R(x) : x < 0
(a) Determine the truth or falsity of the following statements, where the universe is all integers.
i. ∀x(P (x) →∼ R(x)) ii. ∀x(Q(x) → R(x)) iii. ∃x(P (x) → R(x))
Solution: First, note that x2 − 7x + 10 = (x − 2)(x − 5) and therefore the truth set of P(x) is {2,5}, and x2 −2x−3 = (x+1)(x−3) and therefore the truth set of Q(x) is {−1, 3}. The truth set of R(x) is the set of negative integers. The answers are: T F T
(b) Find the answers to part (a) when the universe consists of all positive integers. Solution: T F T
(c) Find the answers to part (a) when the universe contains only the integers 2 and 5. Solution: T T F
why is statement iii ture for question a, and becomes false in question c. Also why is statement ii is true for question c
Remember that an implication $\varphi \to \psi$ is equivalent to $\lnot \varphi \lor \psi$. In other words, an implication is true iff its left part is false or its right part is true. Depending on the definitions you are using, this either is the direct definition of $\to$, or it follows from its truth table.
Why is iii) true in a), but becomes false in c)?
The formula iii) says "There is an x, such that if Px holds, then also Rx holds."
In a), you are asked to interpret the formula over the domain of all integers. Here the formula is true: As stated in your question $Px$ only holds for $x \in \{2,5\}$. Take some $x \notin \{2,5\}$, for example $x=3$. Then for this $x$, the formula $Px$ is false. Therefore, for this $x$ also $Px \to Rx$ is true. Therefore, $\exists x (Px \to Rx)$ is true.
In c), you are asked to interpret the formula over the domain {2,5}. Here you cannot use $x = 3$, which is not in the domain. You only have two choices, setting $x = 2$ or $x=5$. For both, $Px$ is true and $Rx$ is false. Therefore, for both $Px \to Rx$ is false. Therefore, $\exists x (Px \to Rx)$ is false.
Why is ii) true in c)?
The formula ii) says "For all x, if Qx holds, then also Rx holds."
You are asked to interpret this formula over the domain $\{2,5\}$. As stated in your question, $Qx$ is only true for for $x \in \{-1,3\}$. Therefore, for all $x \in \{2,5\}$, $Qx$ is false, and therefore $Qx \to Rx$ is true. Therefore, $\forall x (Qx \to Rx)$ is true.